Question

6h^{+}(aq)+2mno_{4}^{-}(aq)+5h_{2}c_{2}o_{4}(aq)→10co_{2}(g)+8h_{2}o(l)+2mn^{2+}(aq)
a student dissolved a 0.139 g sample of oxalic acid, h_{2}c_{2}o_{4}, in water in an erlenmeyer flask. then the student titrated the h_{2}c_{2}o_{4} solution in the flask with a solution of kmno_{4}, which has a dark purple color. the balanced chemical equation for the reaction that occurred during the titration is shown above.
(a) identify the species that was reduced in the titration reaction. justify your answer in terms of oxidation numbers.
note on your ap exam, you will handwrite your responses to free - response questions in a test booklet

Explanation:

In the reaction \(6H^{+}(aq)+2MnO_{4}^{-}(aq)+5H_{2}C_{2}O_{4}(aq)\to10CO_{2}(g)+8H_{2}O(l)+2Mn^{2 +}(aq)\), we look at the oxidation numbers.

For \(MnO_{4}^{-}\), the oxidation number of \(Mn\) is calculated as follows: Let the oxidation number of \(Mn\) be \(x\). In \(MnO_{4}^{-}\), since \(O\) has an oxidation number of \(- 2\), we have \(x+4\times(-2)=-1\). Solving \(x - 8=-1\) gives \(x = +7\).

In \(Mn^{2+}\), the oxidation number of \(Mn\) is \(+2\).

Since the oxidation number of \(Mn\) decreases from \(+7\) in \(MnO_{4}^{-}\) to \(+2\) in \(Mn^{2+}\), \(MnO_{4}^{-}\) is reduced.

Answer:

The species that was reduced is \(MnO_{4}^{-}\). The oxidation number of \(Mn\) decreases from \(+7\) in \(MnO_{4}^{-}\) to \(+2\) in \(Mn^{2+}\).