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:
E ° M g 2 + ∕ M g = – 2.37 V ,
E ° C u 2 + ∕ C u = + 0.34 V
E ° F e 2 + ∕ F e = – 0.44 V ,
E ° S n 2 + ∕ S n = – 0.14 V
E ° B r 2 ∕ B r – = + 1.08 V
(i) M g ( s ) | M g 2 + ( 0.001 M ) | | C u 2 + ( 0.0001 M ) | C u ( s )
(ii) F e ( s ) | F e 2 + ( 0.001 M ) | | H + ( 1 M ) | H 2 ( g ) ( 1 b a r ) | P t ( s )
(iii) S n ( s ) | S n 2 + ( 0.050 M ) | | H + ( 0.020 M ) | H 2 ( g ) ( 1 b a r ) | P t ( s )
(iv) P t ( s ) | B r 2 ( l ) | B r – ( 0.010 M ) | | H + ( 0.030 M ) | H 2 ( g ) ( 1 b a r ) | P t ( s )
Given :
Solution:
(i) The electrode reactions are
At anode :M g ( s ) → M g 2 + ( 0.001 M ) + 2 e –
At cathode :C u 2 + ( 0.0001 M ) + 2 e – → C u ( s )
Net reaction :M g ( s ) + C u 2 + ( 0.001 M ) → M g 2 + ( 0.0001 M ) + C u ( s )
The Nernst equation for this cell at 25°C
E c e l l = E ° c e l l −
log
WhereE ° a n o d e = − 2.37 V ; E ° c a t h o d e = + 0.34 V
∴ E ° c e l l = E ° c a t h o d e − E ° a n o d e
= ( + 0.34 V ) − ( − 2.37 V )
= + 2.71 V
The cell emf is then given by
E c e l l = 2.71 −
log (
)
E c e l l = ( 2.71 −
log 10 ) V
= 2.71 – 0.03 = 2.68 V
(ii) The electrode reactions are
At anode :F e ( s ) → F e 2 + ( 0.001 M ) + 2 e –
At cathode :2 H + ( 1 M ) + 2 e – → H 2 ( 1 b a r )
Net reaction :F e ( s ) + 2 H + ( 1 M ) → F e 2 + ( 0.001 M ) + H 2 ( 1 b a r )
The Nernst equation of this cell at 25°C
E c e l l = E ° c e l l −
log
E ° c e l l = E ° c a t h o d e − E ° a n o d e
= E ° H + | H 2 − E ° F e 2 + | F e
= 0.000 V – ( – 0.44 V )
= + 0.44 V
The cell emf is then given by
E c e l l = 0.44 −
log
= 0.44 − 0.0296 log (
)
= 0.44 − 0.0296 log ( 10 − 3 )
= 0.44 + ( 3 × 0.0296 )
= 0.44 + 0.0888
Therefore,E c e l l = + 0.53 V
(iii) The electrode reactions are
At anode :S n ( s ) → S n 2 + ( 0.05 M ) + 2 e –
At cathode :2 H + ( 0.02 M ) + 2 e – → H 2 ( 1 b a r )
Net reaction :S n ( s ) + 2 H + ( 0.02 M ) → S n 2 + ( 0.05 M ) + H 2 ( 1 b a r )
The Nernst equation of this cell at 25°C
E c e l l = E ° c e l l −
log
E ° c e l l = E ° H + | H 2 − E ° S n 2 + | S n
= 0.000 V – ( – 0.14 V )
= + 0.14 V
or,E c e l l = E ° c e l l − 0.0296 log
= E ° c e l l − 0.0296 log (
)
= E ° c e l l − 0.0296 ( log 125 )
= E ° c e l l − 0.0296 × 2.0969
= E ° c e l l − 0.06
E c e l l = 0.14 – 0.06 = 0.08 V
(iv) The electrode reactions are
At anode :2 B r – ( 0.01 M ) → B r 2 ( l ) + 2 e –
At cathode :2 H + ( 0.03 M ) + 2 e – → H 2 ( 1 b a r )
Net reaction :2 H + ( 0.03 M ) + 2 B r − ( 0.01 M ) → B r 2 ( l ) + H 2 ( 1 b a r )
The Nernst equation of this cell at 25°C is
E c e l l = E ° c e l l −
log
E c e l l 0 = E ° H + ∣ H 2 − E ° B r 2 ∣ B r −
= 0 − 1.08
= − 1.08 − 0.0296 × log
= − 1.08 − 0.0296 log [
] V = − 1.08 − 0.0296 × log (
)
= − 1.08 − 0.0296 ( log 10 8 − log 9 ) V
= − 1.08 − 0.0296 ( 8 − 0.9542 ) V
= – 1.08 – 0.0296 ( 7.0457 )
E c e l l = – 1.08 – 0.21 = – 1.29 V
At anode :
At cathode :
Net reaction :
The Nernst equation for this cell at 25°C
Where
The cell emf is then given by
(ii) The electrode reactions are
At anode :
At cathode :
Net reaction :
The Nernst equation of this cell at 25°C
The cell emf is then given by
Therefore,
(iii) The electrode reactions are
At anode :
At cathode :
Net reaction :
The Nernst equation of this cell at 25°C
or,
(iv) The electrode reactions are
At anode :
At cathode :
Net reaction :
The Nernst equation of this cell at 25°C is
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