Posted by Julie on Tuesday, May 11, 2010 at 2:32pm.
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
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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