Calculate Ered and Eox at 25°C for the ClO3- ion in neutral solution, at pH 7.00, assuming all other species are at standard concentration (E°red = +1.442 V; E°ox = -1.226 V). Will the ClO3- ion disproportionate at pH 7.00?

Question

Calculate [latex]E_{red}[/latex] and [latex]E_{ox}[/latex] at 25°C for the [latex]ClO_{3}^{-}[/latex] ion in neutral solution, at pH 7.00, assuming all other species are at standard concentration ([latex]E°_{red}[/latex] = +1.442 V; [latex]E°_{ox}[/latex] = -1.226 V). Will the [latex]ClO_{3}^{-}[/latex] ion disproportionate at pH 7.00?

ANALYSIS
[latex]ClO_{3}^{-}[/latex]: [latex]E°_{red}[/latex] (1.442 V); [latex]E°_{ox}[/latex] (-1.226 V) pH (7.00); T (25°C) All species besides [latex]ClO_{3}^{-}[/latex] are at 1.00 M.	Information given:
Will [latex]ClO_{3}^{-}[/latex] disproportionate at the given pH?	Asked for:

STRATEGY

1. Write a half-equation for the reduction of [latex]ClO_{3}^{-}[/latex] .

2. Calculate [latex]E_{red}[/latex] by substituting into the Nernst equation for 25°C.

[latex]E_{red}[/latex] = [latex]E°_{red} - \frac{0.0257}{n}lnQ[/latex]

3. Write a half-equation for the oxidation of [latex]ClO_{3}^{-}[/latex] .

4. Calculate [latex]E_{ox}[/latex] by substituting into the Nernst equation for 25°C.

[latex]E_{ox}[/latex] = [latex]E°_{ox} - \frac{0.0257}{n}lnQ[/latex]

5. Find E
E = [latex]E_{red}[/latex] + [latex]E_{ox}[/latex]

If E > 0, [latex]ClO_{3}^{-}[/latex] will disproportionate.

Accepted Answer

[latex]ClO_{3}^{-}[/latex](aq) + 6[latex]H^{+}[/latex](aq) + 6[latex]e^{-}[/latex] → [latex]Cl^{-}[/latex](aq) + 3[latex]H_{2}O[/latex]	1. Reduction half-reaction
[latex]E_{red}[/latex] = 1.442 - [latex]\frac{0.0257}{6}ln\frac{[Cl^{-}]}{[ClO_{3}^{-}][H^{+}]^{6}} = 1.442 - \frac{0.0257}{6}ln\frac{(1)}{(1)(1.00 × 10^{-7})^{6}}[/latex] [latex]E_{red}[/latex] = 1.028 V	2. [latex]E_{red}[/latex]
[latex]ClO_{3}^{-}[/latex](aq) + [latex]H_{2}O[/latex] → [latex]ClO_{4}^{-}[/latex](aq) + 2[latex]H^{+}[/latex](aq) + 2 [latex]e^{-}[/latex]	3. Oxidation half-reaction
[latex]E_{ox}[/latex] = - 1.226 - [latex]\frac{0.0257}{2}ln\frac{[ClO_{4}^{-}][H^{+}]}{[ClO_{3}^{-}]} = -1.226 - \frac{0.0257}{2}ln\frac{(1)(1 × 10^{-7})²}{(1)}[/latex] [latex]E_{ox}[/latex] = -0.812 V	4. [latex]E_{ox}[/latex]
E = [latex]E_{red}[/latex] + [latex]E_{ox}[/latex] = 1.028 + (-0.812) = 0.216 V Disproportionation should occur.	5. E

$ClO_{3}^{-}$ (aq) + 6 $H^{+}$ (aq) + 6 $e^{-}$ → $Cl^{-}$ (aq) + 3 $H_{2}O$	1. Reduction half-reaction
$E_{red}$ = 1.442 – $\frac{0.0257}{6}ln\frac{[Cl^{-}]}{[ClO_{3}^{-}][H^{+}]^{6}} = 1.442 – \frac{0.0257}{6}ln\frac{(1)}{(1)(1.00 × 10^{-7})^{6}}$ $E_{red}$ = 1.028 V	2. $E_{red}$
$ClO_{3}^{-}$ (aq) + $H_{2}O$ → $ClO_{4}^{-}$ (aq) + 2 $H^{+}$ (aq) + 2 $e^{-}$	3. Oxidation half-reaction
$E_{ox}$ = – 1.226 – $\frac{0.0257}{2}ln\frac{[ClO_{4}^{-}][H^{+}]}{[ClO_{3}^{-}]} = -1.226 – \frac{0.0257}{2}ln\frac{(1)(1 × 10^{-7})²}{(1)}$ $E_{ox}$ = -0.812 V	4. $E_{ox}$
E = $E_{red}$ + $E_{ox}$ = 1.028 + (-0.812) = 0.216 V Disproportionation should occur.	5. E

Question 21.8: Calculate Ered and Eox at 25°C for the ClO3- ion in neutral ...

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