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E^@ value of Ni^(2+)//Ni is -0.25 V and...

`E^@` value of `Ni^(2+)//Ni` is -0.25 V and `Ag^+ //Ag` is +0.80 V . If a cell is made by taking the two electrodes what is the feasibility of the reaction?

A

Since `E^@` value for the cell will be positive, redox reaction is feasible.

B

Since `E^@` value for the cell will be negative, redox reaction is not feasible.

C

Ni cannot reduce `Ag^+` to Ag hence reaction is not feasible.

D

Ag can reduce `Ni^(2+)` to Ni hence reaction is feasible.

Text Solution

Verified by Experts

The correct Answer is:
A,B
The cell reaction will be `Ni_((s))+2Ag_((aq))^+ to Ni_((aq))^(2+) + 2Ag_((s))`
`E_"cell"^@=E_"cathode"-E_"anode"=0.80-(-0.25)= pm 1.05` V
`DeltaG^@= -nFE_"cell"^@ As E_"cell"^@ = +ve`,
`DeltaG^@=-ve` , hence reaction is feasible .
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