Answer:
[tex]\large \boxed{\text{1.14 $\times 10^{-3}$ mol/L}}[/tex]
Explanation:
We must use the Nernst equation
[tex]E = E^{\circ} - \dfrac{RT}{zF}\ln Q[/tex]
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
[tex]\begin{array}{rcl}0.23 & = & 0.146 - \dfrac{8.314\times 298}{2 \times 96 485} \ln Q\\\\0.084& = & -0.01284 \ln Q\\\ln Q & = & -6.542\\Q & = & e^{-6.542}\\ & = &1.14 \times 10^{-3}\\\end{array}[/tex]
3. Calculate [Cd²⁺]
[tex]\begin{array}{rcl}Q & = & \dfrac{\text{[Cd$^{2+}$]}}{\text{[Ni}^{2+}]}\\\\1.14 \times 10^{-3} & = & \dfrac{x}{1.00}\\\\x& = & 1.14 \times 10^{-3}\\\end{array}\\\text{The concentration of Cd$^{2+}$ is $\large \boxed{\textbf{1.14 $\times \mathbf{10^{-3}}$ mol/L}}$}[/tex]
