Calculate the standard cell potential of galvanic cell in which the following reaction take place :
(Given `E_(OP)^(@)Cr, Cd, Fe^(2+), Ag` are `0.74V, 0.40V, -0.77V` and `-0.80V` respectively)
`(a) 2Cr_((s))+3Cd_((aq.))^(2+) rarr 2Cr_((aq.))^(3+)+3Cd`
(b) `Fe_((aq.))^(2+)+Ag_((aq.))^(+) rarr Fe_((aq.))^(3+)+Ag_((s))`
Calculate the `Delta_(r)G^(@)` and equilibrium constant of the reactions.

Explanation

(a) `E_(Cell)^(@) = E_(OP_(Cr//Cr^(3+)))^(@)+E_(RP_(Cd^(2+)//Cd))^(@)`
`[2Cr rarr 2Cr^(3+)+6e, 3Cd^(2+)+6e rarr3Cd]`
`= 0.74 + (-0.40) = +0.34V`
Since electrons `(n = 6)` are used in redox change.
`-Delta_(r)G^(@) = nE^(@)F = 6 xx 0.34 xx 96500J`
`196860J`
or `Delta_(r)G^(@) = -196.86kJ`
Also `-Delta_(r)G^(@) = 2.303 xx 8.314 xx 298logK`
`K = 3.17 xx 10^(34)`
(b) `E_(Cell)^(@) = E_(OP_(Fe^(2+)//Fe))^(@)+E_(RP_(Ag^(+)//Ag))^(@)`
`[Fe^(2+) rarr Fe^(3+) + e, Ag^(+)+e rarr Ag]`
`= -0.77 + 0.80 = 0.03 V`
Also `-Delta_(r)G^(@) = nE^(@)F = 1 xx 0.03 xx 96500`
or `Delta_(r)G^(@) = -2895J`
Also `-Delta_(r)G^(@) = 2.303 RTlogK`
`2895 = 2.303 xx 8314 xx 298logK`
`K = 3.22`