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At 25^(@)C equilibrium constant for the ...

At `25^(@)C` equilibrium constant for the reaction
`Ni_((s))+2Ag_((aq))^(+) too Ni_((aq))^(++)+2Ag_((s))`
is `2 xx 10^(35)`, then emf of the cell is

A

1.247 V

B

1.043 V

C

3.559 V

D

1.430 V

Text Solution

Verified by Experts

The correct Answer is:
B
`E_("cell")^(0) =(0.059)/(n)log KC`
`E_("cell")^(0) =(0.59)/(2)log 2 xx 10^(35)`
`=(0.59)/(2)xx35.3010=1.043 V`
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