PLEASE HELP GENERAL CHEM QUESTION Electric Cells !!

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The8

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I don't understand this all kaplan's explanation is absolutely horrendous.

For a cell with the following half reactions:
Anode: so2 + 2h20 ---> (so4)2- + 4h+ + 2e-
Cathode: pd2+ + 2e- ---> pd

How would decreasing the pH of the solution inside the cell affect the electromotive force (emf)?
A. Emf decreases
B. Emf remains the same
C. Emf increases
D. Emf becomes 0

The answer according to Kaplan is A. I think it should be C. Kaplan says that the reasoning is "a change in the pH correspond to an increase in H plus meaning oxidation product is increased." Isn't this the opposite of what happens with Le chatliers..... Please help!!!!

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For electricity to flow the anode must be able to liberate its electrons as drawn. According to Le Chatlier's principle, if you increase the protons in solution, you will drive the reaction to the left; this makes the anode more reluctant to give up electrons. This should intuitively tell you that the emf will be reduced.

Or you can use the nernst equation to see the relation ship.
 
So the easy way to approach this is just to imagine Le Chatelier's principle. The reactions need to proceed as written in order for the cell to work. Adding protons favors the reverse reaction in the anode and thus makes the cell less likely to work, meaning that EMF logically decreases.

Mathematically, E(cell) = E(cathode) - E(anode) where E refers to reduction potentials. E(cathode) stays the same. As you increase proton concentration, the reverse reaction is favored in the anode, which is a reduction. So reduction at the anode becomes more favored and thus the reduction potential of the anode, E(anode), increases. Thus, E(cell) decreases.
 
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