Question

A voltaic cell employs the following redox reaction:
Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq)
The half-reactions are as follows:
Sn2+(aq)+2e−→Sn(s) E∘=−0.14
Mn2+(aq)+2e−→Mn(s) E∘=−1.18
Calculate the cell potential at 25 ∘C under each set of conditions.
[Sn2+]= 1.73×10−2 M and [Mn2+]= 1.75 M
[Sn2+]= 1.75 M and [Mn2+]= 1.73×10−2 M

Answer 1

Ered Sn = Eored - (0.0592/2)*log(Sn/Sn^2+)
Ered Sn = -0.14 - (0.0592/2)*log (1/1.73E-2 M). You solve it.
Ered Mn = -1.18 - (0.0591/2)*log (1/1.75 M). You solve it.
Then Ecell = Eored + Eoox.
Eored is Sn and that goes in as calculated above.
Eoox is Mn and you turn the reaction above around AND change sign to +
Add the two to get the cell reaction and the Ecell.
Post your work if you get stuck.
Personally, I think it is easier to do it as follows and I recommend it to you.
Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq)
Sn^2+(aq) + 2e => Sn(s) ..........Eored = -0.14 v
Mn(s) ==> Mn^2+ + 2e ..............Eoox = +1.18 v
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Sn^2+(aq) + Mn(s) ==> Sn(s) + Mn^2+(aq) Eocell = 1.04 v
That is the equation you started with except now is standard Eo cell potential.
Now use the Nernst equation to correct for the fact it isn't standard concentrations.
Ecell = Eo cell - (0.0592/2)*log (Qrxn) and plug in as follows;
For Eo cell you have that above as +1.04 v. For Qrxn it is (products)/(reactants)
Qrxn = [Sn(s)][Mn^2+]/[Sn^2+][Mn(s)] = (1)(1.75)/(1.73E-2)(1)]
You see I plugged in the first set of conditions. Where did the 1 come from? For the pure element we plug in 1 for any pure element which is standard. Hope this helps.