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Emf of the cell Ni|Ni^(2+) (0.1M)||Au^(3...

Emf of the cell `Ni|Ni^(2+) (0.1M)||Au^(3+) (1.0M)|Au` will be (Given `E_(Ni^(2+)//Ni)^(@)= 0.25V, E_(Au^(2+)//Au)^(@)= 1.5V`)

A

`+1.75V`

B

`+1.7795V`

C

`+0.7795V`

D

`-1.7795V`

Text Solution

Verified by Experts

The correct Answer is:
B
Cell reaction : `3Ni+ 2Au^(3+) rarr 3Ni^(2+) + 2Au`
`E_("cell") = E_("cell")^(@) - (0.0591)/(6) "log " ([Ni^(2+)]^(3))/([Au^(3+)]^(2))= (0.25+1.5) - (0.0591)/(6) "log" ((0.1)^(2))/((1)^(2))`
`=1.75 + (0.0591)/(2) xx log (10) = 1.75 + 0.0295 = +1.7795V`
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