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E^(@)(Ni^(2+)//Ni) =- 0.25 volt, E^(@)(A...

`E^(@)(Ni^(2+)//Ni) =- 0.25` volt, `E^(@)(Au^(3+)//Au) = 1.50` volt. The emf of the voltaic cell `Ni|Ni^(2+) (1.0M)||Au^(3+)(1.0M)|Au` is:-

A

`1.25` volt

B

`-1.75` volt

C

`1.75` volt

D

`4.0`volt

Text Solution

Verified by Experts

The correct Answer is:
C
`E_("cell")^(@)= E_("cathod")^(@)-E_("anode")^(@)=1.50 - (-0.25)= 1.75 V`
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