JEE MAIN - Chemistry (2019 - 11th January Morning Slot - No. 5)

For the cell Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s), different half cells and their standard electrode potentials are given below :

Mx+ (aq)/M(s) Au3+(aq)/Au(s) Ag+(aq)/Ag(s) Fe3+(aq)/Fe2+ (aq) Fe2+(aq)/Fe(s)
E0Mx+/M/(V) 1.40 0.80 0.77 $$-$$0.44


If $$E_{z{n^{2 + }}/zn}^0$$ = $$-$$ 0.76 V, which cathode will give maximum value of Eocell per electron transferred?
Ag+/Ag
Fe3+/Fe2+
Au3+/Au
Fe2+/Fe

Explanation

Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s)
---------------------------------------
        Anode              Cathode

Eocell = Eocathode – Eoanode

(i) For Ag+/Ag :

Eocell = 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

$$ \therefore $$ Eocell per electron = $${{1.56} \over 2}$$ = 0.78

(ii) For Fe3+/Fe2+ :

Eocell = 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

$$ \therefore $$ Eocell per electron = $${{1.53} \over 2}$$ = 0.76

(iii) For Au3+/Au :

Eocell = 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

$$ \therefore $$ Eocell per electron = $${{2.16} \over 6}$$ = 0.36

(iv) For Fe2+/Fe :

Eocell = –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

$$ \therefore $$ Eocell per electron = $${{0.32} \over 2}$$ = 0.16

Eocell is maximum for EoAg+(aq)/Ag(s) .

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