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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)` |
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Answer» `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)` |
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