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- May 22, 2016
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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!!!!
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!!!!