The reduction potential of hudrogen electrode is`- 118 mV`. The concentration of `H^(+)` in the solution is .

Options

A. 0.001 M
B. 2 M
C. `10^(-4)`
D. 1 M

Answer

A

Explanation

`-118 mv = -0.118v`
`E = (0.059)/nlog [H^(+)]`
`-0.118 = 0.059/1 log [H^(+)]`
`log[H^(+)] =-2`
`[H^(+)] = 10^(-2)`