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

Answer:

(a)  Cu²⁺ +2e⁻ ⇌ Cu

(c) 0.07 V  

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

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