Question

A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached to the positive terminal of a potentiometer and the calomel electrode was attached to the negative terminal.(a) Write a half-reaction for the Cu electrode. (Use the lowest possible coefficients. Omit states-of-matter.)

(c) Calculate the cell voltage.

Answer 1

(a) Cu²⁺ +2e⁻ ⇌ Cu

(c) 0.07 V

Step-by-step explanation:

(a) Cu half-reaction

Cu²⁺ + 2e⁻ ⇌ Cu

(c) Cell voltage

The standard reduction potentials for the half-reactions are+

E°/V

Cu²⁺ + 2e⁻ ⇌ Cu; 0.34

Hg₂Cl₂ + 2e⁻ ⇌ 2Hg + 2Cl⁻; 0.241

The equation for the cell reaction is

E°/V

Cu²⁺(0.1 mol·L⁻¹) + 2e⁻ ⇌ Cu; 0.34

2Hg + 2Cl⁻ ⇌ Hg₂Cl₂ + 2e⁻; -0.241

Cu²⁺(0.1 mol·L⁻¹) + 2Hg + 2Cl⁻ ⇌ Cu + Hg₂Cl₂; 0.10

The concentration is not 1 mol·L⁻¹, so we must use the Nernst equation

(ii) Calculations:

T = 25 + 273.15 = 298.15 K

`Q = \frac{\text{[Cl}^(-)]^(2)}{ \text{[Cu}^(2+)]} = (1)/(0.1) = 10\\\\E = 0.10 - \left ((8.314 * 298.15 )/(2 * 96485)\right ) \ln(10)\\\\=0.010 -0.01285 * 2.3 = 0.10 - 0.03 = \textbf{0.07 V}\\\text{The cell potential is }\large\boxed{\textbf{0.07 V}}`