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Calculate the standard cell potentails o...

Calculate the standard cell potentails of the galvanic cells in which the following reactions take place.
(a) `" " 2Cr(s)+3Cd^(2+) to 2Cr^(3+)(aq)+3Cd(s)`
Given `E_(Cr^(3+)//Cr)^(@)=-0.74" V" , E_(Cd^(2+)//Cd)^(@)=-0.40" V"`
(b) `" " Fe^(2+)(aq)+Ag^(+)(aq) to Fe^(3+)(aq)+Ag(s)`
Gievn `E_(Ag^(+)//Ag)^(@)=0.80" V" ,E_(Fe^(3+)//Fe^(2+))^(@)=0.77 " V"`
Also calculate `DeltaG^(@)` and equilibrium constant for the reaction.

Text Solution

Verified by Experts

(i) `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=-0.40V-(-0.74V)=+0.34V`
`Delta_(r)G^(@)=-nFE_(cell)^(@)=-6molxx96500" C "mol^(-1)xx0.34V`
`=-196860" CV "mol^(-1)=-196860J" "mol^(-1)=-196.86" kJ "mol^(-1)`
`-Delta_(r)G^(@)=2.303" RT "logK`
`196860=2.303xx8.314xx298logK` or `logK=34.5014`
K=Antilog 34.5014=3.192`xx10^(34)`
(ii) `E_(cell)^(@)=+0.80V-0.77V=+0.03V`
`Delta_(r)G^(@)=-nFE_(cell)^(@)=-(1mol)xx(96500" C "mol^(-1))xx(0.03V)`
`=-2895" CV "mol^(-1)=-2895" J "mol^(-1)` ltBrgt `=-2.895" kJ "mol^(-1)`
`Delta_(r)G^(@)=-2.303" RT "logK`
`-2895=-2.303xx8.314xx298xxlogK`
or log K`=0.5974` or K=Antilog (0.5974)=3.22.
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