Determine the spontaneous direction at pH

5 [-(aq) I2O3(aq) + 6 H+(aq) → 3 [I-(s) + 9 H2O(l) (a) Determine the spontaneous direction at pH = 2.00 and [I2O3] = 0.100 M. (b) Repeat the calculation at pH = 1.00. (c) At what pH is this redox reaction at equilibrium at these concentrations of [I2O3] and [I-]? E°(I2O3/I-) = 0.5355 V, E°(I2O3/H+) = 1.195 V, E = E° – (0.0592/n) log Q E = 0 at equilibrium

The Correct Answer and Explanation is:

To solve this redox problem, we use the Nernst equation:E=E∘−0.0592nlog⁡QE = E^\circ – \frac{0.0592}{n} \log QE=E∘−n0.0592​logQ

Given:

• Balanced half-reaction:
  I2O3(aq)+6H+(aq)+6e−→2I−(s)+3H2O(l)\text{I}_2\text{O}_3(aq) + 6H^+(aq) + 6e^- \rightarrow 2I^-(s) + 3H_2O(l)I2​O3​(aq)+6H+(aq)+6e−→2I−(s)+3H2​O(l)
• E∘=0.5355 VE^\circ = 0.5355 \text{ V}E∘=0.5355 V (for the I₂O₃/I⁻ couple)
• n=6n = 6n=6
• [I2O3]=0.100 M[I_2O_3] = 0.100 \text{ M}[I2​O3​]=0.100 M
• Solid I⁻ is pure, so activity = 1
• Water is a liquid, so also not included in Q
• Q=1[I2O3][H+]6Q = \frac{1}{[I_2O_3][H^+]^6}Q=[I2​O3​][H+]61​

(a) At pH = 2.00

[H+]=10−2=0.010[H^+] = 10^{-2} = 0.010[H+]=10−2=0.010Q=1(0.100)(0.010)6=1(0.100)(1×10−12)=1×1013Q = \frac{1}{(0.100)(0.010)^6} = \frac{1}{(0.100)(1 \times 10^{-12})} = 1 \times 10^{13}Q=(0.100)(0.010)61​=(0.100)(1×10−12)1​=1×1013E=0.5355−0.05926log⁡(1×1013)=0.5355−0.05926⋅13E = 0.5355 – \frac{0.0592}{6} \log(1 \times 10^{13}) = 0.5355 – \frac{0.0592}{6} \cdot 13E=0.5355−60.0592​log(1×1013)=0.5355−60.0592​⋅13E=0.5355−(0.0592⋅2.1667)=0.5355−0.1282=0.4073 VE = 0.5355 – (0.0592 \cdot 2.1667) = 0.5355 – 0.1282 = 0.4073 \text{ V}E=0.5355−(0.0592⋅2.1667)=0.5355−0.1282=0.4073 V

Since E > 0, the reaction is spontaneous in the forward direction.

(b) At pH = 1.00

[H+]=0.10[H^+] = 0.10[H+]=0.10Q=1(0.100)(0.10)6=1(0.100)(1×10−6)=1×107Q = \frac{1}{(0.100)(0.10)^6} = \frac{1}{(0.100)(1 \times 10^{-6})} = 1 \times 10^7Q=(0.100)(0.10)61​=(0.100)(1×10−6)1​=1×107E=0.5355−0.05926⋅log⁡(1×107)=0.5355−(0.0592⋅1.1667)=0.5355−0.0691=0.4664 VE = 0.5355 – \frac{0.0592}{6} \cdot \log(1 \times 10^7) = 0.5355 – (0.0592 \cdot 1.1667) = 0.5355 – 0.0691 = 0.4664 \text{ V}E=0.5355−60.0592​⋅log(1×107)=0.5355−(0.0592⋅1.1667)=0.5355−0.0691=0.4664 V

Again, E > 0, so the reaction is spontaneous in the forward direction.

(c) At equilibrium, E = 0

Set up the Nernst equation and solve for [H⁺]:0=0.5355−0.05926log⁡Q0 = 0.5355 – \frac{0.0592}{6} \log Q0=0.5355−60.0592​logQlog⁡Q=0.5355⋅60.0592=54.27\log Q = \frac{0.5355 \cdot 6}{0.0592} = 54.27logQ=0.05920.5355⋅6​=54.27Q=1054.27Q = 10^{54.27}Q=1054.27

Recall that Q=1(0.100)[H+]6Q = \frac{1}{(0.100)[H^+]^6}Q=(0.100)[H+]61​(0.100)[H+]6=10−54.27⇒[H+]6=10−54.27/0.100=10−53.27(0.100)[H^+]^6 = 10^{-54.27} \Rightarrow [H^+]^6 = 10^{-54.27} / 0.100 = 10^{-53.27}(0.100)[H+]6=10−54.27⇒[H+]6=10−54.27/0.100=10−53.27log⁡[H+]=−53.276=−8.88⇒pH=8.88\log [H^+] = \frac{-53.27}{6} = -8.88 \Rightarrow \text{pH} = 8.88log[H+]=6−53.27​=−8.88⇒pH=8.88

Final Answers:

(a) Reaction is spontaneous forward, E = 0.407 V
(b) Reaction is spontaneous forward, E = 0.466 V
(c) Reaction is at equilibrium when pH = 8.88

Explanation

This redox reaction involves the reduction of iodic oxide (I₂O₃) to iodide ion (I⁻) in acidic solution. The spontaneity of a redox reaction can be assessed by calculating the cell potential (E) using the Nernst equation. A positive E indicates a spontaneous reaction in the forward direction.

To apply the Nernst equation, we first identify the half-reaction and its standard reduction potential. The balanced reaction shows that 6 electrons are transferred, and water is formed from hydrogen ions. The reaction quotient (Q) includes only aqueous species, so solids like iodide and liquids like water are excluded.

In part (a), at pH 2.00, the hydrogen ion concentration is 0.010 M. Substituting values into the Nernst equation, we find a positive E of 0.407 V, indicating that the forward reaction is spontaneous. In part (b), at a more acidic pH of 1.00, the higher hydrogen ion concentration further favors the forward direction, giving a slightly larger E value of 0.466 V.

In part (c), we find the pH where the reaction is at equilibrium by setting E = 0 and solving for [H⁺]. This gives a very low hydrogen ion concentration, corresponding to a pH of 8.88. At this point, the forward and reverse reactions occur at the same rate, and no net change occurs in concentrations.

This problem highlights how pH affects redox equilibria. As pH increases (fewer H⁺ ions), the forward redox reaction becomes less favorable, and equilibrium is reached at basic pH.