So a galvanic cell is constructed with 0.1 AgNO3 solution on one side with a sil

Question

So a galvanic cell is constructed with 0.1 AgNO3 solution on one side with a silver wire as the electrode, and the other cell consists of a 0.1 M Mg(NO3)2 solution with magnesium wire as the electrode. What is the cell potential for the cell?

I've looked up the potential for the half reactions at 298K, pH= 0, where

Ag (s) --> Ag+ + e-, +.8 V (anode)

Mg2+ + 2e- --> Mg(s), -2.372 V (cathode)

the overall reaction comes out to be Ag(s) + Mg+2 + e- --> Mg(s) + Ag+ .

I am trying to solve for the E using the following equation : E = -.0592/n (log anode/cathode). But what is n in this case if there is a e- in the reaction? Or is this equation even the correct one to use in this case?

Explanation / Answer

you have to multiply eqn 1 by 2 and then add eqns..n factor 2..in molarity turm you have to square conc.of Ag+ final eqn=-2.372-(0.8*2)-E = -.0592/2 (log anode)^2/(cathode)

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