The EMF of the cell `M|M^(n+)` (0.02 M) | H^(+) (1 M) | H_(2)(g) (1 atm), Pt at `25^(@)`C is 0.81 V. Calculate the valency of the metal if the standard oxidation potential of the metal is 0.76 V

Answer

`n=2`

Explanation

`(MtoM_((0.02M))^(n+)+ne^(-))xx2`
`(2H^(+)+2e^(-)toH_(2))xxn`
`2M+2nH^(+)to2M^(n+)+nH_(2)`
`0.81=(0.76+0)-(0.0591)/(2n)log(((0.02)^(2))/((1)^(2n)))implies(0.81-0.76)=(0.0591)/(2n)log4xx10^(-4)`
`n=-(0.0591)/(2xx0.5)xxlog4xx10^(-4)=-0.591(-4+0.6)=2`.