QJEEM19Che15f: JEE Main 2019 Chemistry - Class 12 JEEM19Che15f Solution: For the cell Zn(s)

Zn(s) |Zn²⁺ (aq)| |M^x+ (aq)| M(s)
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        Anode              Cathode

E^o_cell = E^o_cathode – E^o_anode

(i) For Ag⁺/Ag :

E^o_cell = 0.80 – (– 0.76) = 1.56 V

No of electrons transferred = LCM of valency factor of two electrode

Valency factor of Zn(s) |Zn²⁺ = 2

Valency factor of Ag⁺/Ag = 1

LCM of 1 and 2 = 2

$$\therefore$$ No of electrons transferred = 2

$$\therefore$$ E^o_cell per electron = $$\frac{1.56}{2}$$ = 0.78

(ii) For Fe³⁺/Fe²⁺ :

E^o_cell = 0.77 – (– 0.76) = 1.53 V

No of electrons transferred = LCM of valency factor of two electrode

Valency factor of Zn(s) |Zn²⁺ = 2

Valency factor of Fe³⁺/Fe²⁺ = 1

LCM of 2 and 1 = 2

$$\therefore$$ No of electrons transferred = 2

$$\therefore$$ E^o_cell per electron = $$\frac{1.53}{2}$$ = 0.76

(iii) For Au³⁺/Au :

E^o_cell = 1.40 – (– 0.76) = 2.16 V

No of electrons transferred = LCM of valency factor of two electrode

Valency factor of Zn(s) |Zn²⁺ = 2

Valency factor of Au³⁺/Au = 3

LCM of 2 and 3 = 6

$$\therefore$$ No of electrons transferred = 6

$$\therefore$$ E^o_cell per electron = $$\frac{2.16}{6}$$ = 0.36

(iv) For Fe²⁺/Fe :

E^o_cell = –0.44 – (– 0.76) = 0.32 V

No of electrons transferred = LCM of valency factor of two electrode

Valency factor of Zn(s) |Zn²⁺ = 2

Valency factor of Fe²⁺/Fe = 2

LCM of 2 and 2 = 2

$$\therefore$$ No of electrons transferred = 2

$$\therefore$$ E^o_cell per electron = $$\frac{0.32}{2}$$ = 0.16

E^o_cell is maximum for E^o_{Ag⁺(aq)/Ag(s)} .