What element is being reduced in the following redox reaction?

MnO4⁻(aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)

Question

What element is being reduced in the following redox reaction?

MnO4⁻(aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)

BlackTom AI · Accepted Answer

Mn in MnO4^- has oxidation state +7, while Mn2+ is +2, so Mn is reduced.

1) Relevant concepts: oxidation numbers determine electron transfer. Reduction is a decrease in oxidation number (gain of electrons). In acidic solution permanganate is commonly reduced: MnO4^- → Mn2+. Oxalic acid (H2C2O4) / oxalate (C2O4^2-) is commonly the reducing agent and is oxidized to CO2.

2) Step-by-step oxidation-number reasoning:
- For Mn in MnO4^-: let Mn = x. Oxygen is -2 each.
  x + 4(-2) = -1 → x - 8 = -1 → x = +7.
- For Mn in Mn2+: oxidation state = +2.
  Change: +7 → +2 (decrease of 5), so Mn gains 5 electrons and is reduced.

3) Oxidation of carbon (shows complementary process):
- For a carbon in H2C2O4: H = +1, O = -2. Let each C = y.
  2y + 2(+1) + 4(-2) = 0 → 2y + 2 - 8 = 0 → 2y = 6 → y = +3.
- In CO2: C + 2(-2) = 0 → C = +4.
  Change: +3 → +4 (increase of 1 per C), so carbon is oxidized (loses electrons). Thus electrons lost by carbon are gained by Mn.

4) Half-reactions (acidic medium) illustrate electron flow:
- Reduction: MnO4^- + 8H+ + 5e^- → Mn2+ + 4H2O (Mn gains electrons).
- Oxidation (oxalate/oxalic acid): C2O4^2- → 2CO2 + 2e^- (oxalate loses electrons).
  Multiplying to equalize electrons (LCM 10) gives the full redox balance, but the key point is Mn’s +7 → +2 reduction.

5) Why other options are incorrect:
- C: carbon increases oxidation state (+3 → +4), so it is oxidized, not reduced.
- O: oxygen stays at -2 in reactants and products (no net reduction).
- H: hydrogen remains +1 (no reduction).

Conclusion: Manganese (Mn) is the element being reduced. This matches the provided correct answer: Mn.

What element is being reduced in the following redox reaction? MnO4⁻(aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)单项选择题

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