Find the `EMF` of the concentration cell represented as given below`:`

Explanation

`E_(cell)` will be positively only when
`[H^(o+)]_(cathode)gt[H^(o+)]_(anode)` and `(p_(H_(2)))_(anode)gt(p_(H_(2)))_(cathode)`
So in the above case, `E_(cell)` will be positive, and cell will be spontaneous or feasible or exergonic.
`[H^(o+)]_(anode)=2xx0.01M=2xx10^(-2)N`
`(n` fact o r fo r H_(2)SO_(4)=2)`
`pH_(anode)=-log(2xx10^(-2))=2-0.3=1.7`
`[H^(o+)]_(cathode)=1xx0.1M" "(n` factor for `HCl=1)`
`pH_(cathode)=-log(10^(-1))=1`
Use the relation,
`E_(cell)=0.059[(pH_(anode)-pH_(cathode ))+(1)/(2)log .((p_(H_(2)))_(anode))/((p_(H_(2)))_(cathode))]`
`=0.059(1.7-1+(1)/(2)log.(2atm)/(1atm))`
`=0.059(0.7+(1)/(2)log 2)`
`=0.059(0.7+(1)/(2)xx0.3)`
`=0.059(0.7+0.15)`
`=0.059xx0.85~~0.06xx0.85=0.05V`