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title: "A student standardizes an aqueous solution of potassium permanganate by titrating it against a solution of oxalic acid under acidic conditions. The reaction that occurs during the titration is represented by the following balanced equation: \\[ 2\\text{MnO}_4^-\\text{(aq)} + 5\\text{H}_2\\text{C}_2\\text{O}_4\\text{(aq)} + 6\\text{H}^+\\text{(aq)} \\rightarrow 2\\text{Mn}^{2+}\\text{(aq)} + 10\\text{CO}_2\\text{(g)} + 8\\text{H}_2\\text{O}\\text{(l)} \\] Which species acts as the reducing agent in this reaction, and why?"
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url: "https://nerd-notes.com/ubq/121413/"
date_modified: "2026-08-23T05:02:09+00:00"
---

# A student standardizes an aqueous solution of potassium permanganate by titrating it against a solution of oxalic acid under acidic conditions. The reaction that occurs during the titration is represented by the following balanced equation: \[ 2\text{MnO}_4^-\text{(aq)} + 5\text{H}_2\text{C}_2\text{O}_4\text{(aq)} + 6\text{H}^+\text{(aq)} \rightarrow 2\text{Mn}^{2+}\text{(aq)} + 10\text{CO}_2\text{(g)} + 8\text{H}_2\text{O}\text{(l)} \] Which species acts as the reducing agent in this reaction, and why?

A student standardizes an aqueous solution of potassium permanganate by titrating it against a solution of oxalic acid under acidic conditions. The reaction that occurs during the titration is represented by the following balanced equation: \[ 2\text{MnO}_4^-\text{(aq)} + 5\text{H}_2\text{C}_2\text{O}_4\text{(aq)} + 6\text{H}^+\text{(aq)} \rightarrow 2\text{Mn}^{2+}\text{(aq)} + 10\text{CO}_2\text{(g)} + 8\text{H}_2\text{O}\text{(l)} \] Which species acts as the reducing agent in this reaction, and why?

- **A.** \(\text{MnO}_4^-\text{(aq)}\), because \(\text{Mn}\) gains electrons as its oxidation number decreases from \(+7\) to \(+2\)
- **B.** \(\text{MnO}_4^-\text{(aq)}\), because \(\text{Mn}\) loses electrons as its oxidation number increases from \(+2\) to \(+7\)
- **C.** \(\text{H}_2\text{C}_2\text{O}_4\text{(aq)}\), because \(\text{C}\) loses electrons as its oxidation number increases from \(+3\) to \(+4\)
- **D.** \(\text{H}_2\text{C}_2\text{O}_4\text{(aq)}\), because \(\text{C}\) gains electrons as its oxidation number decreases from \(+4\) to \(+3\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121413/*