Efficiency of the following cell is 84 %. A(s)+B2+(aq.)⇌A2+(aq.)+B(s);ΔHo = -285kJ. Then the standard EMF of the cell will be:
2.2 V
2.40 V
0.9 V
1.24 v
A
2.2 V
B
2.40 V
C
0.9 V
D
1.24 v
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Solution
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A+B+2⇌A+2+Bn=2
Δ4=Δ−TΔS
Δs=0
Δ4=ΔH=−2gskg
Δ4=−nFEocell
Eocell=Δ4−nF=−285×100096500×2
Ecell=1.4767
Ecell=1.4767×84100=1.24V
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