Calculate the standard cell potential of the galvanic cell in which the following reaction takes place:
`2Cr(s)+3Cd^(2+)(aq)to2Cr^(3+)(aq)+3Cd(s)`
Also calcuate the `triangle_(r)G^(ɵ)` value of the reaction
(given `E_(cr^(3+)//Cr)^(ɵ)=-0.74V,E_(Cd^(3+)//Cd)^(ɵ)=-0.40V` and `F=96500Cmol^(-1)`

Explanation

`E_(cell)^(ɵ)=E_(Cd^(2+)//Cd)^(@)-E_(Cr^(3-)//Cr)^(@)`
`=-0.40-(-0.74)`
`=-0.40+0.74=+0.34V`
the half cell reaction are
`2Cr(s)to2Cr^(3+)+6e^(-)`
`underline(3Cd^(2+)+6e^(-)to3Cd(s))`
`underline(2Cr(s)+3Cd^(2+)to2Cr^(3+)+3Cd(s))`
`therefore` No. of electrons `n=6`
`triangle_(r)G^(ɵ)=nFE^(ɵ)`
`=-6xx96500xx0.34`
`=-196.86kJmol^(-1)`