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Exercises · 7.21

Q.The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.

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The key idea is that Mn³⁺ disproportionates — one Mn³⁺ is oxidised to MnO₂ and another is reduced to Mn²⁺ — and the balanced ionic equation is 2Mn3++2H2O→Mn2++MnO2+4H+2\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + \text{MnO}_2 + 4\text{H}^+.

Disproportionation is a special kind of redox reaction where a single species (here, Mn³⁺) acts as both the oxidising agent and the reducing agent. One part of it gets oxidised (loses electrons, oxidation number increases) and another part gets reduced (gains electrons, oxidation number decreases). The trick is to figure out the two products and then balance atoms and charge.

Let’s work through it.

  1. Identify the oxidation states. In Mn³⁺, manganese is in the +3 oxidation state. In Mn²⁺, it’s +2 — that’s a decrease of 1 electron per ion (reduction). In MnO₂, oxygen is −2 each, so Mn must be +4 — that’s an increase of 1 electron per ion (oxidation). So the disproportionation is:

Mn3+→Mn2+(reduction, gain of 1 e−)\text{Mn}^{3+} \rightarrow \text{Mn}^{2+} \quad (\text{reduction, gain of 1 e}^-)

Mn3+→MnO2(oxidation, loss of 1 e−)\text{Mn}^{3+} \rightarrow \text{MnO}_2 \quad (\text{oxidation, loss of 1 e}^-)

  1. Balance the electron transfer.

    Each Mn³⁺ that becomes Mn²⁺ gains 1 electron. Each Mn³⁺ that becomes MnO₂ loses 1 electron. So the electrons already cancel if we take one of each — but we also need to balance atoms, especially oxygen. That’s where water and H⁺ come in.

  2. Write the half-reactions in acidic medium.

    Reduction half:

Mn3++e−→Mn2+\text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+}

Oxidation half: Mn³⁺ to MnO₂. Start with:

Mn3+→MnO2\text{Mn}^{3+} \rightarrow \text{MnO}_2

Balance oxygen by adding water:

Mn3++2H2O→MnO2\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2

Balance hydrogen by adding H⁺:

Mn3++2H2O→MnO2+4H+\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+

Now balance charge: left side has +3, right side has +4 (from 4H⁺). So add 1 electron to the right:

Mn3++2H2O→MnO2+4H++e−\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+ + e^-

  1. Combine the half-reactions. The reduction half gives 1 electron, the oxidation half gives 1 electron — they cancel directly. Add them:

(Mn3++e−→Mn2+)+(Mn3++2H2O→MnO2+4H++e−)(\text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+}) + (\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+ + e^-)

Cancel the electrons: …

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