Question

Calculate pH of following half-cell. Pt, \(H_2\) / \(H_2\)\(SO_4\) , if its electrode potential is 0.03V.
pH = - log [H^+]
½ \(H^+\) e- → \(H_2\) (g)
\(E_{cell}\) = \(E°_{cell}\) - 0.059/n log 1/[\(H^+\)]
0.03 = 0- 0.059 /1 log 1/[\(H^+\)]
pH = 5.07 V

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