Question

The following reaction is spontaneous as written when the components are in their standard states:

Answer 1

Answer:

$\large \boxed{\text{1 mol/L}}$

Explanation:

We must use the Nernst equation

E_{\text{cell}} = E_{\text{cell}}^{\circ} - \dfrac{RT}{zF}\ln Q

We want Ecell < 0 for a reverse reaction, so assume it is 0.

Step 1. Calculate E°cell

Anode:     3Zn ⟶ 3Zn²⁺(4 mol·L⁻¹) + 6e⁻;                              E° = +0.7618 V

Cathode: 2Cr³⁺ (x mol·L⁻¹) + 6e⁻ ⟶ 2Cr;                               E° = -0.744   V

Overall:   3Zn + 2Cr³⁺(x mol·L⁻¹) ⟶ 3Zn²⁺(4 mol·L⁻¹) + 2Cr;  E° = +0.018   V

Step 2. Calculate Q

$\begin{array}{rcl}0 & = & 0.018 - \dfrac{8.314\times 298}{6 \times 96 485} \ln Q\\\\-0.18& = & -0.00428 \ln Q\\\ln Q & = & 4.16\\Q & = & e^{4.16}\\ & = & \mathbf{64.0}\\\end{array}$

3. Calculate [Cr³⁺]

$\begin{array}{rcl}Q & = & \dfrac{\text{[Zn ^{2+} ] ^{3} }}{\text{[Cr}^{3+}]^{2}}\\\\64.0 & = & \dfrac{4^{3}}{\text{[Cr}^{3+}]^{2}}\\\\\text{[Cr}^{3+}]^{2}& = & \dfrac{64}{64.0}\\\\& = & 1\\\text{[Cr}^{3+}] & = & \textbf{1 mol/L}\\\end{array}\\\text{[Cr ^{3+} ] must be less than \large \boxed{\textbf{1 mol/L}} for the reaction to be spontaneous in the reverse}\\\text{direction.}$