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For the redox reaction: Zn(s) +Cu^(2+)...

For the redox reaction:
`Zn(s) +Cu^(2+) (0.1M) rarr Zn^(2+) (1M)+Cu(s)` taking place in a cell,
`E_(cell)^(@) is 1.10` volt. `E_(cell)` for the cell will be `(2.303 (RT)/(F) = 0l.0591)`

A

2.14volt

B

1.80volt

C

1.07volt

D

0.82volt

Text Solution

Verified by Experts

The correct Answer is:
C
`E=E^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
[Here n=2]
`E=1.1V-(0.0591V)/(2)"log"(1)/(0.1)`
`=(1.1-0.02955)V`
=1.07045=1.07V
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