The following two cells with initial concentration as given are connected in parallel with each other ltbr (1). `Fe(s)|Fe(NO_(3))_(2)(aq.)(1M)||SnCl_(2)(aq.)(1M)|Sn(s)`
(2). `Zn(s)ZnSO_(4)(aq.)(1M)||Fe(NO_(3))_(2)(aq.)(1M)|Fe(s)`
After sufficient time equilibrium is establised in the circuit. What will be the concentrations (in mmoles/L) of `Fe^(2+)` ions in first and second cells respectively.
[Take `E_(Sn^(2+)//Sn)^(0)=-0.14V,E_(Zn^(2+)//Zn)^(0)=-0.76V,E_(Fe^(2+)//Fe)^(0)=-0.44V,2.3xxRT=6433,log2=0.3]`

Answer

`[Fe^(+)]` in the first cell `=[Fe^(+)]` in the second cell `=(2)/(3)M=667"mmoles"//L`

Explanation

`Fe(s)+Sn^(2+)(aq.)hArrFe^(2+)(aq.)+Sn(s)" "E_(cell)^(0)=0.30V`
`Zn(s)+Fe^(2+)(aq.)hArrZn^(2+)(aq.)+Fe(s)" "E_(cell)^(0)=0.32V`
if above cell are connected in parallel then first cell will get charged up and second cell will get discharged so net cell reaction will be.
`underset(underset(t(aq.))(t=0)" "underset((1-x))(1M)" "underset((1+x))(1M)" "underset((1+x))(1M))(Fe_(2)^(2+)(aq.)+Fe_(1)^(2+)(aq.)hArrZn^(2+)(aq.)+Sn^(2+)(aq.))`" "E_("net")^(0)=0.02`
`triangleG_("net")^(0)=-nFE_("net")^(0)`
`E=0.3-(0.059)/(2)log(([Fe^(2+)])/([Sn^(2+)]))impliesE=0.32-(0.059)/(2)log(([Z^(2+)])/([Fe^(2+)]))=-2xx96500xx0.02J//"mole"`
So, `-2.30RTlogK_(eq)=-2xx96500xx0.02`
`logK_(eq)=(2xx96500xx0.02)/(6433)=0.6impliesK_(eq)=4`
`((1+x)^(2))/((1-x)^(2))=4implies((1+x))/((1-x))=2impliesx=(1)/(3)`
So, `[Fe^(2+)]` in the first cell `=[Fe^(2+)]` in the second cell `=(2)/(3)M=667"mmoles"//L`