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Chapter 10: Problem 105

Assign an oxidation number to each atom and identify the oxidizing agent and reducing agent: \(3 \mathrm{Na}_{2} \mathrm{SO}_{3}+2 \mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow\) \(3 \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{MnO}_{2}+2 \mathrm{KOH}\)

Short Answer

Expert verified
In the given reaction, the oxidation numbers for each atom are as follows: Na is always +1, S in Na2SO3 is +4 and in Na2SO4 is +6, O is -2, H is +1, K is +1, and Mn in KMnO4 is +7 and in MnO2 is +4. Comparing the changes in oxidation numbers, we find that S in Na2SO3 is oxidized (increasing from +4 to +6), whereas Mn in KMnO4 is reduced (decreasing from +7 to +4). Thus, Na2SO3 is the reducing agent and KMnO4 is the oxidizing agent.

Step by step solution

01

Assign oxidation numbers to reactants

For the reactants, we have Na2SO3, KMnO4, and H2O. For each compound, let's assign oxidation numbers: 1. In Na2SO3, Na is +1, O is -2 and S is +4 (since the overall charge of the compound is 0, and 2Na = +2, 3O = -6, so S must be +4). 2. In KMnO4, K is +1, Mn is +7 (since the overall charge of the compound is 0, and K = +1, 4O = -8, so Mn must be +7) and O is -2. 3. In H2O, O is -2 and H is always +1.
02

Assign oxidation numbers to products

For the products, we have Na2SO4, MnO2 and KOH. For each compound, let's assign oxidation numbers: 1. In Na2SO4, Na is +1, O is -2 and S is +6 (since the overall charge of the compound is 0, and 2Na = +2, 4O = -8, so S must be +6). 2. In MnO2, Mn is +4 (since the overall charge of the compound is 0, and 2O = -4, so Mn must be +4) and O is -2. 3. In KOH, K is +1, O is -2 and H is +1.
03

Identify the oxidizing agent and reducing agent

Now let us compare the change in oxidation numbers for atoms in reactants and products. 1. Na: No change, remains +1. 2. S: Oxidation number increases from +4 to +6 so S in Na2SO3 is being oxidized. 3. H: No change, remains +1. 4. O: No change, remains -2. 5. K: No change, remains +1. 6. Mn: Oxidation number decreases from +7 to +4 so Mn in KMnO4 is being reduced. Since S in Na2SO3 is being oxidized, Na2SO3 is the reducing agent. KMnO4 is the oxidizing agent as Mn in KMnO4 is being reduced.

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