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

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

Step-by-step explanation:

We must use the Nernst equation

`E = E^(\circ) - (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 - (8.314* 298)/(2 * 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 & = & \frac{\text{[Cu$^(+)$]}^(2)}{\text{[Pd]}}\\\\4.392 & = & \frac{{x}^(2)}{0.498}\\\\x^(2)& = & 2.187\\x & = & 1.48\\\end{array}\\\text{The concentration of Cu$^(+)$ is $\large \boxed{\textbf{1.48 mol/L}}$}`