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 :
If $$E_{z{n^{2 + }}/zn}^0$$ = $$-$$ 0.76 V, which cathode will give maximum value of Eocell per electron transferred?
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)
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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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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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