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Question

Find the G of a galvanic cell with Ag and Al.

Standard Reduction Potentials at 25C
Half-ReactionE(V)
F2+2e2F2.87
Ag2+eAg+1.99
Co3++eCo2+1.82
H2O2+2H++2e2H2O1.78
MnO4+4H++3eMnO2+2H2O1.68
2e+2H++IO4IO3+H2O1.60
MnO4+8H++5eMn2++4H2O1.51
Au3++3eAu1.50
Cl2+2e2Cl1.36
O2+4H++4e2H2O1.23
MnO2+4H++2eMn2++2H2O1.21
Br2+2e2Br1.09
NO3+4H++3eNO+2H2O0.96
ClO2+eClO20.954
2Hg2++2eHg2+20.91
Ag++eAg0.80
Hg2+2+2e2Hg0.80
Fe3++eFe2+0.77
O2+2H++2eH2O20.68
MnO4+eMnO240.56
I2+2e2I0.54
Cu++eCu0.52
O2+2H2O+4e4OH0.40
Cu2++2eCu0.34
AgCl+eAg+Cl0.22
Cu2++eCu+0.16
2H++2eH20.00
Fe3++3eFe0.036
Pb2++2ePb0.13
Ni2++2eNi0.23
Cd2++2eCd0.40
Fe2++2eFe0.44
Cr3++eCr2+0.50
Cr3++3eCr0.73
Zn2++2eZn0.76
2H2O+2eH2+2OH0.83
Mn2++2eMn1.18
Al3++3eAl1.66
H2+2e2H2.23
Mg2++2eMg2.37
Na++eNa2.71
Ca2++2eCa2.76
Ba2++2eBa2.90
K++eK2.92
Li++eLi3.05

A
712 kJ
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B
240 kJ
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C
250 kJ
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D
730 kJ
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Solution

The correct option is A 712 kJ
We have,
E0 for Al=1.66 V and E0 for Ag=0.80 V

E0cell=0.8(1.66)=2.46 V

Now,
ΔG0=nFE0cell

ΔG0=3×96500×2.46

ΔG0=712170 J=712 kJ.

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