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Calculate the cell e.m.f. and DeltaG for...

Calculate the cell e.m.f. and `DeltaG` for the cell reaction at 298K for the cell.
`Zn(s) | Zn^(2+) (0.0004M) ||Cd^(2+) (0.2M)|Cd(s)`
Given, `E_(Zn^(2+)//Zn)^(@) =- 0.763 V, E_(Cd^(+2)//Cd)^(@) = - 0.403 V` at `298K`.
`F = 96500 C mol^(-1)`.

Text Solution

Verified by Experts

`E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)=-0.403-(-0.763)=0.360V`
`E_(cell)=E_(cell)^(@)-(0.059)/(n)log(([Zn^(2+)(aq)])/([Cd^(2+)(aq)]))`
`E_(cell)=0.36-(0.059)/(2)log((0.1)/(0.01))=0.33V`
`triangle_(r)G^(@)=-nFE_(cell)^(@)=-2xx96500xx0.36=-69480Jmol^(-1)=-69.48kJmol^(-1)`
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