Cu⁺ + e^-⇾ Cu; ∆G^o₃=?
(i)------(ii) gives the required result, i.e.,ΔG₃^° = ∆G⁰₁ - ΔG₂^° = [-0.674 - (- 0·153)] F = - 0·521 F
\(-nFE^0_{Cu^+/CU}=-0.5621F\ or\ E^0_{CU^+/Cu}=0.521V\)

Comparing oxidation potential of iron with those of metals A, B and C, iron has higher oxidation potential than C only. Hence, when coating of C is broken, rusting will become faster.

Tarnish will be removed if the following reaction takes place :
\(Al+{ { Ag }_{ 2 } }S\longrightarrow { Al }^{ 3+ }+2Ag+{ S }^{ 2- }\)
EMF of this cell reaction = \({ E }_{ { Ag }_{ 2 }S/{ 2Ag },{ S }^{ 2- } }^{ 0 }-{ E }_{ { Al }^{ 3+ }/{ Al } }^{ 0 }=-0.71-(-1.66)V=+0.95V\)
As EMF is positive, the reaction will take place and tarnish will be removed.

For the cell as represented, \({ E }_{ cell }^{ 0 }={ E }_{ { Ce }^{ 4+ },{ Ce }^{ 3+ } }^{ 0 }-{ E }_{ { Fe }^{ 3+ },{ Fe }^{ 2+ } }^{ 0 }=1.61-0.77=0.84V\)
As \({ E }_{ cell }^{ 0 }>0\) , left hand electrode is anode while right hand electrode is cathode. Hence, current will flow from right to left. The current will decrease with time ultimately the reaction  stops.

Fe➝ Fe²⁺+ 2 e^- , - ΔG^o₁ = 2 E⁰₁F
Fe²⁺➝ Fe³⁺+ e^-, - ΔG^o₂ = E⁰₂F
Fe ➝ Fe³⁺+ 3 e^-, -ΔG^o₃ = 3 E⁰₃F
Subtracting eqn. (i) from equation (iii), we get
0➝ Fe³⁺- Fe²⁺+ e^-, - ΔG^o₃+ ΔG^o₁ = F (3 E⁰₃- 2 E⁰₁)
Fe²⁺➝ Fe³⁺+ e^-, - ΔG^o₃+ ΔG^o₁ = (3 E⁰₃- 2 E⁰₁)F
Comparing eqn. (ii) and (iv), we get
- ΔG^o₂ = -ΔG^o₃+ ΔG^o₁
E⁰₂F = (3 E⁰₃- 2 E^O₁)F or E⁰₂ = 3 E⁰₃- 2 E^o₁ or 3 E⁰₃ = E⁰₂+ 2 E⁰_l