Calculate the potential of hydrogten electrode placed in a solution of pH 10.

Explanation

For hydrogen electrode, `H^(+)+e^(-) to 1/2H_(2)`
Applying Nernst equation,
`E_((H^(+)//1//2H_(2)))=E _((H^(+)//1//2H_(2)))-(0.0591)/(n)log "'(1)/([H^(+)])`
`=0-(0.0591)/(1)log ""(1)/((10^(-10)))[pH=10, [H^(+)]=10^(-10)M]`
`=0-(0.0591)/(1)xx(10 log 10) =-0.591V`
`E_((H^(+)//1//2 H_(2)))=-0.591V`