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

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`.