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

3 Zn(s) +2 Cr3+(aq) →3 Zn2+(aq) +2 Cr(s)
If the [Zn2+] is 4 molL−1, determine the value of [Cr3+] below which the reaction will be spontaneous in the opposite direction.

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

3. Calculate [Cr³⁺]

i believe it is 40–40–20%

explanation: