Experiment 6 lecture video

00:41:02
https://www.youtube.com/watch?v=cHmdHhPabSQ

Summary

TLDRLa vidéo traite des réactions de dismutation, aussi appelées auto-oxydation-réduction, où une molécule agit simultanément comme agent oxydant et réducteur, conduisant à différents produits. Elle explique les principes des réactions redox, la fonction des agents oxydants et réducteurs, et les étapes pour équilibrer ces réactions de manière méthodique en utilisant des demi-équations. Le peroxyde d'hydrogène est utilisé comme exemple de molécule de dismutation typique. L'importance du spectrophotomètre est aussi abordée, cet instrument étant utilisé pour mesurer quantitativement l'absorbance de la lumière, fonction de la concentration d'une substance dans une solution. Enfin, l'expérience sur les états d'oxydation du vanadium est présentée, chaque état exhibant une couleur spécifique qui change lors des réactions, ce qui permet d'étudier son état d'oxydation grâce à des couleurs bien définies.

Takeaways

  • 🔄 La dismutation implique une auto-oxydation et réduction.
  • ⚖️ L'équilibrage des réactions redox suit des étapes précises.
  • 💡 Le peroxyde d'hydrogène est un exemple clé de dismutation.
  • 📊 Un spectrophotomètre mesure l'absorbance pour déterminer des concentrations.
  • 🌈 Les couleurs changent avec les états d'oxydation du vanadium.
  • 🧪 Les réactions redox affectent les états de charge et d'oxydation.
  • 📏 Lambda Max est crucial pour identifier le pic d'absorption.
  • 🔍 Analyser les spectres d'absorption prépare à des mesures quantitatives.
  • 🔬 L'expérience utilise des agents chimiques spécifiques à un milieu acide.
  • ☀️ La conservation appropriée du peroxyde d'hydrogène empêche sa décomposition par la lumière.

Timeline

  • 00:00:00 - 00:05:00

    L'expérimentation 6 commence par l'exploration de la réaction d'auto-oxydation et de réduction, appelée disproportionation, où une molécule agit à la fois comme agent oxydant et réducteur. Le processus se résume à une compétition pour les électrons entre ces agents. L'agent réducteur perd des électrons, augmentant son nombre d'oxydation, tandis que l'agent oxydant les gagne, réduisant son nombre d'oxydation. Exemples : le peroxyde d'hydrogène, où l'oxygène se balance entre -2 et 0, et se transforme en eau et oxygène.

  • 00:05:00 - 00:10:00

    L'exemple du zinc est utilisé pour démontrer comment prédire les produits dans un milieu acide ou neutre où le zinc passe à l'état d'oxydation deux plus. En milieu basique, il se transforme en hydroxyde de zinc, illustrant que le zinc ne peut pas être un agent oxydant car il ne peut être réduit. Concernant le brome, il peut être réduit en ions bromure et potentiellement oxydé dans certains composés tels que le bromate.

  • 00:10:00 - 00:15:00

    Le permanganate en milieu acide transforme en ion Mn2+, en milieu basique en MnO2, et en milieu très basique en ion MnO4-2, chaque produit ayant ses propres indicateurs visuels distincts (Mn2+ est incolore, MnO2 est brun noir, MnO4-2 est vert) ce qui permet de déterminer le milieu réactionnel basé sur des observations expérimentales.

  • 00:15:00 - 00:20:00

    Pour équilibrer les réactions redox : identifier les changements d'oxydation, réduire les demi-réactions séparément, équilibrer les charges avec H+ ou OH- selon le milieu, et ajouter de l'eau pour équilibrer l'oxygène et l'hydrogène. Lors de la recombinaison des demi-réactions, ajuster le nombre d'électrons pour qu'ils se neutralisent. Exemple expliqué avec la réaction entre le dichromate et le fer.

  • 00:20:00 - 00:25:00

    Exemple de disproportion avec le chlore dans une solution acide, montrant comment Cl2 se transforme en Cl- et en ClO3-, chaque processus étant détaillé pour équilibrer les charges et les électrons. Thiosulfate en solution acide se disproportionne également pour fournir du soufre élémentaire et de l'acide sulfureux, illustrant l'intérêt des spectrophotomètres dans la mesure des changements de couleur avec précision.

  • 00:25:00 - 00:30:00

    La spectrophotométrie est introduite comme une technique quantitative pour mesurer l'absorbance de la lumière par une solution. L'appareil détecte spécifiquement quelle portion de lumière est absorbée, donnant un spectre d'absorption correspondant. Le pic d'absorption maximal (Lambda Max) correspond à la couleur la plus absorbée, la couleur visible étant son complémentaire.

  • 00:30:00 - 00:35:00

    Analyser le spectre d'absorption révèle quelle partie du spectre est la plus absorbée, exemple donné avec un spectre présentant une absorption maximale dans le bleu, donnant une couleur apparente orange. Les instructions indiquent que le spectre absorbe des régions invisibles peuvent encore influencer la couleur observée. Importance de comprendre ces principes avant le laboratoire.

  • 00:35:00 - 00:41:02

    L'expérience se concentre sur la réduction progressive du dioxovanadium où différents états d'oxydation se manifestent par des changements de couleur, allant du jaune (état +5) au bleu (état +4), vert (état +3) et violet (état +2). Identifier quel élément est oxydé ou réduit et la signification de ces réactions, surtout lorsque les oxydations plus basses ne sont pas atteignables en utilisant uniquement le zinc.

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Mind Map

Video Q&A

  • Qu'est-ce qu'une réaction de dismutation ?

    Une réaction de dismutation est une auto-oxydation-réduction dans laquelle une molécule agit à la fois comme agent oxydant et réducteur, produisant des produits différents.

  • Comment équilibrer une réaction redox ?

    Pour équilibrer une réaction redox, identifiez les substances oxydées et réduites, séparez-les en demi-équations, et équilibrez-les séparément selon l'ordre : atomes autres qu'hydrogène et oxygène, électrons, charges, puis oxygène et hydrogène avec l'eau.

  • Qu'est-ce que le spectrophotomètre mesure ?

    Le spectrophotomètre mesure l'absorbance de la lumière par une solution, ce qui est proportionnel à la concentration de la substance absorbante.

  • Pourquoi le peroxyde d'hydrogène est-il efficace pour traiter les plaies ?

    Le peroxyde d'hydrogène est efficace car il produit de l'oxygène, qui est puissant pour tuer les bactéries.

  • Comment le changement de couleur indique-t-il l'état d'oxydation du vanadium ?

    Chaque état d'oxydation du vanadium est associé à une couleur spécifique, permettant de déduire l'état d'oxydation du vanadium dans une solution en observant sa couleur.

  • Pourquoi le peroxyde d'hydrogène doit-il être conservé dans une bouteille sombre ?

    Il doit être conservé à l'abri de la lumière pour ralentir sa décomposition, car la lumière peut faciliter la réaction de dismutation.

  • Qu'est-ce que la longueur d'onde Lambda Max ?

    Lambda Max est la longueur d'onde où une substance absorbe le maximum de lumière.

  • Quelles sont les étapes pour le spectre d'absorption ?

    On obtient un spectre d'absorption en mesurant l'absorbance en fonction de la longueur d'onde, indiquant où une solution absorbe le plus.

  • Quels ions permettent de balancer la charge en solution acide ?

    Dans une solution acide, on utilise les ions H+ pour équilibrer les charges.

  • Quels éléments ne doit-on pas balancer tout de suite dans une demi-équation redox ?

    Il ne faut pas balancer l'oxygène et l'hydrogène en premier, mais plutôt les autres éléments.

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  • 00:00:00
    in this video we'll look at everything
  • 00:00:02
    you need for experiment 6.
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    the first topic that we'll look at is a
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    self-oxidation reduction reaction this
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    is also called disproportionation and
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    overall what it means is that one
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    molecule will act as both the oxidizing
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    agent and the reducing agent so that you
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    get one product that goes one way and
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    one product that goes the other way
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    so remember that a redox reaction is a
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    tug of war for electrons
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    between the reducing agent and the
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    oxidizing agent so the reducing agent is
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    the thing that gets oxidized so it's
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    going to lose electrons
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    it starts out with the electron and then
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    whichever one wants that electron more
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    we'll end up with the electron
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    so our reducing agent gets oxidized by
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    losing electrons and its oxidation
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    number increases
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    our oxidizing agent gets reduced it
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    gains electrons and its oxidation number
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    decreases
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    when an atom is at its lowest oxidation
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    number it can only be a reducing agent
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    so what this means is that there is a
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    range of possible oxidation numbers for
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    every atom and if it has already given
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    up all of its electrons it doesn't have
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    any more to give up there's no way that
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    it could be oxidized
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    and when an atom is at its highest
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    oxidation number it can only be an
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    oxidizing agent so if an atom
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    has
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    no electrons it will want to get the
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    electrons so if something is at its
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    lowest oxidation number it cannot get
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    any more electrons all right
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    so some substances are in the middle
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    because there's a range of possible
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    oxidation numbers that an atom can have
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    and when it's in the middle it can act
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    as a reducing agent or an oxidizing
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    agent
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    so for example in hydrogen peroxide
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    which is H2O2 it has an oxidation number
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    of negative one
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    now the range of oxidation numbers for
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    oxygen the most negative that it can go
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    is negative two
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    and the most positive it can go is zero
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    so if you start with a negative one
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    oxidation number it could gain one
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    electron and be reduced
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    and a molecule in which oxygen has
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    negative two as an oxidation number
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    would be water
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    and if it loses an electron
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    the oxygen could have an oxidation
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    number of zero which it does when it is
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    molecular oxygen so in O2
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    so hydrogen peroxide is a molecule that
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    can disproportionate and it does so by
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    having one of the oxygens become
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    oxidized to O2 and one of the oxygens
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    become reduced to H2O
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    so this is a self-oxidation reduction
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    reaction
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    and the way that it works out once
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    everything is balanced is that two
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    molecules of hydrogen peroxide will turn
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    into two molecules of water and one
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    molecule of o2 and the hydrogen peroxide
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    that you buy at any drugstore
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    um that's why it's in that dark bottle
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    so that it does this a little bit less
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    readily because sunlight can sometimes
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    help with that but this is also why
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    hydrogen peroxide is great for treating
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    wounds the O2 the oxygen is uh pretty
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    good at killing bacteria
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    so again the oxygen going to in peroxide
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    going to O2 is oxidation the oxygen
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    going to negative 2 as it is in water is
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    reduction
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    okay so predicting products of redox
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    reactions it is as cool as it sounds and
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    it is not as difficult as it sounds
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    so we will give you all of the
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    information that you need for this to
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    predict the products in a redox reaction
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    it will be based on the chemical
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    properties of the reactants
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    and your observations in lab
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    so for example zinc metal zinc will turn
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    into zinc two plus plus two electrons
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    all right some things that you know
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    about need to know about zinc this will
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    happen in an acidic or a neutral medium
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    so
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    one of the things that you might look at
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    in your experimental conditions is was
  • 00:05:23
    there any acid present
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    or was their base present was it just in
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    water
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    well other things were in the solution
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    that would give you a clue about the
  • 00:05:35
    medium that it's in
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    so what about if it's in a basic medium
  • 00:05:41
    well we're going to tell you first of
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    all so zinc will turn into a zinc
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    hydroxide plus two electrons well this
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    is
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    this is still zinc two plus it's in a
  • 00:05:54
    basic medium
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    it's not
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    exactly the same the way that it's
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    listed in this reaction but in both of
  • 00:06:04
    these cases zinc is oxidized to zinc two
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    plus
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    the way that you know that zinc would be
  • 00:06:12
    oxidized to zinc 2 plus is from the
  • 00:06:15
    stuff that we did in experiment one
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    so in those naming tables you will see
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    the common charges which are also listed
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    as oxidation States and that will give
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    you a clue about whether or not a metal
  • 00:06:30
    can have more than one oxidation state
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    and what those oxidation states are for
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    zinc there's only one option of zinc two
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    plus so if zinc gets oxidized the only
  • 00:06:43
    thing it could turn into is a zinc two
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    plus now what about reduction
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    could zinc gain an electron to turn into
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    a zinc one minus
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    or a zinc two minus could have gained
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    two electrons
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    what do you think
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    definitely not so you will not see
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    Metals with negative oxidation States so
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    zinc cannot be an oxidizing agent
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    because it cannot be reduced
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    okay let's look at bromine so br2 is how
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    the element bromine exists
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    we could give it some electrons so we
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    could reduce bromine and turn it into
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    two BR minuses right the br2 has a zero
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    and then we get two bromines that each
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    have a negative one
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    this can happen
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    what about oxidation
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    could you take br2
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    remove some electrons and get a product
  • 00:07:52
    that has bromine in a higher oxidation
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    number
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    here again you you might wonder how
  • 00:08:00
    could I figure this out
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    you could look at things that have
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    bromine in it and check their oxidation
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    number and in this case I'm going to
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    tell you what those things are
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    which is all of these polyatomic ions
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    now you may not have seen the bromine
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    version written out we usually deal with
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    the chlorine version but this is
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    hypobromite bromite bromate and per
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    bromate it parallels perfectly what you
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    would see with
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    hypochlorite chloride chlorate and
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    perchlorate so check the oxidation
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    numbers for Bromine in each of these
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    molecules and you should see that this
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    could happen if br2 is oxidized you
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    should end up with a positive oxidation
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    number for those
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    now
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    which is it kind of has that similar
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    pattern to what you just saw in this
  • 00:09:01
    other polyatomic ion permanganate it's
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    kind of special we don't really see the
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    pattern that you see for the halogens we
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    really just see mno for minus as the
  • 00:09:14
    permanganate polyatomic ion now
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    permanganate has three very specific
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    things that it'll turn into
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    if you are in an acidic medium
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    medium just means
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    solution mixture whatever you have this
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    ion in for us it's going to be a
  • 00:09:35
    solution
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    if you're in an acidic solution
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    permanganate will turn into mn2 Plus
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    so check the oxidation number of
  • 00:09:46
    manganese in permanganate and then in
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    all of these products to see whether
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    it's been oxidized or reduced
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    if you are in a basic medium
  • 00:09:56
    permanganate will turn into mno2
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    which is also a solid so you in addition
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    to
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    knowing that the solution is basic
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    um some other things you're going to see
  • 00:10:09
    mno2 is a solid that would precipitate
  • 00:10:13
    and if you're in a very basic medium
  • 00:10:16
    will give you in the reading some
  • 00:10:18
    criteria for this off the top of my head
  • 00:10:21
    I think the pH has to be something like
  • 00:10:24
    14. so we mean extremely basic
  • 00:10:27
    and you will get mno4 2 minus
  • 00:10:31
    so watch that really closely because the
  • 00:10:33
    only thing we've changed is the charge
  • 00:10:37
    and the experimental observations will
  • 00:10:40
    help you determine the product so this
  • 00:10:42
    will be a case where you'll be able to
  • 00:10:45
    identify based on what you put into a
  • 00:10:48
    reaction whether the solution is acidic
  • 00:10:51
    basic or very basic
  • 00:10:53
    and
  • 00:10:55
    all of these things appear different
  • 00:10:59
    from each other so it should be obvious
  • 00:11:02
    and if you have any issues seeing colors
  • 00:11:05
    someone should tell you the color so
  • 00:11:07
    that it can be obvious that a reaction
  • 00:11:10
    has or has not happened
  • 00:11:13
    so permanganate is this magenta e
  • 00:11:18
    purpley
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    extremely vibrant color
  • 00:11:24
    mn2 plus is colorless
  • 00:11:27
    so The Disappearance of the purple color
  • 00:11:31
    would be your clue that you have formed
  • 00:11:33
    mn2 plus as long as you're in an acidic
  • 00:11:36
    solution
  • 00:11:38
    mno2 I mentioned that it's a solid it's
  • 00:11:41
    also this like brownish black color so
  • 00:11:45
    you will get a brown black precipitate
  • 00:11:47
    if you form mno2
  • 00:11:50
    and mno42 minus is green so our options
  • 00:11:55
    are bright magenta e purple colorless
  • 00:11:59
    black brown precipitate or green so
  • 00:12:04
    between the conditions is it acidic is
  • 00:12:08
    it basic and the color and appearance of
  • 00:12:11
    what is in your reaction you will be
  • 00:12:13
    able to predict the product of this
  • 00:12:16
    redox reaction
  • 00:12:19
    okay so balancing redox reactions we are
  • 00:12:23
    going to need it for this if you have
  • 00:12:26
    already done this in class this part
  • 00:12:28
    will be a review it was also in the
  • 00:12:30
    previous video but it's okay if you did
  • 00:12:32
    not watch it in the previous video
  • 00:12:34
    because it was not required for the last
  • 00:12:36
    experiment
  • 00:12:39
    for balancing a redox reaction I'm going
  • 00:12:42
    to give you these rules they're in order
  • 00:12:45
    they're in priority order so follow them
  • 00:12:49
    in order every time and you'll be good
  • 00:12:51
    to go so the first thing you need to do
  • 00:12:54
    is identify what is oxidized and what is
  • 00:12:57
    reduced you'll have to do this by
  • 00:13:00
    assigning oxidation numbers and looking
  • 00:13:02
    at what changed
  • 00:13:05
    then the oxidation part and the
  • 00:13:08
    reduction part get separated we call
  • 00:13:11
    this a half reaction because it only
  • 00:13:13
    shows half of the reaction and we're
  • 00:13:15
    going to balance those half reactions
  • 00:13:18
    separately from each other before we put
  • 00:13:21
    them back together
  • 00:13:23
    within a half reaction the first thing
  • 00:13:25
    you do is balance all of the elements
  • 00:13:27
    but not oxygen and hydrogen
  • 00:13:31
    of course there's always going to be one
  • 00:13:33
    of those caveats now if we have the
  • 00:13:37
    hydrogen peroxide that we talked about
  • 00:13:39
    earlier in the video as one of the
  • 00:13:41
    things in our reaction we're probably
  • 00:13:44
    going to have to deal with oxygen
  • 00:13:46
    toward the beginning of this but in all
  • 00:13:49
    other cases leave oxygen and hydrogen
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    for later otherwise you are immediately
  • 00:13:54
    setting yourself up for failure so
  • 00:13:57
    balance everything else
  • 00:13:59
    then balance the number of electrons
  • 00:14:02
    gained or lost
  • 00:14:07
    then you'll balance charges so you will
  • 00:14:11
    need to know if the solution is acidic
  • 00:14:14
    or basic if you are just being given a
  • 00:14:17
    reaction out of context you will be told
  • 00:14:20
    whether it's acidic or basic
  • 00:14:22
    if it's related to an experiment you're
  • 00:14:24
    doing you might have to look through the
  • 00:14:26
    steps to figure it out
  • 00:14:29
    and then the very last step is to
  • 00:14:31
    balance oxygen and hydrogen by adding
  • 00:14:34
    water
  • 00:14:35
    when you get to this step the number of
  • 00:14:38
    water molecules that you add should be
  • 00:14:42
    one
  • 00:14:44
    obvious or very easy to calculate
  • 00:14:48
    and two should balance perfectly
  • 00:14:52
    you should just be able to add a whole
  • 00:14:55
    number of water molecules to one of the
  • 00:14:57
    sides of the reaction to make it balance
  • 00:15:00
    and if you can't that means that
  • 00:15:03
    something went wrong in an earlier step
  • 00:15:07
    with the two half reactions now balanced
  • 00:15:10
    you will set it up so that the number of
  • 00:15:12
    electrons gained will cancel with the
  • 00:15:15
    number of electrons lost when you add
  • 00:15:17
    them together
  • 00:15:19
    and then you will add them together once
  • 00:15:21
    you add them together you may have water
  • 00:15:25
    on both sides and you'll need to do some
  • 00:15:27
    canceling you may have
  • 00:15:30
    h plus or o h minus on both sides and
  • 00:15:33
    need to do some canceling but you should
  • 00:15:36
    not see any of the electrons so this is
  • 00:15:39
    one of the things that's challenging
  • 00:15:41
    about balancing a redox reaction we
  • 00:15:43
    never show the electrons because they
  • 00:15:46
    never Exist by themselves in the
  • 00:15:48
    solution they get transferred from one
  • 00:15:51
    thing to another
  • 00:15:54
    all right let's do an example so in an
  • 00:15:57
    acidic solution we're going to balance
  • 00:15:59
    dichromate reacting with iron two plus
  • 00:16:02
    and the products are chromium 3 plus and
  • 00:16:05
    iron three plus
  • 00:16:07
    we call this the skeleton of the
  • 00:16:10
    reaction
  • 00:16:11
    or the skeleton equation it has the
  • 00:16:15
    major reactants in products but it is
  • 00:16:17
    not balanced it doesn't have all of the
  • 00:16:19
    other things that are going to make it
  • 00:16:21
    balance
  • 00:16:23
    so assign oxidation numbers chromium and
  • 00:16:26
    dichromate is a plus six and then
  • 00:16:29
    everything else was a monoatomic ion so
  • 00:16:32
    the charge in the oxidation number are
  • 00:16:34
    the same
  • 00:16:36
    my chromium goes from a plus six to a
  • 00:16:40
    plus three this will happen if it gains
  • 00:16:43
    electrons so my chromium is reduced
  • 00:16:46
    and iron lost electrons it's oxidized
  • 00:16:52
    so when we're talking about half
  • 00:16:53
    reactions
  • 00:16:55
    the oxidation half reaction just shows
  • 00:16:59
    what happens to iron it's the thing
  • 00:17:02
    that's oxidized
  • 00:17:04
    and my reduction half reaction
  • 00:17:08
    is dichromate turning into chromium 3
  • 00:17:11
    plus and the first part of balancing the
  • 00:17:15
    half reactions is also done on here I
  • 00:17:18
    have two chromiums on the left
  • 00:17:21
    I need to balance the chromium on the
  • 00:17:23
    right
  • 00:17:24
    forget the oxygen for now we'll come
  • 00:17:26
    back to it it'll get fixed up but I need
  • 00:17:29
    to know that my elements other than
  • 00:17:32
    oxygen and hydrogen are balanced so I
  • 00:17:34
    needed to in front of that chromium
  • 00:17:37
    feeling
  • 00:17:39
    that step the next step is to balance
  • 00:17:41
    electrons so iron goes from a plus two
  • 00:17:44
    to a plus three so it lost one electron
  • 00:17:49
    and when we set up chemical equations we
  • 00:17:54
    never subtract anything right I can't
  • 00:17:56
    write iron two plus minus one electron
  • 00:17:59
    gives me iron three plus
  • 00:18:01
    but I can
  • 00:18:04
    can imagine that and then write it in a
  • 00:18:07
    way that is positive so I will write
  • 00:18:09
    plus one electron on the reactant side
  • 00:18:12
    to show that my or on the product side
  • 00:18:15
    to show that my reactant kicked out one
  • 00:18:18
    electron
  • 00:18:19
    for chromium
  • 00:18:21
    I am going from my plus six to a plus
  • 00:18:24
    three
  • 00:18:25
    that's three electrons
  • 00:18:28
    but that is three electrons
  • 00:18:31
    per chromium
  • 00:18:34
    I have two chromiums that need to go
  • 00:18:36
    from a plus six to a plus three
  • 00:18:38
    so I need six electrons to be added to
  • 00:18:43
    my reactant
  • 00:18:45
    after this it is balancing not charges
  • 00:18:48
    so if we look at the oxidation reaction
  • 00:18:53
    on the left side the only thing I have
  • 00:18:55
    is iron two plus so the overall charge
  • 00:18:59
    on that side is plus two
  • 00:19:01
    on the product side I have an iron three
  • 00:19:03
    plus and an electron
  • 00:19:06
    if I have something with a three plus
  • 00:19:07
    charge and I add a negative one to it
  • 00:19:11
    that is an overall charge of plus two
  • 00:19:16
    so I have plus two on both sides that
  • 00:19:19
    matches I don't need to do anything to
  • 00:19:21
    balance the overall charges
  • 00:19:23
    for the reduction half reaction
  • 00:19:26
    dichromate is a negative two I've added
  • 00:19:29
    six electrons to it so this is a
  • 00:19:32
    negative eight total on the left
  • 00:19:36
    and on the right I have two things with
  • 00:19:39
    a plus three so I have a plus six charge
  • 00:19:42
    on the right and a negative eight on the
  • 00:19:44
    left not balanced
  • 00:19:47
    so I am in an acidic solution which
  • 00:19:51
    means that h plus is my only option for
  • 00:19:55
    balancing the charges in this half
  • 00:19:58
    reaction
  • 00:19:59
    which I can balance them if I add 14 h
  • 00:20:03
    plus it seems like a lot and it will
  • 00:20:07
    seem like a lot sometimes but
  • 00:20:09
    we only had h plus the only way to get
  • 00:20:13
    the same charge on both sides was to add
  • 00:20:16
    14 of them so I now have a plus sex
  • 00:20:19
    charge on both sides
  • 00:20:22
    and my last step is to balance oxygen
  • 00:20:25
    and hydrogen with water well I I have 14
  • 00:20:30
    H pluses that's only going to balance if
  • 00:20:33
    I can add seven water
  • 00:20:35
    and very conveniently
  • 00:20:38
    I have seven oxygens that need to be
  • 00:20:41
    balanced so if I had seven water
  • 00:20:43
    molecules to the right side it's perfect
  • 00:20:47
    my half reaction is now perfectly
  • 00:20:49
    balanced the number of waters I needed
  • 00:20:51
    to add just worked because the other
  • 00:20:54
    steps were done correctly
  • 00:20:56
    in order to add these together and have
  • 00:20:59
    my electrons cancel
  • 00:21:03
    I need to multiply the top reaction by
  • 00:21:05
    six this is what it means to have the
  • 00:21:08
    number gained equal the number lost
  • 00:21:11
    the electrons that are being given to
  • 00:21:14
    chromium have to come from Iron so I
  • 00:21:18
    need enough irons to give the right
  • 00:21:20
    number of electrons for the chromiums
  • 00:21:23
    that I have and vice versa so if I
  • 00:21:27
    multiply by 6 on the top reaction I'll
  • 00:21:30
    have six electrons in both of them and
  • 00:21:33
    they'll cancel each other out
  • 00:21:35
    so add these together and my final
  • 00:21:39
    result My overall is 14h plus plus one
  • 00:21:43
    dichromate plus six iron two plus will
  • 00:21:47
    turn into iron six iron three plus two
  • 00:21:51
    chromium three plus and seven water
  • 00:21:54
    molecules
  • 00:21:57
    let's do another example and this time
  • 00:22:00
    let's do something that's a
  • 00:22:01
    disproportionation so we have chlorine
  • 00:22:05
    turning into CL minus and clo3 minus it
  • 00:22:10
    is in an acidic solution
  • 00:22:14
    so putting oxidation numbers on here
  • 00:22:17
    chlorine and cl2 is a zero it's the
  • 00:22:20
    element Chlorine
  • 00:22:22
    CL minus is negative 1 and the chlorine
  • 00:22:25
    in clo3 minus is plus five
  • 00:22:29
    so chlorine is changing oxidation
  • 00:22:32
    numbers here
  • 00:22:34
    and the oxidation half reaction is that
  • 00:22:39
    cl2 is turning into clo3 minus right
  • 00:22:43
    going from a zero to a plus five that's
  • 00:22:46
    going to be losing electrons so let's
  • 00:22:50
    balance this half reaction first and
  • 00:22:52
    then look at the reduction side so
  • 00:22:55
    balancing this half reaction I need to
  • 00:22:57
    balance elements other than oxygen and
  • 00:22:59
    hydrogen which means my chlorines need
  • 00:23:02
    to be balanced and I need a 2 in front
  • 00:23:05
    of the clo3 minus
  • 00:23:08
    now each chlorine is going from a zero
  • 00:23:12
    to a plus five so that's five electrons
  • 00:23:17
    for each chlorine so it's kicking out a
  • 00:23:21
    total of 10 electrons
  • 00:23:24
    it's it's a lot of electrons but if our
  • 00:23:27
    count is right our count is right so
  • 00:23:29
    it's kicking out 10 electrons
  • 00:23:32
    the charge balancing in The Next Step
  • 00:23:36
    I'm in an acidic solution so I only have
  • 00:23:39
    h plus is available
  • 00:23:41
    on the left side My overall charge is
  • 00:23:44
    zero because all I have is that neutral
  • 00:23:46
    molecule
  • 00:23:48
    on the right side I have two things with
  • 00:23:51
    a negative one and then ten more things
  • 00:23:53
    that have a negative one I have a
  • 00:23:55
    negative 12 charge on the right side so
  • 00:23:59
    I need 12 h plus to balance this out
  • 00:24:05
    and now I have a zero charge on both
  • 00:24:09
    sides
  • 00:24:10
    and I have to balance this with water on
  • 00:24:14
    the other side well I have 12 H pluses
  • 00:24:17
    on the right if I make six water
  • 00:24:19
    molecules I know my H pluses will
  • 00:24:21
    balance
  • 00:24:22
    uh the question is do my oxygens balance
  • 00:24:26
    I have six oxygens on the left now
  • 00:24:28
    because of my water molecule
  • 00:24:30
    uh the clo3 minus has three oxygens
  • 00:24:35
    but I have two of them so I do have six
  • 00:24:38
    oxygens on the right so this is done
  • 00:24:41
    for the reduction half reaction it's
  • 00:24:46
    something turning into CL minus right
  • 00:24:49
    it's the other product
  • 00:24:51
    the reactant is also cl2 right so my cl2
  • 00:24:56
    has a zero oxidation number it is
  • 00:25:00
    disproportionating so this one molecule
  • 00:25:03
    I mean we'll right it will end up
  • 00:25:05
    reacting with another molecule but
  • 00:25:07
    there's always more than one molecule in
  • 00:25:09
    the solution
  • 00:25:10
    the cl2 is both being oxidized and being
  • 00:25:16
    reduced in this
  • 00:25:19
    reaction overall that we have
  • 00:25:22
    so the cl2 turning into CL minus
  • 00:25:26
    if you go through the steps to balance
  • 00:25:28
    it this is where you should end up cl2
  • 00:25:31
    plus two electrons turning into two CL
  • 00:25:34
    minus
  • 00:25:35
    so give this a try and make sure that
  • 00:25:38
    you end up in the same place remember
  • 00:25:39
    you're going to balance the chlorines
  • 00:25:41
    because it's not oxygen or hydrogen
  • 00:25:44
    then you're going to balance the
  • 00:25:46
    electrons and
  • 00:25:48
    once you do that I mean based on how
  • 00:25:49
    this turned out the charges will be the
  • 00:25:52
    same on both sides so you won't need to
  • 00:25:54
    add h plus
  • 00:25:55
    and there are no oxygens in here so we
  • 00:25:57
    don't need to balance anything out with
  • 00:25:59
    water molecules
  • 00:26:01
    before I add them together I do need to
  • 00:26:04
    make the number of electrons equal in
  • 00:26:07
    both which means that my reduction
  • 00:26:09
    reaction gets multiplied by 5.
  • 00:26:12
    this means my electrons will cancel when
  • 00:26:14
    I add it together
  • 00:26:16
    and my overall reaction is 6 H2O plus 6
  • 00:26:21
    cl2 turning into two clo3 minus
  • 00:26:25
    plus 12h plus plus 10 CL minus
  • 00:26:30
    now I don't have
  • 00:26:32
    h plus on both sides I don't have water
  • 00:26:35
    on both sides I don't have to worry
  • 00:26:36
    about that
  • 00:26:38
    but
  • 00:26:40
    all of my coefficients are divisible by
  • 00:26:42
    2. so you do still have to reduce the
  • 00:26:48
    coefficients in your reaction to get to
  • 00:26:51
    whole numbers
  • 00:26:54
    there is another example of
  • 00:26:56
    disproportionation in this experiment
  • 00:26:59
    and thiosulfate is the molecule that can
  • 00:27:04
    disproportionate
  • 00:27:05
    if you are in a highly acidic solution
  • 00:27:09
    it can turn into sulfur so Elemental
  • 00:27:14
    sulfur
  • 00:27:15
    just solid sulfur I know that many of
  • 00:27:18
    you know that there are different
  • 00:27:20
    combinations of sulfur that exist if you
  • 00:27:23
    need to write Elemental sulfur you can
  • 00:27:25
    just write the letter and then call it a
  • 00:27:27
    solid
  • 00:27:28
    and the other product that it turns into
  • 00:27:32
    is sulfurous acid
  • 00:27:35
    so I would take a minute here assign the
  • 00:27:39
    oxidation number to sulfur and
  • 00:27:42
    thiosulfate which is what this s203 two
  • 00:27:45
    minus is
  • 00:27:46
    the h plus is just telling you it's an
  • 00:27:48
    acidic solution obviously this is not
  • 00:27:51
    balanced but assign your oxidation
  • 00:27:53
    numbers in solid sulfur
  • 00:27:56
    and in the sulfurous acid to convince
  • 00:27:59
    yourself that this is disproportionation
  • 00:28:03
    at this point in the semester we will
  • 00:28:05
    start using the spectrophotometers so
  • 00:28:09
    the spectrophotometers are instruments
  • 00:28:11
    that will give you information about a
  • 00:28:14
    solution related to the color of light
  • 00:28:18
    it absorbs so let's get into it
  • 00:28:22
    so absorbance of light white light is a
  • 00:28:26
    combination of many colors of light
  • 00:28:29
    the visible region of the spectrum goes
  • 00:28:33
    from around 400 nanometers where the
  • 00:28:35
    light appears Violet to 700 nanometers
  • 00:28:38
    where the light appears red
  • 00:28:40
    the regions just outside the visible
  • 00:28:42
    region have names related to the colors
  • 00:28:45
    they are adjacent to so ultraviolet is
  • 00:28:49
    next to Violet and infrared is next to
  • 00:28:52
    Red
  • 00:28:53
    the energy of a photon of light is also
  • 00:28:57
    related to the wavelength and then in
  • 00:29:00
    turn to the frequency of that light
  • 00:29:03
    when a substance absorbs energy from the
  • 00:29:06
    photons of light the energy causes it to
  • 00:29:09
    go from a ground state to an excited
  • 00:29:11
    state where the difference in energy
  • 00:29:14
    between those two two states is exactly
  • 00:29:17
    equal to the energy of that photon
  • 00:29:22
    if we have a container of a solution and
  • 00:29:25
    white light hits the solution
  • 00:29:28
    the solution absorbs photons of a
  • 00:29:32
    specific energy so when the light comes
  • 00:29:35
    out the other side the spectrum is
  • 00:29:37
    missing the portion that was absorbed
  • 00:29:40
    and in our example the solution looks
  • 00:29:43
    blue and the reason it looks blue is
  • 00:29:47
    because it absorbed orange light so the
  • 00:29:51
    color absorbed and the color scene are
  • 00:29:54
    not the same
  • 00:29:56
    the color that we see is the
  • 00:29:58
    complementary color of the absorbed
  • 00:30:01
    light
  • 00:30:02
    you might remember using a color wheel
  • 00:30:04
    to match up complementary colors so blue
  • 00:30:08
    and orange are complementary which is
  • 00:30:10
    what we saw in this example
  • 00:30:11
    red and green are complementary and it's
  • 00:30:15
    not marked here but yellow and violet
  • 00:30:18
    are complementary
  • 00:30:21
    we are going to use an instrument called
  • 00:30:24
    a spectrophotometer and this is what we
  • 00:30:26
    will use to quantitatively measure the
  • 00:30:30
    amount of light absorbed so this this is
  • 00:30:33
    a quantitative
  • 00:30:35
    measurement quantitative process we
  • 00:30:39
    won't have to use it quantitatively just
  • 00:30:42
    yet but we will in the future and in
  • 00:30:45
    next semester too
  • 00:30:47
    um so to do this we actually take two
  • 00:30:49
    measurements
  • 00:30:51
    for both of the measurements we shine
  • 00:30:54
    white light through a container
  • 00:30:56
    any light that is not absorbed is passed
  • 00:30:59
    through a prism which separates it into
  • 00:31:02
    its component colors this is exactly
  • 00:31:04
    like the kind of prism that you might
  • 00:31:06
    put up to
  • 00:31:08
    um like in the window to make sure that
  • 00:31:11
    you get cute rainbows on your walls when
  • 00:31:13
    the light shines through so it's the
  • 00:31:14
    same type of prism probably a little bit
  • 00:31:17
    fancier
  • 00:31:19
    um one of the containers that we're
  • 00:31:22
    gonna pass this white light through
  • 00:31:24
    contains what we call a blank solution
  • 00:31:29
    um this is shown in the top sequence on
  • 00:31:30
    the slide right now
  • 00:31:32
    the measurement from the blank sample is
  • 00:31:34
    the Baseline it's like a background
  • 00:31:36
    measurement it makes sure that anything
  • 00:31:40
    we measure later
  • 00:31:42
    is only from the thing we are interested
  • 00:31:45
    in
  • 00:31:47
    so the light
  • 00:31:48
    that is passed through the sample which
  • 00:31:52
    is the container that has our colored
  • 00:31:54
    complex is also passed through a prism
  • 00:31:57
    and split into its component colors and
  • 00:32:00
    what the instrument does is it measures
  • 00:32:02
    how much of a specific color is missing
  • 00:32:06
    because it was absorbed
  • 00:32:09
    the way that these are related the
  • 00:32:12
    absorbance is the amount of light
  • 00:32:14
    absorbed by the sample
  • 00:32:16
    where this becomes useful quantitatively
  • 00:32:19
    is that the absorbance is proportional
  • 00:32:22
    to the concentration of the substance
  • 00:32:24
    that is absorbing the light
  • 00:32:26
    and if we plot that absorbance as a
  • 00:32:30
    function of wavelength we call the
  • 00:32:33
    result and absorption Spectrum so this
  • 00:32:36
    is all we're going to do right now is
  • 00:32:38
    look at absorbance versus wavelength
  • 00:32:41
    and pull some information out of it
  • 00:32:45
    so here's one example of what an
  • 00:32:48
    absorbent Spectrum might look like
  • 00:32:51
    from the shape we can tell there's one
  • 00:32:54
    specific wavelength at which the
  • 00:32:56
    solution absorbed the most light right
  • 00:32:59
    the tallest if absorbance is increasing
  • 00:33:02
    on the y-axis the tallest point of our
  • 00:33:05
    Peak
  • 00:33:06
    the wavelength that corresponds to this
  • 00:33:10
    highest level of absorbance is what we
  • 00:33:13
    call the Lambda Max right Lambda is what
  • 00:33:16
    we use for wavelength and then Max
  • 00:33:18
    because it is the wavelength of Maximum
  • 00:33:21
    absorption
  • 00:33:24
    the Lambda Max for this peak is in the
  • 00:33:28
    blue region of the visible spectrum
  • 00:33:30
    which we can tell from how our range
  • 00:33:34
    goes from 400 being Violet 700 being red
  • 00:33:38
    and if this absorption is in the blue
  • 00:33:42
    region of the spectrum then the solution
  • 00:33:44
    would appear Orange
  • 00:33:47
    an absorbent Spectrum with a Lambda Max
  • 00:33:49
    in the yellow green region would appear
  • 00:33:53
    violet
  • 00:33:55
    and some solutions have a Lambda Max
  • 00:33:57
    outside the visible region
  • 00:34:00
    so for example a solution of dichromate
  • 00:34:04
    has a Lambda Max around 330 nanometers
  • 00:34:07
    which is outside of the visible region
  • 00:34:11
    however because the tail of the
  • 00:34:14
    absorbent Spectrum
  • 00:34:16
    shows that dichromate absorbs some light
  • 00:34:20
    in the Violet region it does appear
  • 00:34:23
    yellow
  • 00:34:24
    and you should carefully read the
  • 00:34:27
    operation instructions in the appendix
  • 00:34:30
    of your lab manual which these are also
  • 00:34:33
    posted on lab archives for the sparta
  • 00:34:36
    photometer before you go to lab and or
  • 00:34:40
    before you watch The Experiment video
  • 00:34:43
    um probably both and if you watch The
  • 00:34:46
    Experiment video twice it's maybe okay
  • 00:34:49
    to read it the first time not read it
  • 00:34:51
    the first time but make sure you have
  • 00:34:52
    looked at the instructions
  • 00:34:55
    and also looked at the video
  • 00:34:57
    in conjunction in some way before going
  • 00:35:00
    into lab
  • 00:35:02
    when you get data for an absorbent
  • 00:35:05
    Spectrum you're going to get two values
  • 00:35:08
    so make sure you know what these values
  • 00:35:12
    mean
  • 00:35:13
    our spectrophotometer will read in the
  • 00:35:18
    visible and near visible region so
  • 00:35:23
    a number between 400 and 700 is going to
  • 00:35:29
    be a wavelength so the the computer so
  • 00:35:33
    well it's a little computer but the
  • 00:35:35
    computer connected to the sparkle
  • 00:35:36
    photometer will find this wavelength for
  • 00:35:39
    you but you need to make sure that
  • 00:35:41
    you're looking for the bigger number as
  • 00:35:44
    your Lambda Max
  • 00:35:46
    and the smaller number as your
  • 00:35:49
    absorbance
  • 00:35:52
    okay
  • 00:35:53
    let's take a look at what's happening in
  • 00:35:56
    this experiment
  • 00:35:58
    so there are a couple of parts to this
  • 00:36:00
    experiment the one that I want to really
  • 00:36:02
    spend time on is part A where you will
  • 00:36:06
    have the progressive reduction of
  • 00:36:08
    dioxovanadium five
  • 00:36:11
    so this is
  • 00:36:14
    um this is really cool the colors are
  • 00:36:16
    really pretty which extra reminder don't
  • 00:36:19
    forget to have someone tell you what
  • 00:36:21
    those colors are if you're having any
  • 00:36:23
    trouble distinguishing them
  • 00:36:25
    and you will see a color change for
  • 00:36:28
    every oxidation state of vanadium that
  • 00:36:31
    it's possible for us to achieve
  • 00:36:34
    so starting with metavanidate which is
  • 00:36:37
    vo3 minus
  • 00:36:40
    in which Vanadium has a plus five
  • 00:36:43
    oxidation state
  • 00:36:45
    if you put this into
  • 00:36:47
    sulfuric acid you get VO2 plus which is
  • 00:36:51
    dioxovanidium
  • 00:36:54
    and we're going to put a 5 after it
  • 00:36:57
    because
  • 00:36:58
    Vanadium can obviously have multiple
  • 00:37:00
    oxidation numbers so we need to label
  • 00:37:03
    that when we're doing naming so VO2 plus
  • 00:37:06
    is dioxovanadium in which Vanadium has a
  • 00:37:10
    plus five oxidation state so just taking
  • 00:37:13
    metavanidate and dissolving it in
  • 00:37:15
    sulfuric acid does not change the
  • 00:37:19
    oxidation number
  • 00:37:20
    it just ends up in a different form that
  • 00:37:24
    gives us this really pretty yellow color
  • 00:37:28
    oh and I guess I do I have a reminder
  • 00:37:30
    here of of what's Happening which is
  • 00:37:34
    really just some h3o Plus
  • 00:37:38
    and or h plus however you like to write
  • 00:37:40
    it reacting with the vo3 minus
  • 00:37:43
    and turning into H2O so we don't have
  • 00:37:47
    any oxidation number changes here
  • 00:37:50
    from the dioxovanadium five if you put
  • 00:37:54
    zinc metal in the solution you can turn
  • 00:37:58
    into
  • 00:37:59
    oxovanadium 4 which is v o
  • 00:38:04
    with a two plus charge which right these
  • 00:38:07
    are really difficult to say because
  • 00:38:09
    there's VO2 with a plush charge and then
  • 00:38:12
    there's vo with a two plus
  • 00:38:14
    in the vo with a two plus the oxidation
  • 00:38:19
    number of vanadium is four and it's blue
  • 00:38:25
    if this continues to react with zinc
  • 00:38:28
    metal you'll go to Vanadium three plus
  • 00:38:31
    just the Vanadium Ion with a plus three
  • 00:38:35
    this is kind of an aqui green color
  • 00:38:39
    and if it reacts with zinc again
  • 00:38:43
    you'll get Vanadium two and Vanadium two
  • 00:38:46
    is purple
  • 00:38:47
    so you will see from yellow to Blue to
  • 00:38:51
    Aqua to purple the oxidation number of
  • 00:38:54
    vanadium change from five to four to
  • 00:38:56
    three to two
  • 00:38:58
    and as long as there is zinc metal in
  • 00:39:01
    here you can maintain this reaction so
  • 00:39:07
    a couple of things to think about and
  • 00:39:10
    make sure you understand
  • 00:39:12
    what does the thing turn into
  • 00:39:14
    and does that mean the zinc is oxidized
  • 00:39:17
    or reduced
  • 00:39:19
    is the Vanadium oxidized or reduced
  • 00:39:24
    what gas could be forming
  • 00:39:30
    the Vanadium 2 this is the
  • 00:39:35
    um lowest number that you can get for a
  • 00:39:37
    Vanadium oxidation number you cannot go
  • 00:39:40
    Vanadium one and
  • 00:39:44
    when mixed with zinc we're not going to
  • 00:39:46
    go to Vanadium zero
  • 00:39:48
    so you will not produce Elemental
  • 00:39:51
    Vanadium so Vanadium two plus is the
  • 00:39:54
    limit
  • 00:39:54
    so that means Vanadium 2 plus can only
  • 00:39:58
    act as a reducing agent
  • 00:40:02
    so if you mixed Vanadium 2 plus with
  • 00:40:04
    something
  • 00:40:06
    the only thing Vanadium could do is
  • 00:40:09
    cause the reduction of that thing
  • 00:40:12
    and Vanadium 5 can only act as an
  • 00:40:15
    oxidizing agent so our yellow solution
  • 00:40:19
    Vanadium with a plus five oxidation
  • 00:40:22
    state you cannot go any higher
  • 00:40:27
    so the only thing that Vanadium 5 could
  • 00:40:31
    do is cause the oxidation of something
  • 00:40:34
    else
  • 00:40:37
    and the color of the solution here tells
  • 00:40:41
    us the oxidation number of vanadium in
  • 00:40:43
    the solution so if you had any one of
  • 00:40:47
    these Vanadium solutions to start with
  • 00:40:50
    from the color you would know the
  • 00:40:52
    beginning oxidation number and after
  • 00:40:55
    reacting it with something
  • 00:40:56
    from the color you would know the ending
  • 00:40:59
    oxidation number
Tags
  • dismutation
  • réaction redox
  • équilibrage
  • spectrophotomètre
  • absorbance
  • Lambda Max
  • oxyde de vanadium
  • chimie analytique
  • peroxyde d'hydrogène
  • équation chimique