What will be standard cell potential of galvanic cell with the following reaction?
`2Cr_((s)) + 3Cd_((aq))^(2+) to 2Cr_((aq))^(3+) + 3Cd_((s))`
[Given: `E_(Cr^(3+)//Cr)^@`=-0.74 V and `E_(Cd^(2+)//Cd)^@`=-0.40 V]

Answer

`a. Delta_(r)G^(c-)=-196.86 kJ mol^(-1),K_(c)=3.192xx10^(34)`
`b. Delta_(r)G^(c-)=-2.895 kJmol^(-1),K_(c)=3.22`

Explanation

a. `E_("cell")^(c-)=E_("cathode")^(c-)-E_("anode")^(c-)`
`=-0.40V-(-0.74)=+0.34V`
`Delta_(r)G^(c-)=-nFE_("cell")^(c-)`
`=-6 molxx96400C mol^(-1)xx0.34V`
`=-196860J mol^(-1)=-196.86k J mol^(-1)`
`Delta_(r)G^(c-)=--2.303RTlogK`
`196860=2.303xx8.314xx298 log K`
`logK=34.5014`
`K="Antilog" (34.5014)=3.192xx10^(34)`
`b`. `E^(c-)``_(cell)=+0.80V-0.77V=+0.03V`
`Delta_(r)G^(c-)=-nFE_(cell)^(c-)`
`=-(1 mol)xx(96500 C mol^(-1))xx(0.03V)`
`=-2895J mol^(-1) Delta_(r)G^(c-)=-2.303RTlogK`
`-2895=-2.303xx8.314xx298xxlogK`
or log `K =0.5074` or `K ="Anitlog"(0.5074)=3.22`