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Question : 21
Total: 30
Calculate the emf of the following cell at 298 K
Fe ( s ) | Fe 2 + ( 0.001 M ) | | H + ( 1 M ) | H 2 ( g ) ( 1 bar ) ,
Pt ( s )
( Given E ° cell = + 0.44 V )
Solution:
According to given equation:
F e ( s ) + 2 H + ( a q ) → F e + 2 ( a q ) + H 2 ( g )
E ° cell = 0.44 V
By applying Nernst equation:
E cell = E ° cell −
log
E cell = 0.44 −
log
E cell = 0.44 − 0.0295 log 10 − 3
E cell = 0.44 − 0.0295 ( − 3 log 10 )
where[ log 10 = 1 ]
= 0.44 − 0.0295 ( − 3 × 1 )
= 0.44 + 0.0885
E cell = 0.528 V
By applying Nernst equation:
where
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