Correct Answer - C
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)]xxp_(H_(2)))/([H^(+)]^(2))`
On adding `H_(2)SO_(4),[H^(+)]` will increase, therefore `E_(cell)` will increase. The equilibrium `Zn+2H^(+)hArrZn^(2+)+H_(2)` will shift towards right on increasing the concentration of `H^(+)` ions.