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Electrolytic cell; [Ni²⁺] from emf of Ni/Cu cell

Short answer

An electrolytic cell uses electrical energy to drive a non-spontaneous reaction. Electrode potential is the interfacial potential difference. With E°cell = 0.59 V and Nernst equation for Ni + Cu²⁺ → Ni²⁺ + Cu, when EMF = 0.601 V, [Ni²⁺] = 0.32 M.

What is meant by electrolytic cell and electrode potential? Calculate the concentration of Ni²⁺ in the cell at 25°C, if the emf is 0.601 V. Ni/Ni²⁺(α=?) // Cu²⁺(0.75M)/Cu Given, E° Ni/Ni²⁺ = 0.25 V and E° Cu²⁺/Cu = 0.34 V

Electrolytic cell

A cell in which electrical energy from an external source is used to bring about a non-spontaneous chemical reaction (electrolysis). Here electrical energy is converted into chemical energy; e.g. electrolysis of molten NaCl or of CuSO₄ solution. (Anode is positive and cathode is negative.)

Electrode potential

The potential difference set up between a metal and the solution of its ions at the interface, due to the tendency to lose or gain electrons. Standard electrode potential is measured at 1 M and 25°C against SHE.

Calculation

Cell: Ni | Ni²⁺ (x) || Cu²⁺ (0.75 M) | Cu. The cell reaction is Ni + Cu²⁺ → Ni²⁺ + Cu (n = 2).

E°cell = 0.25 + 0.34 = 0.59 V
E = E° − (0.0591/2) log ([Ni²⁺]/[Cu²⁺])
0.601 = 0.59 − 0.02955 log (x/0.75)
log (x/0.75) = (0.59 − 0.601)/0.02955 = −0.372
x/0.75 = 10⁻⁰·³⁷² = 0.4246
x = 0.318 M

Answer: [Ni²⁺] ≈ 0.32 M.

✓ Answer checked by Claude Opus 5.5 (high effort)

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