Sukanya Mohapatra 2G
Posts: 100
Joined: Sat Aug 17, 2019 12:18 am
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How does assigning oxidation numbers help you balance chemical equations? I saw my TA use it for a problem during discussion but wasn't able to get clarification. Thanks.

Jessica Castellanos
Posts: 102
Joined: Sat Aug 24, 2019 12:17 am
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I think your TA was balancing a redox reaction which is a different method of balancing reactions that uses oxidation numbers to balance chemical equations due to the amount of electrons lost and gained in the reaction that need to be accounted for and balanced. If that's the case and you need help balancing redox reactions, there are plenty of videos online to help explain the process. I hope this helps!

John Arambulo 1I
Posts: 58
Joined: Sat Jul 20, 2019 12:15 am

I’m not exactly sure what kind of chemical reaction it was when your TA was balancing an equation for one of your discussion problems, but in general, oxidation numbers are essential for oxidation-reduction (redox) reactions. The reason for this is because, in redox reactions, electrons are stripped/lost from an element (oxidation) while another element gains the electrons (reduction). Because both oxidation and reduction occur, there are redox half-reactions to show the distinction between the electrons being lost and being gained. Here is an example from Fundamentals K Redox Reactions on Section K.4 p.F85:

$Cu_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Cu_{(aq)}^{+2} + 2Ag_{(s)}$
The half-reactions would be:
Oxidation
$Cu_{(s)}\rightarrow Cu_{(aq)}^{+2}+2e^{-}$
Reduction
$2Ag_{(s)}^{+}+2e^{-}\rightarrow 2Ag_{(s)}$

These half-reactions represent the gain and loss of electrons in the reaction, but since the electrons are balanced out, they aren't shown in the overall equation. It must be noted that before working on any chemical reaction including redox reactions, the equation must be balanced. The half-reaction for the oxidation of Cu shows that $Cu_{(s)}$ loses two electrons to have the copper ion $Cu_{(aq)}^{+2}$ and $2e^{-}$ in the products. The oxidation number of Cu increases from the neutral 0 as $Cu_{(s)}$ to a positive +2 as $Cu_{(aq)}^{+2}$, and $2e^{-}$ is there to balance the charge. The importance of the balanced chemical equation is represented in the reduction half-reaction. The stoichiometric coefficients of $2Ag_{(s)}^{+}$ and $2Ag_{(s)}$ are 2 because the oxidation number of $2Ag_{(s)}^{+}$ is +1, but the $2e^{-}$ lost from $Cu_{(s)}$ needs to be accounted for to have $2e^{-}$ gained in the reduction half-reaction, hence the stoichiometric coefficient of 2. In this half-reaction, the oxidation number of $2Ag_{(s)}^{+}$ changes from a positive 2(+1)=+2, to a neutral 2(0)=0 as $2Ag_{(s)}$ because of the gained electrons. When the oxidation numbers are assigned to the reactants and products in the chemical equation, you will know that the equation is balanced because the charges are balanced and neutral in the overall chemical equation.

If you want to know more about assigning oxidation numbers and redox reactions, you can go to Fundamentals K starting on p.F78.

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