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
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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
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
1 answer
1 answer