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The emf of the cell, Ni|Ni^(2+)(1.0M)||A...

The emf of the cell, `Ni|Ni^(2+)(1.0M)||Ag^(+)(1.0M)|Ag [E^(@)` for `Ni^(2+)//Ni =- 0.25` volt, `E^(@)` for `Ag^(+)//Ag = 0.80` volt] is given by : `[E^(@)` for `Ag^(+)//Ag = 0.80` volt]

A

`-0.25 +0.80 = 0.55` volt

B

`-0.25 -(+0.80) =- 1.05` volt

C

`0 +0.80 -(-0.25)=+ 1.05` volt

D

`-0.80 -(+0.25) =- 0.55` volt

Text Solution

Verified by Experts

The correct Answer is:
C
`E_("cell")^(@)= E_("cathod")^(@)-E_("anode")^(@)=0.80 - (-0.25)= 1.05 V`
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