JEE Mains Model Paper 14

The electrode reaction written as reduction reaction is
Cu2+ + 2e− → Cu ∴ n = 2
Applying Nerst equation E = E° -

0.0591

2

log

1

[Cu2+]

[∵ [Cu (s) ] = 1]
= 0.34 -

0.0591

2

log

1

0.1

= 0.34 - 0.0295 × 1 = 0.3105 V