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For the above cell :...

For the above cell :

A

`E_(cell)lt0,DeltaGgt0`

B

`E_(cell)lt0, DeltaG lt0`

C

`E_(cell) lt 0, DeltaG =^(@) gt 0`

D

`E_(cell)gt 0,DeltaG^(@) lt 0`

Text Solution

Verified by Experts

The net cell reaction is
`M_((s))+M_(((aq,0.05M)))^(+)+M_((s))`
according to Nernst equation,
`E_(cell)=E_(cell)^(@)-(0.059)/nlog.([M^(+)]_((0.05M)))/([M^(+)]_(1M))=E_(cell)^(@)-(0.059)/1log(0.05)=0-0.059 log(5xx10^(-2))=-0.059[-2+log5]`
`therefore E_(cell)=+ve or E_(cell)gt 0 and` Hence `DeltaGgt0` as `Deltag=-nFE_(cell)`
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