Question

Calculate the Cd^2+ concentration in the following cell if Ecell = 0.23

Cd | Cd^2+ (x M) || Ni^2+ (1.00 M) | Ni

Answer 1

`\large \boxed{\text{1.14 $* 10^(-3)$ mol/L}}`

Step-by-step explanation:

We must use the Nernst equation

`E = E^(\circ) - (RT)/(zF)\ln Q`

Step 1. Calculate E°

Anode: Cd ⟶ Cd²⁺(x mol·L⁻¹) + 2e⁻; E° = +0.4030 V

Cathode: Ni²⁺ (1.00 mol·L⁻¹) + 2e⁻ ⟶ Ni; E° = - 0.257 V

Overall: Ni²⁺(1.00 mol·L⁻¹) + Cd ⟶ Ni + Cd²⁺ (x mol·L⁻¹); E° = 0.146 V

Step 2. Calculate Q

`\begin{array}{rcl}0.23 & = & 0.146 - (8.314* 298)/(2 * 96 485) \ln Q\\\\0.084& = & -0.01284 \ln Q\\\ln Q & = & -6.542\\Q & = & e^(-6.542)\\ & = &1.14 * 10^(-3)\\\end{array}`

3. Calculate [Cd²⁺]

`\begin{array}{rcl}Q & = & \frac{\text{[Cd$^(2+)$]}}{\text{[Ni}^(2+)]}\\\\1.14 * 10^(-3) & = & (x)/(1.00)\\\\x& = & 1.14 * 10^(-3)\\\end{array}\\\text{The concentration of Cd$^(2+)$ is $\large \boxed{\textbf{1.14 $* \mathbf{10^(-3)}$ mol/L}}$}`