HW 6K.3

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Hannah Lee 2F
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Joined: Thu Jul 11, 2019 12:15 am

HW 6K.3

Postby Hannah Lee 2F » Wed Feb 19, 2020 11:33 pm

When balancing redox reactions, I thought that the # of e- transferred should correspond to the change in the oxidation number.

In 6K.3(a), S is oxidized from 2+ --> 3+, which means each S atom "lost" an e-; however, in the balanced oxidation half-reaction, it says that S2O3^2- actually lost 8 e-.

Does the # of e- transferred not always correspond to change in oxidation number?

Junwei Sun 4I
Posts: 125
Joined: Wed Oct 02, 2019 12:16 am

Re: HW 6K.3

Postby Junwei Sun 4I » Thu Feb 20, 2020 12:31 am

The change in oxidation number is not the change in number of electrons. S2O3 2- actually lost 8 electrons because when you write out the half reactions, you have to make sure the oxidation half and the reduction half reactions have the same number of electrons that can cancel out when adding them together.

805303639
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Re: HW 6K.3

Postby 805303639 » Thu Feb 20, 2020 2:37 pm

In S2O3 2-(aq), S has an oxidation number of +2. In SO4 2-(aq), S has an oxidation number of +6. After balancing the oxidation half-reaction, SO4 2-(aq) has a stoichiometric coefficient of 2. Since S2O3 2-(aq) has 2 S atoms, each with an oxidation number of +2, those S atoms together have an oxidation number of +4. Meanwhile, 2 SO4 2-(aq) has 2 S atoms, each with an oxidation number of +6, so those S atoms together have an oxidation number of +12. The sulfur atoms thus lost 12-4 = 8 electrons in this reaction.


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