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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

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2 answers
  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.

    Ecell = Eocell - (0.0592/n)*log Q.
    Q is where you substitute the concns given.

    Then delta Go = nFEocell

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  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.

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