The emf of cell containing Zn and hydrogen electrodes represented as below `(E_(Zn)^(0) = -0.76V)`
`Zn|Zn^(2+) (0.2 M) || H^(+) (0.4M) | H_(2 (g)) ` Pt at 298 K is

Options

A. `0.742` V
B. `-0.472` V
C. `0.769` V
D. `-0.769` V

Answer

C

Explanation

`E_("cell")^(@) = E_((RHS))^(@) - E_((LHS))^(@)`
`= E_(H_(2))^(@) - E_(ZN)^(@)`
`= 0 - (-0.76) = 0.76` V
`E_("cell") = E_("cell")^(@) - (0.0591)/(n) "log"_(10) ([P])/([R])`
`= E_("cell")^(@) - (0.0591)/(n) "log" ([Zn^(2+)])/([H^(+)])`
`= 0.76 - (0.0591)/(2) "log" (0.2)/(0.4)`
Reversing and changing the sign
`E_("cell") = 0.76 + (0.0591)/(2) "log" (0.4)/(0.2)`
` = 0.76 + (0.0591)/(2) "log" 2`
`= 0.76 + (0.0591 xx 0.3010)/(2)`
`= 0.769` V