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Question

Calculate the cell potential and Gibb's energy change for the following galvanic cell at 298 K.
Cd(s)|Cd(aq)(0.001M)||H+(aq)(0.01M)|H2(g), 1atm| Pt

A
0.38
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B
-0.38
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C
0.36
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D
-0.36
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Solution

The correct option is A 0.38
Cd2++2eCdE0=0.41V
H++2eH2E0=0.0V
Cd+2H+Cd2++H2
E0cell=0.0V(0.41)V
=0.41V
Now, nernst equation:
Ecell=E00.059nlogCd2+(H+)2
Ecell=0.410.0592log[0.001(0.01)2]
Ecell=0.3805V

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