The cell notation for a redox reaction is given as the following at (T= 298 K). Calculate the cell potential for the reaction at 25ºC.
Zn (s) | Zn²⁺ (aq, 0.37 M) || Ni²⁺ (aq, 0.059 M) | Ni (s)
Standard Reduction Potentials
Zn²⁺ (aq) + 2 e^– → Zn (s)     E°_red = - 0.7621
Ni²⁺ (aq) + 2 e^–­ → Ni (s)      E°_red = - 0.2300

• A. 0.3130 V
• B. 0.4033 V
• C. 0.5085 V
• D. 0.1199 V

Answer: (C)

Verified step by step guidance

1. Identify the half-reactions from the cell notation: the anode (oxidation) is Zn (s) → Zn²⁺ (aq) + 2 e⁻, and the cathode (reduction) is Ni²⁺ (aq) + 2 e⁻ → Ni (s).
2. Write down the standard reduction potentials (E°) for both half-reactions: E°(Zn²⁺/Zn) = -0.7621 V and E°(Ni²⁺/Ni) = -0.2300 V.
3. Determine the standard cell potential (E°_cell) by subtracting the anode potential from the cathode potential: \(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\).
4. Use the Nernst equation to calculate the cell potential under non-standard conditions: \(E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln Q\), where \(Q\) is the reaction quotient, \(n\) is the number of electrons transferred, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(F\) is the Faraday constant.
5. Calculate the reaction quotient \(Q\) using the concentrations given: \(Q = \frac{[\text{Zn}^{2+}]}{[\text{Ni}^{2+}]}\), then plug all values into the Nernst equation to find the cell potential at 25ºC (298 K).