14.11 e, calculation of standard cell potential  [ENDORSED]

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Yiling Liu 1N
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Joined: Sat Jul 09, 2016 3:00 am

14.11 e, calculation of standard cell potential

Postby Yiling Liu 1N » Mon Feb 06, 2017 9:29 pm

Pt(s) | Sn4+(aq), Sn2+(aq) || Cl-(aq) | Hg2Cl2(s) | Hg(l)

I looked up the standard potential for the reduction of Sn4+ to Sn2+, which is -0.14V.

However, we know that in this problem since Sn4+ and Sn2+ are on the left side of the diagram, they are at the anode and will be oxidized, not reduced. Thus the potential for oxidation should be +0.14V.

The reduction potential should stay the same for the right hand side (cathode) from Hg2Cl2 to Hg and Cl-, which is +0.27V.

Therefore I would think that the standard cell potential for this reaction is 0.27V + 0.14V = 0.41V. However, the answer key writes the answer as 0.27V - 0.15V.

What am I getting wrong? Thanks.

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Re: 14.11 e, calculation of standard cell potential  [ENDORSED]

Postby Chem_Mod » Mon Feb 06, 2017 10:11 pm

Remember that the equation

Eo(cathode) - Eo(anode) = Eocell

is only true when we plug in Eo for half reactions written as standard reduction potentials.

Don't flip signs when using this equation, use the signs as you find them in standard reduction potential tables and plug directly into the equation.

However, you appear to be mistaken in your statement "standard potential for the reduction of Sn4+ to Sn2+, which is -0.14V". It is +0.15 V. Check your answer again.


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