The correct choice is (a) 5.08
Best explanation: Given,
Standard oxidation potential of Ni/Ni^2+ electrode, E°OP = 0.3 V,
Ni → Ni^2+ + 2e
2H^+ + 2e → H2
E°cell = E° (OP) + E° (RP)
E° (cell) = 0.3 + 0.0 = 0.3 V
According to Nernst equation, E (cell) = E°(cell) + \(\frac{0.059}{2}\) log10([H^+]^2 / [Ni]^+2)
0 = 0.3 + \(\frac{0.059}{2}\)log10([H^+]^2)
-log ([H^+]) = 5.08
pH = 5.08.