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.

Moles taken:

In basic medium permanganate is reduced $\text{MnO}_4^- \to \text{MnO}_2$ (gain of $3e^-$), and bromide is oxidised $\text{Br}^- \to \text{BrO}_3^-$ (loss of $6e^-$).

Total electrons accepted by permanganate:

Bromide oxidised to bromate: …

Concept & Intuition

The reaction involves iodide ($I^-$) being oxidized to iodine ($I_2$) and permanganate ($MnO_4^-$) being reduced to $Mn^{2+}$ in acidic medium. The key is to balance the electron transfer: each $I^-$ loses 1 electron (two $I^-$ give $I_2$ and lose 2 electrons total), while each $MnO_4^-$ gains 5 electrons (Mn goes from +7 to +2). The ratio of electrons lost to gained must be equal, so the number of $MnO_4^-$ to $I^-$ is in the ratio of electrons lost per $I^-$ pair to electrons gained per $MnO_4^-$. Then we balance atoms and charge with $H^+$ and $H_2O$.

Step-by-step balancing

1. Write the half-reactions

   Oxidation: $2I^- \rightarrow I_2 + 2e^-$

   Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

2. Equalize electrons transferred

   The oxidation half gives 2 electrons, reduction gives 5. LCM = 10.

   Multiply oxidation by 5: $10I^- \rightarrow 5I_2 + 10e^-$

   Multiply reduction by 2: $2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$

3. Add the half-reactions

   $10I^- + 2MnO_4^- + 16H^+ \rightarrow 2Mn^{2+} + 8H_2O + 5I_2$

   So the coefficients are: $x=10$, $y=2$, $z=16$, $a=2$, $b=8$, $c=5$. …

The key idea here is that equivalent weight depends on the number of electrons gained or lost by one formula unit of the substance in the balanced redox reaction. That number is called the n-factor (or valence factor). For a compound acting as a reducing agent, n-factor = total increase in oxidation number per molecule; for an oxidising agent, it’s the total decrease.

Let’s work through it step by step.

1. Assign oxidation numbers to find the change for chromium.

   In $\mathrm{Cr(OH)_3}$, oxygen is –2 and hydrogen is +1, so Cr must be +3 to balance: $x + 3(-2+1) = 0 \Rightarrow x = +3$.

   In $\mathrm{CrO_4^{2-}}$, each O is –2, so Cr is $x + 4(-2) = -2 \Rightarrow x = +6$.

   The change per Cr atom is $+6 - (+3) = +3$. Since there is one Cr per $\mathrm{Cr(OH)_3}$, the total increase in oxidation number is 3. This is the n-factor for $\mathrm{Cr(OH)_3}$ as a reducing agent.

2. Find the change for iodine.

   In $\mathrm{IO_3^-}$, O is –2, so I is $x + 3(-2) = -1 \Rightarrow x = +5$.

   In $\mathrm{I^-}$, I is –1.

   The change per I atom is $-1 - (+5) = -6$. That’s a decrease of 6, so the n-factor for $\mathrm{IO_3^-}$ as an oxidising agent is 6.

3. Apply the equivalent weight formula. …

The concept first: why we balance electrons before atoms

A redox equation is really two stories happening at once — something is being reduced and something is being oxidised — and the two are locked together by a single conservation law: every electron lost by one species must be gained by another. So the smart order of work is:

1. Find the oxidation-number changes (this tells you the electron count).
2. Balance the electrons by scaling the two half-reactions.
3. Only then patch up O with $\mathrm{H_2O}$ and H with $\mathrm{H^+}$ (in acidic medium).

If you try to balance atoms first, you are guessing. If you balance electrons first, the coefficients are forced.

Step-by-step

1. Assign oxidation numbers.

• In $\mathrm{Cr_2O_7^{2-}}$: let Cr be $x$. Then $2x + 7(-2) = -2 \Rightarrow x = +6$. In $\mathrm{Cr^{3+}}$, Cr is $+3$. So each Cr gains 3 electrons, and there are 2 Cr atoms $\Rightarrow$ 6 electrons per dichromate ion. Dichromate is the oxidant (it is reduced).
• In $\mathrm{NO_2^-}$: $x + 2(-2) = -1 \Rightarrow x = +3$. In $\mathrm{NO_3^-}$: $x + 3(-2) = -1 \Rightarrow x = +5$. Nitrogen goes $+3 \rightarrow +5$, losing 2 electrons. Nitrite is the reductant (it is oxidised).

2. Write the two half-reactions (acidic medium).

Reduction:

Oxidation:

3. Equalise the electrons. The reduction consumes 6 $e^-$; each oxidation supplies 2 $e^-$. So multiply the oxidation half by $\mathbf{3}$:

4. Add the halves and cancel. …

We need to balance the reaction:

Concept & Intuition

This is a classic redox reaction in acidic solution. Permanganate ($\mathrm{MnO_4^-}$) is reduced to $\mathrm{Mn^{2+}}$, while oxalate ($\mathrm{C_2O_4^{2-}}$) is oxidized to $\mathrm{CO_2}$. The key is to balance electrons transferred: each Mn gains 5 electrons, each oxalate loses 2 electrons (since each carbon goes from +3 to +4, two carbons per oxalate). The least common multiple of 5 and 2 is 10, so we need 2 permanganates and 5 oxalates. Then balance the remaining atoms and charge with $\mathrm{H^+}$ and $\mathrm{H_2O}$.

Step‑by‑step balancing

1. Write the two half‑reactions

   Reduction: $\mathrm{MnO_4^- \rightarrow Mn^{2+}}$

   Oxidation: $\mathrm{C_2O_4^{2-} \rightarrow CO_2}$

2. Balance atoms other than H and O in each half

   Reduction: Mn is already balanced (1 Mn each side).

   Oxidation: 2 C on left, so put 2 $\mathrm{CO_2}$ on right: $\mathrm{C_2O_4^{2-} \rightarrow 2\,CO_2}$.

3. Balance oxygen by adding $\mathrm{H_2O}$

   Reduction: left has 4 O, right has 0 O → add 4 $\mathrm{H_2O}$ to right:

   $\mathrm{MnO_4^- \rightarrow Mn^{2+} + 4\,H_2O}$

   Oxidation: left has 4 O, right has 4 O (in 2 $\mathrm{CO_2}$) → already balanced.

4. Balance hydrogen by adding $\mathrm{H^+}$

   Reduction: right has 8 H (from 4 $\mathrm{H_2O}$), left has 0 → add 8 $\mathrm{H^+}$ to left:

   $\mathrm{MnO_4^- + 8\,H^+ \rightarrow Mn^{2+} + 4\,H_2O}$

   Oxidation: no H atoms, so no $\mathrm{H^+}$ needed.

5. Balance charge by adding electrons

   Reduction: left charge: $-1 + 8(+1) = +7$; right charge: $+2$ → need 5 electrons on left:

   $\mathrm{MnO_4^- + 8\,H^+ + 5e^- \rightarrow Mn^{2+} + 4\,H_2O}$

   Oxidation: left charge: $-2$; right charge: $0$ (neutral $\mathrm{CO_2}$) → need 2 electrons on right:

   $\mathrm{C_2O_4^{2-} \rightarrow 2\,CO_2 + 2e^-}$

6. Equalize electrons transferred

   LCM of 5 and 2 is 10. Multiply reduction half by 2, oxidation half by 5:

   $2\mathrm{MnO_4^-} + 16\mathrm{H^+} + 10e^- \rightarrow 2\mathrm{Mn^{2+}} + 8\mathrm{H_2O}$

   $5\mathrm{C_2O_4^{2-}} \rightarrow 10\mathrm{CO_2} + 10e^-$

7. Add the half‑reactions

   Electrons cancel:

   $2\mathrm{MnO_4^-} + 5\mathrm{C_2O_4^{2-}} + 16\mathrm{H^+} \rightarrow 2\mathrm{Mn^{2+}} + 10\mathrm{CO_2} + 8\mathrm{H_2O}$

8. Account for spectator ions (potassium) …

We are balancing the reaction:

Concept & Intuition

This is a classic redox reaction: permanganate ($\mathrm{MnO_4^-}$) is reduced to $\mathrm{Mn^{2+}}$, and oxalate ($\mathrm{C_2O_4^{2-}}$) is oxidized to $\mathrm{CO_2}$. In acidic solution, we balance each half‑reaction for atoms and charge using $\mathrm{H^+}$ and $\mathrm{H_2O}$, then multiply so electrons cancel. The coefficients $x$, $y$, $z$ come directly from that process.

1. Write the reduction half‑reaction (permanganate → Mn²⁺) $\mathrm{MnO_4^-}$ contains Mn in +7 oxidation state; it gains 5 electrons to become $\mathrm{Mn^{2+}}$ (Mn in +2). Balance oxygen with water, then hydrogen with $\mathrm{H^+}$:

2. Write the oxidation half‑reaction (oxalate → CO₂) Each $\mathrm{C_2O_4^{2-}}$ has two carbons in +3 state; each becomes $\mathrm{CO_2}$ (carbon in +4). That’s a loss of 1 electron per carbon, so 2 electrons per oxalate ion. Balance oxygen (already balanced: 4 O on each side) and charge:

3. Equalize electrons transferred Reduction gains 5 electrons; oxidation loses 2 electrons. The least common multiple is 10. Multiply reduction by 2:

Multiply oxidation by 5:

4. Add the half‑reactions Electrons cancel:

5. Account for spectator ions (potassium) The reactants are given as $\mathrm{KMnO_4}$ and $\mathrm{K_2C_2O_4}$, not as free ions.
   • $2\,\mathrm{KMnO_4}$ provides 2 $\mathrm{MnO_4^-}$ and 2 $\mathrm{K^+}$. …

The key is to track the electron transfer — not the coefficients of the full balanced equation, but the stoichiometry of oxidation and reduction per mole of each reactant. KMnO$_4$ in acidic medium is a powerful oxidising agent; it gets reduced to Mn$^{2+}$, and the change in oxidation state tells you exactly how many electrons it takes up per mole. Oxalic acid, on the other hand, gets oxidised to CO$_2$, and each molecule of oxalic acid loses a fixed number of electrons. The number of CO$_2$ molecules produced per mole of KMnO$_4$ is simply the ratio of electrons transferred, scaled by the fact that each oxalic acid molecule yields two CO$_2$ molecules.

1. Determine the electron change for KMnO$_4$ In acidic medium, the half-reaction is:

Manganese in MnO$_4^-$ has an oxidation state of +7; in Mn$^{2+}$ it is +2. The gain of 5 electrons per Mn atom is clear. So 1 mole of KMnO$_4$ accepts 5 moles of electrons.

2. Determine the electron change for oxalic acid Oxalic acid is H$_2$C$_2$O$_4$. Each carbon atom has an oxidation state of +3 (since H is +1, O is –2, and the molecule is neutral: $2(+1) + 2x + 4(-2) = 0 \Rightarrow 2x = 6 \Rightarrow x = +3$). In CO$_2$, carbon is +4. So each carbon atom loses 1 electron, and since there are two carbons per oxalic acid molecule, each mole of H$_2$C$_2$O$_4$ loses 2 moles of electrons when fully oxidised to CO$_2$. The balanced half-reaction is:

3. Find how many moles of oxalic acid are oxidised per mole of KMnO$_4$ Electrons lost by oxalic acid must equal electrons gained by KMnO$_4$. If $n$ moles of H$_2$C$_2$O$_4$ react per mole of KMnO$_4$, then:

So 2.5 moles of oxalic acid are oxidised by 1 mole of KMnO$_4$.

4. Convert moles of oxalic acid to moles of CO$_2$ Each mole of oxalic acid produces 2 moles of CO$_2$ (from the half-reaction above). Therefore: …