1.

Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Answer»

SOLUTION :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^(+)])`
Substituting the values, we get
`E_(H^(+)//1/2H_(2))^(@) =0-(0.0591)/1 log 1/(10^(-10))` [pH = 10 means `[H^(+)]=10^(-10) M]`
`=-0.0591 xx 10 = -0.591` V
Thus, potential of hydrogen electrode `=-0.591` V


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