(a) The e.m.f. of the following cell at 298 K is 0.1745 V
`Fe(s) //Fe^(2+) (0.1 M) ////H^(+)(x M)//H_(2)(g) ("1 bar")//Pt (s)`
Given : `E_(Fe^(2+)//Fe)^(0) = - 0.44 V`
Calculate the `H^(+)` ions concentration of the solution at the electrode where hydrogen is being produced.
(b) Aqueous solution of copper sulphate and silver nitrate are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different? Explain your answer.

Explanation

(a) `Fe+2H^(+)toH_(2)+Fe^(2+)`
`E_(cell)=E_(cell)^(0)-(2.303RT)/(nF)"log"(Fe^(2+))/([H^(+)]^(2))`
`E_(cell)^(0)=E_(H+//H_(2))^(0)-E_(Fe^(2+)//Fe)^(0)`
`=0-(-0.44)=0.44V`
`0.1745=0.44-(0.0591)/(2)"log"([0.1])/([x]^(2))`
`Logx=-5`
`Log[H^(+)]=-5`
`[H^(+)]=10^(-5)`
(b) The mass of copper and silver deposited at the cathode will be different.
The amount of different sunstances deposited by the same quantity of electricity passing through the electrolytic solution are directly proportional to their chemical equivalents.