Question 5.IE: Sodium dithionite, Na2S2O4, is an important reducing agent. ......
Sodium dithionite, Na_{2}S_{2}O_{4}, is an important reducing agent. One interesting use is the reduction of chromate ion to insoluble chromium(III) hydroxide by dithionite ion, S_{2}O_{4}^{2−}, in basic solution. Sulfite ion is another product. The chromate ion may be present in wastewater from a chromium-plating plant, for example. What mass of Na_{2}S_{2}O_{4} is consumed in a reaction with 100.0 L of wastewater having [CrO_{4}^{2−}] = 0.0148 M?
Analyze
The phrase “reduction of chromate” tells us that the reaction between CrO_{4}^{2−} and S_{2}O_{4}^{2−} is a redox reaction. We must obtain a balanced chemical equation for the reaction by using the method summarized in Table 5.6, and then convert 100.0 L of wastewater into grams of Na_{2}S_{2}O_{4}. The necessary conversions are as follows:
100.0 L wastewater → mol CrO_{4}^{2−} → mol S_{2}O_{4}^{2−} → mol Na_{2}S_{2}O_{4} → g Na_{2}S_{2}O_{4}
Solve
1. Write an ionic expression representing the reaction. CrO_{4}^{2−}(aq) + S_{2}O_{4}^{2−}(aq) + OH^{−}(aq) → Cr(OH)_{3}(s) + SO_{3}^{2−}(aq)
2. Balance the redox equation. Begin by writing skeleton half-equations. CrO_{4}^{2−} → Cr(OH)_{3}
S_{2}O_{4}^{2−} → SO_{3}^{2−}
Balance the half-equations for Cr, S, O, and H atoms as if the half-reactions occur in acidic solution. CrO_{4}^{2−} + 5 H^{+} → Cr(OH)_{3} + H_{2}O
S_{2}O_{4}^{2−} + 2 H_{2}O → 2 SO_{3}^{2−} + 4 H^{+}
Balance the half-equations for charge, and label them as oxidation and reduction. Oxidation: S_{2}O_{4}^{2−} + 2 H_{2}O → 2 SO_{3}^{2−} + 4 H^{+} + 2 e^{−}
Reduction: CrO_{4}^{2−} + 5 H^{+} + 3 e^{−} → Cr(OH)_{3} + H_{2}O
Combine the half-equations into an overall equation.
3. Change the conditions to basic solution. Add 2 OH^{−} to each side of the equation for acidic solution, and combine 2 H^{+} and 2 OH^{−} to form 2 H_{2}O on the right. 3 S_{2}O_{4}^{2−}+ 2 CrO_{4}^{2−} + 4 H_{2}O + 2 OH^{−} \longrightarrow 6 SO_{3}^{2−} + 2 Cr(OH)_{3} + 2 H_{2}O
Subtract 2 H_{2}O from each side of the equation to obtain the final balanced equation. 3 S_{2}O_{4}^{2−}(aq) + 2 CrO_{4}^{2−}(aq) + 2 H_{2}O(l) + 2 OH^{−}(aq) \longrightarrow 6 SO_{3}^{2−}(aq) + 2 Cr(OH)_{3}(s)
4. Complete the stoichiometric calculation. The conversion pathway is
100.0 L waste water → mol CrO_{4}^{2−} → mol S_{2}O_{4}^{2−} → mol Na_{2}S_{2}O_{4} → g Na_{2}S_{2}O_{4}. ? g Na_{2}S_{2}O_{4} = 100.0 L × \frac{0.0148 mol CrO_{4}^{2−}}{1 L} × \frac{3 mol S_{2}O_{4}^{2−}}{2 mol CrO_{4}^{2−}} × \frac{1 mol Na_{2}S_{2}O_{4}}{1 mol S_{2}O_{4}^{2−}} × \frac{174.1 g Na_{2}S_{2}O_{4}}{1 mol Na_{2}S_{2}O_{4}} = 387 g Na_{2}S_{2}O_{4}
Assess
In solving this problem the major effort was to balance a redox equation for a reaction under basic conditions. This allowed us to find the molar relationship between dithionite and chromate ions. The remainder of the problem was a stoichiometry calculation for a reaction in solution, much like Example 4-10 (page 127). A quick check of the final result
involves (1) ensuring that the redox equation is balanced, and (2) noting that the number of moles of CrO_{4}^{2−} is about 1.5 (i.e, 100 × 0.0148), that the number of moles of S_{2}O_{4}^{2−} is about 2.25 (i.e, 1.5 × 3/2), and that the mass of Na_{2}S_{2}O_{4} is somewhat more than 350 (i.e, 2.25 × 175).
| TABLE 5.5 Balancing Equations for Redox Reactions in Acidic Aqueous Solutions by the Half-Equation Method: A Summary | ||||||
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(1) Balance atoms of all the elements except H and O (2) Balance oxygen by using H_{2}O (3) Balance hydrogen by using H^{+} (4) Balance charge by using electrons |
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| TABLE 5.6 Balancing Equations for Redox Reactions in Basic Aqueous Solutions by the Half-Equation Method: A Summary | ||||||
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