KEAM 2016 Physics and Chemistry Paper

Given,
Zn2+∕Zn‌‌=−0.76V
Ni2+∕Ni‌‌=−0.23V
Fe2+∕Fe‌‌=−0.44V
and, relation
X+Y2+⟶X2++Y, means
X is oxidised to X2+ (acts as anode)
y2+is reduced to Y (actsas cathode)
Also. E‌cell ‌∘=ECo−EA∘
Now, for (a) X=Ni and Y=Fe
∴‌‌E‌cell ‌∘=EY∘−Ex∘
Ec‌cill ‌∘=−0.44−(−0.23)
‌‌=−0.44+0.23=−0.21V
∆G∘‌‌=−nFE∘
Thus. ‌‌=(−2×−0.21)F
‌‌=+0.42F(n=2)
(b) X= Ni, Y=Zn
∴ 0 Ecell∘=−0.76−(−0.23)=−0.53V
∴ ∆G∘=−nFE∘=(−2×−0.53)F
∆G∘=+1.06F
(c) X=Fe:Y=Zn
Ecell∘=−0.76−(−0.44)=−0.32V
∆G∘=−nFE∘=(−2×−0.32)F=+0.64F
(d) X=Zn,Y‌‌=Ni
E‌cell ‌∘‌‌=−0.23−(−0.76)=+0.53
∆G∘=−nFE∘=(−2×0.53)F
∴‌‌∆O=−1.06F