The electrode potential of the following half cell at 298 K
X|X²⁺(0.001 M) ∥ Y²⁺(0.01 M)| Y is _______ × 10⁻² V (Nearest integer)
Given: E⁰ X²⁺ | X=-2.36 V
E_(Y²⁺ | Y)⁰=+0.36 V
(2.303 RT)/F=0.06 V

Solution

Chemistry ma’am · 4 steps

Find E° of the cell first, then let the Nernst term nudge it using the ratio [X²⁺]/[Y²⁺].

1. Standard cell potential. X is the anode on the left and Y is the cathode on the right. So E°_(cell) = E°_(cathode) - E°_(anode) = 0.36 - (-2.36) = 2.72 V.
2. Cell reaction. The overall reaction is X + Y²⁺ → X²⁺ + Y, with n = 2. The reaction quotient is Q = [X²⁺]/[Y²⁺] = 0.001/0.01 = 10⁻¹.

   E_(cell) = E°_(cell) - 0.06/n log Q = 2.72 - 0.06/2 log 10⁻¹

3. Put in numbers. log 10⁻¹ = -1, so the Nernst term is -0.03 × (-1) = +0.03 V. Then E_(cell) = 2.72 + 0.03 = 2.75 V.
4. So. E_(cell) = 2.75 V = 275 × 10⁻² V, so the answer is 275.

The common slip: 269. Writing Q upside down as [Y²⁺]/[X²⁺] = 10 subtracts 0.03 V instead of adding it, giving 2.69 V, that is 269.

The idea to keep. In the Nernst equation, Q is products over reactants of the cell reaction: anode ion on top, cathode ion below. A smaller Q than 1 makes the cell potential larger than E°.

Answer from the NTA final answer key. Solution by PrepWiser.