1.

At what pH the oxidation potential of hydrogen electrode will be -0.413 V ?

Answer»


Solution :`E = (-0.0591)/(1) LOG""(1)/([H^(OPLUS)]), (-(E xx 1)/(0.0591)) = - log(H^(oplus)) = P^(H) , P^(H) = 6.988 ~= 7`


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