For `Pt(H_(2))|H_(2)O` , reduction potential at `298 K` and `1 atm` is `:`

Options

A. `-0.207`
B. `-0.414` V
C. 0.207 V
D. 0.414 V

Answer

B

Explanation

For water at 298 K , `[H^(+)] = 10^(-7)` M
Reduction reaction is `H^(+) + e^(-) to (1)/(2) H_(2)`
`E = (-0.592)/(n) log_(10) (P_(H_(2))^(1//2))/([H^(+)]) (because E_(H_(2))^(@) = 0)`
`= -0.0592 log_(10) ((1)/(10^(-7))) = -0.414 V `