Exactly 0.4 faraday of charge is passed through three electrolytic cells connected in series, first containing `AgNO_(3)`, second `CuSO_(4)` while third containing `FeCl_(3)` solution. How many grams of each metal will be deposited assuming only cathodic reaction in each cell ?

Answer

Ag=43.2 g; Cu=12.7 g; Fe=7.47 g

Explanation

`Ag^(+)(aq)+underset(1F)underset(1" mol")(e^(-)) to underset(108" g")underset(1" mol")(Ag (s))`
1 Faraday of charge deposit Ag=108 g
0.4 Faraday of charge deposit Ag `=((108g))/((1F))xx(0.4F)=43.2 g`
`underset(2F)underset(3" mol")(Cu^(2+)(aq))+2e^(-) to underset(63.5 g)underset(1" mol")(Cu(s))` 2 Faraday of charge deposit Cu=63.5 g
0.4 Faraday of charge deposit `Cu=((63.5 g))/((2F))xx(0.4F)=12.7 g`
`underset(3F)underset(3" mol")(Fe^(3+)(aq))+3e^(-)to underset(56" g")underset(1" mol")(Fe(s))`
3 Faraday of charge deposit Fe=56 g
0.4 Faraday of charge deposit `Fe=((56 g))/((3 F))xx(0.4F)=7.47 g`