NCERT Class XII Chemistry
Chapter - Electrochemistry
Questions with Solutions

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Question : 5
Total: 32
Write the Nernst equation and emf of the following cells at 298 K:
(i) Mg(s)|Mg2+(0.001M)||Cu2+(0.0001M)|Cu(s)
(ii) Fe(s)|Fe2+(0.001M)||H+(1M)|H2(g)(1bar)|Pt(s)
(iii) Sn(s)|Sn2+(0.050M)||H+(0.020M)|H2(g)(1bar)|Pt(s)
(iv) Pt(s)|Br2(l)|Br(0.010M)||H+(0.030M)|H2(g)(1bar)|Pt(s)
Given :
E°Mg2+Mg=2.37V,
E°Cu2+Cu=+0.34V
E°Fe2+Fe=0.44V,
E°Sn2+Sn=0.14V
E°Br2Br=+1.08V
Solution:  
(i) The electrode reactions are
At anode : Mg(s)Mg2+(0.001M)+2e
At cathode : Cu2+(0.0001M)+2eCu(s)
Net reaction : Mg(s)+Cu2+(0.001M) Mg2+(0.0001M)+Cu(s)
The Nernst equation for this cell at 25°C
E cell=E° cell
0.0591
2
log
[ Mg2+]
[ Cu2+ ]

Where E° anode=2.37V;E° cathode=+0.34V
E° cell=E° cathodeE° anode
=(+0.34 V)(2.37V)
=+2.71 V
The cell emf is then given by
Ecell=2.71
0.0591
2
log
(
0.001
0.0001
)

E cell= (2.71
0.0591
2
log
10
)
V

=2.710.03=2.68V
(ii) The electrode reactions are
At anode : Fe(s)Fe2+(0.001M)+2e
At cathode : 2H+(1M)+2eH2(1bar)
Net reaction : Fe(s)+2H+(1M)Fe2+(0.001M)+H2(1bar)
The Nernst equation of this cell at 25°C
E cell=E° cell
0.0591
2
log
[ Fe2+] (pH2 )
[ H+ ]2

E° cell=E° cathodeE° anode
=E° H+| H2E° Fe2+|Fe
=0.000V(0.44V)
=+0.44V
The cell emf is then given by
Ecell=0.44
0.0591
2
log
0.001×1
(1)2

=0.440.0296log(
1
1000
)

=0.440.0296log(103)
=0.44+(3×0.0296)
=0.44+0.0888
Therefore, Ecell=+0.53V
(iii) The electrode reactions are
At anode : Sn(s)Sn2+(0.05M)+2e
At cathode : 2H+(0.02M)+2eH2(1bar)
Net reaction :Sn(s)+2H+(0.02M)Sn2+(0.05M)+H2(1bar)
The Nernst equation of this cell at 25°C
E cell=E° cell
0.0591
2
log
[ Sn2+ ] ( pH2 )
[ H+ ]2

E° cell=E° H+|H2E° Sn2+|Sn
=0.000V(0.14V)
=+0.14V
or, E cell=E° cell0.0296log
0.05×1
(0.02)2

=E° cell0.0296log (
0.05
0.0004
)

=E° cell0.0296(log125)
=E°cell0.0296×2.0969
=E° cell0.06
Ecell=0.14 0.06=0.08V
(iv) The electrode reactions are
At anode :2Br(0.01M)Br2(l)+2e
At cathode : 2H+(0.03M)+2eH2(1bar)
Net reaction : 2H+(0.03M)+2Br(0.01M)Br2(l)+H2(1bar)
The Nernst equation of this cell at 25°C is
E cell=E°cell
0.0591
2
log
[Br2(l)] ( pH2 )
[ H+ ]2[ Br]2

Ecell0=E° H+H2E°Br2Br
=01.08
=1.080.0296×log
1
(0.03)2(0.01)2

=1.080.0296log[
1
(9×104 ) (1×104 )
]
V
=1.080.0296×log (
108
9
)

=1.080.0296 (log108log9 )V
=1.080.0296(80.9542) V
=1.080.0296(7.0457)
Ecell=1.080.21=1.29V
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