Chem 1046

An electrochemical cell is based on these two half-reactions:

Ox: Sn(s) → Sn2+(aq, 1.74 M) + 2 e-
Red: ClO2(g, 0.120 atm) + e- → ClO2-(aq, 1.44 M)

Calculate the cell potential at 25°C

E Sn>Sn2+ = -.14
E ClO2 = .95

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  1. There are a couple of ways of doing this. I think the easier of the two is as follows:
    ClO2 + e ==> ClO2^- Eo = +0.95v
    Sn(s) ==> Sn^2+ + 2e Eo = -0.14
    2ClO2 + Sn(s) ==>2ClO2^- + Sn^2+ which is the cell rxn at standard states 1 atm, 1M, etc. Eocell = 0.95+(-0.14) = ?

    Then Ecell = Eocell -(0.0592/n)*log Q
    where Q = (ClO2^-)^2*(Sn^2+)/p^2ofClO2*(Sn)(s)(1.74)(1.44)^2/(0.120)^2 * 1

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