Question

A voltaic electrochemical cell consists of a copper electrode in a Cu2SO4(aq) solution, and a palladium electrode in a PdSO4(aq) solution at 25°C. The salt bridge consists of a solution of KCl(aq).

What is the concentration of the Cu+if the concentration of the PdSO4 is 0.498 M and the measured cell potential is 0.447 V?

Given: Cu+(aq) + e- ↔ Cu(s) E°=+0.521 V

and Pd2+(aq) + 2e- ↔ Pd(s) E°=+0.987 V

Answer 1

Answer:

$\large \boxed{\textbf{1.48 mol/L}}$

Explanation:

We must use the Nernst equation

$E = E^{\circ} - \dfrac{RT}{zF}lnQ$

1. Calculate E°

Anode:     Pd²⁺ (0.498 mol·L⁻¹) + 2e⁻ ⇌ Pd;                               E° = +0.987 V

Cathode: Cu ⇌ Cu⁺ (x mol·L⁻¹) + e⁻;                                          E°=  - 0.521 V

Overall:   Pd²⁺(0.498 mol·L⁻¹) + 2Cu ⟶ Pd + 2Cu⁺ (x mol·L⁻¹); E° =   0.466 V

2. Calculate Q

$\begin{array}{rcl}0.447 & = & 0.466 - \dfrac{8.314\times 298}{2 \times 96 485} \ln Q\\\\-0.019& = & -0.01284 \ln Q\\\ln Q & = & 1.480\\Q & = & e^{1.480}\\ & = & 4.392\\\end{array}$

3. Calculate [Cu⁺]

$\begin{array}{rcl}Q & = & \dfrac{\text{[Cu ^{+} ]}^{2}}{\text{[Pd]}}\\\\4.392 & = & \dfrac{{x}^{2}}{0.498}\\\\x^{2}& = & 2.187\\x & = & 1.48\\\end{array}\\\text{The concentration of Cu ^{+} is \large \boxed{\textbf{1.48 mol/L}} }$