Zinc granules are added in excess to a 500 ml of 1.0 M nickel nitrate solution at 25°C until the equilibrium is reached. If the standard reduction potential of Zn^2+\|Zn and Ni^2+\|Ni are –0.75 V and

Question

Zinc granules are added in excess to a 500 ml of 1.0 M nickel nitrate solution at 25°C until the equilibrium is reached. If the standard reduction potential of Zn^2+\|Zn and Ni^2+\|Ni are –0.75 V and –0.24 V, respectively, the concentration of Ni^2+ in solution at equilibrium is (2.303RT/F = 0.06)

Prashant Kotian · Accepted Answer

Correct answer: A. The reaction is Zn + Ni^2+ <=> Zn^2+ + Ni. E°_cell = E°_Ni - E°_Zn = -0.24 - (-0.75) = 0.51 V. At equilibrium, E_cell = 0, so E°_cell = (0.06/2) * log K. 0.51 = 0.03 * log K, meaning log K = 17, so K = 10^17. Since K is huge, almost all Ni^2+ is reduced to Ni, making [Zn^2+] ≈ 1.0 M. [Zn^2+]/[Ni^2+] = 10^17. 1.0 / [Ni^2+] = 10^17, yielding [Ni^2+] = 1.0 × 10^–17 M. Therefore, correct answer is A.

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