Question

Calculate E, E^o and (delta)G^o for the following cell reaction:
Mg(s) + Sn2+(aq) (yields) Mg2+(aq) + Sn(s)
[Mg2+] = 0.045 M, [Sn2+] = 0.035 M

Answer 1

Mg ==> Mg^+2 + 2e Eo = 2.37 = E1
Sn^+2 + 2e ==> Sn Eo = -0.14 = E2
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Mg + Sn^+2 ==> Mg^+2 + Sn Eo = E1+E2 = 2.23 v.
Note that you should look up these values. My tables are 20 years old.

E_cell = E^o_cell - (0.0592/n)*log Q.
Q is where you substitute the concns given.

Then delta Go = nFE^o_cell

Answer 2

I was taught that E^o of the cell was the cathode minus anode, but when I do this after changing the signs for reversing the reaction, I get the wrong answer. I only get the right answer when I don't change the sign which doesn't seem right.