Q.Permanganate(VII) ion, MnO4–, in basic medium, oxidises iodide ion (I–) to produce molecular iodine (I2) and manganese dioxide (MnO2). Write a balanced ionic equation to represent this reaction.

Concept & Intuition

This is a classic redox reaction. To decide which statement is correct, we need to track oxidation numbers (the charge an atom would have if all bonds were ionic). The key idea: an increase in oxidation number = oxidation (loss of electrons); a decrease = reduction (gain of electrons). The species that causes oxidation is itself reduced — it’s the oxidising agent. Here, dichromate ($\ce{Cr2O7^{2-}}$) contains chromium in a high oxidation state, so it tends to gain electrons and get reduced, while sulfide ($\ce{S^{2-}}$) is electron-rich and tends to lose electrons (get oxidised). Let’s verify step by step.

1. Find the oxidation number of Cr in $\ce{Cr2O7^{2-}}$

   Oxygen is almost always $-2$ (except in peroxides). Let Cr’s oxidation number be $x$.

   For $\ce{Cr2O7^{2-}}$: $2x + 7(-2) = -2$

   $2x - 14 = -2 \implies 2x = +12 \implies x = +6$

   So Cr is $+6$ in dichromate, not $+7$. This already eliminates option (B).

2. Find the oxidation number of Cr in $\ce{Cr^{3+}}$

   Simple: it’s $+3$. So Cr goes from $+6$ to $+3$ — a decrease of 3 per Cr atom. That means each Cr gains 3 electrons; Cr is reduced. Therefore $\ce{Cr2O7^{2-}}$ acts as an oxidising agent.

3. Find the oxidation number of S in $\ce{S^{2-}}$

   The ion $\ce{S^{2-}}$ has sulfur at $-2$ (since it’s a simple monatomic ion).

   In the product, sulfur appears as elemental sulfur $\ce{S(s)}$, where the oxidation number of an element in its standard state is $0$.

   So S goes from $-2$ to $0$ — an increase of 2. That means each S loses 2 electrons; S is oxidised. Therefore $\ce{S^{2-}}$ acts as a reducing agent.

4. Interpret the statements

   • (A) “$\ce{Cr2O7^{2-}}$ reduces the $\ce{S^{2-}}$” — This would mean dichromate causes reduction of sulfide, but reduction means gaining electrons. Here sulfide loses electrons (is oxidised), so this is false. …

The key to this question lies in understanding how the medium (acidic, neutral, or alkaline) dictates the reduction product of KMnO$_4$ and, consequently, which reactions are feasible. In acidic medium, MnO$_4^-$ is reduced to Mn$^{2+}$ (colourless), gaining 5 electrons. In neutral or faintly alkaline medium, it reduces to MnO$_2$ (brown precipitate), gaining 3 electrons. In strongly alkaline medium, it reduces to MnO$_4^{2-}$ (green), gaining 1 electron. Each reaction must be compatible with the medium for the oxidation to proceed.

Let’s examine each reaction one by one.

1. Oxidation of oxalic acid (H$_2$C$_2$O$_4$) by KMnO$_4$

   This is a classic redox titration performed in acidic medium (dilute H$_2$SO$_4$). The half-reactions are:

   • Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
   • Oxidation: $\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2e^-$ The reaction requires H$^+$ ions, so it occurs in acidic medium. Reaction I is valid.

2. Oxidation of iodide (I$^-$) to iodate (IO$_3^-$)

   This conversion involves a large change in oxidation state: I$^-$ (oxidation state -1) to IO$_3^-$ (oxidation state +5), a loss of 6 electrons per iodine atom. In acidic medium, KMnO$_4$ would oxidize I$^-$ to I$_2$ (not IO$_3^-$), because the strong acidic conditions favour the formation of iodine. To push the oxidation all the way to iodate, an alkaline medium is used, where MnO$_4^-$ reduces to MnO$_2$ or MnO$_4^{2-}$ and the reaction proceeds as:

   $\text{I}^- + 6\text{OH}^- \rightarrow \text{IO}_3^- + 3\text{H}_2\text{O} + 6e^-$

   This is not an acidic medium reaction. Reaction II does NOT occur in acidic medium. …

The oxidation state of an element in a compound or ion represents the hypothetical charge an atom would have if all bonds were ionic. It's a useful concept for tracking electron transfer in redox reactions. To determine the change in oxidation state, we first need to identify the initial and final compounds containing the element in question, and then calculate its oxidation state in each.

In this problem, we are looking at the oxidation of the thiosulphate ion ($S_2O_3^{2-}$). When thiosulphate is oxidized by a strong oxidizing agent like potassium permanganate (KMnO$_4$) in a neutral or alkaline solution, it is converted into the sulphate ion ($SO_4^{2-}$). This is a standard reaction outcome under these conditions.

Let's calculate the oxidation state of sulphur in both the reactant and the product.

1. Identify the reactant and product containing sulphur.

   • The reactant is the thiosulphate ion, $S_2O_3^{2-}$.
   • The product, under these specific reaction conditions (oxidation by KMnO$_4$ in neutral/alkaline solution), is the sulphate ion, $SO_4^{2-}$.

2. Calculate the oxidation state of sulphur in the thiosulphate ion ($S_2O_3^{2-}$).

   We assign standard oxidation states to oxygen, which is typically $-2$. The sum of the oxidation states of all atoms in an ion must equal the charge of the ion.

   Let the oxidation state of sulphur be $x$.

   There are two sulphur atoms and three oxygen atoms. The overall charge is $-2$.

So, the *average* oxidation state of sulphur in $S_2O_3^{2-}$ is $+2$. …

The question tests your understanding of the redox behaviour of KMnO$_4$ in acidic medium. Potassium permanganate is a powerful oxidizing agent, and its reduction product depends on the pH of the solution. In acidic medium, the half-reaction is:

So the manganese ends up as Mn$^{2+}$ (colourless), not as MnO$_2$ (brown precipitate, which forms in neutral or alkaline medium). The question asks which change is not correct — meaning which transformation does NOT actually happen when KMnO$_4$ oxidizes something in acid.

Let’s examine each option.

1. Option (A): S$^{2-}$ $\rightarrow$ S

   Sulphide ion is oxidized to elemental sulphur. In acidic medium, KMnO$_4$ can oxidize S$^{2-}$ to S (or further to SO$_4^{2-}$ depending on conditions, but S is a valid product). This change is correct.

2. Option (B): Mn$^{2+}$ $\rightarrow$ MnO$_2$

   Here, Mn$^{2+}$ is being oxidized to MnO$_2$. But in acidic medium, KMnO$_4$ itself gets reduced to Mn$^{2+}$ — it does not oxidize Mn$^{2+}$ further. In fact, Mn$^{2+}$ is the final reduced form of manganese in acid. So this change is backwards: KMnO$_4$ cannot turn Mn$^{2+}$ into MnO$_2$ in acidic conditions. This is the incorrect change.

3. Option (C): SO$_3^{2-}$ $\rightarrow$ SO$_4^{2-}$ …

Concept & Intuition

Equivalent weight is defined as the formula weight (M) divided by the n‑factor — the number of moles of electrons transferred (for redox) or the number of moles of H⁺/OH⁻ exchanged (for acid‑base). The trick is to identify the change in oxidation state (or charge) per formula unit in the given medium. Each substance here behaves differently depending on whether the environment is acidic, neutral, or (for Na₂CO₃) purely acid‑base.

Step‑by‑step reasoning

1. A: Na₂CO₃ (sodium carbonate)

   • This is an acid‑base reaction, not redox. In water, CO₃²⁻ accepts two protons to become H₂CO₃ (or CO₂ + H₂O).
   • Each CO₃²⁻ ion reacts with 2 H⁺. Hence the n‑factor = 2.
   • Equivalent weight = M/2 → matches III.

2. B: KMnO₄ in acidic medium (H⁺)

   • Mn in KMnO₄ has oxidation state +7. In acidic solution, it reduces to Mn²⁺ (oxidation state +2).
   • Change in oxidation number = 7 – 2 = 5 electrons gained per Mn atom.
   • n‑factor = 5 → Equivalent weight = M/5 → matches I.

3. C: K₂Cr₂O₇ in acidic medium (H⁺)

   • Cr in K₂Cr₂O₇ is +6. In acid, it reduces to Cr³⁺ (oxidation state +3). …

The concept here is equivalent weight for a redox reaction. Equivalent weight of an oxidising agent is its molar mass divided by the number of electrons it gains per formula unit (the n-factor). KMnO$_4$ has a molar mass of 158 g/mol. The medium — acidic, neutral, or basic — determines how MnO$_4^-$ is reduced, and therefore how many electrons it takes up.

In acidic medium, MnO$_4^-$ reduces to Mn$^{2+}$ — a change from +7 to +2 oxidation state, a gain of 5 electrons. In neutral or faintly alkaline medium, it reduces to MnO$_2$ (oxidation state +4), gaining 3 electrons. In strongly basic medium, it reduces to MnO$_4^{2-}$ (oxidation state +6), gaining just 1 electron.

Let’s work through each case.

1. Acidic medium The half-reaction is:

n-factor = 5.

Equivalent weight = $\frac{158}{5} = 31.6$.

2. Neutral medium The half-reaction is:

n-factor = 3.

Equivalent weight = $\frac{158}{3} \approx 52.67$, usually rounded to 52.6.

3. Basic medium …

Concept & Intuition

Equivalent weight of an oxidizing agent is its molar mass divided by the number of electrons gained per mole (the “n-factor”). KMnO₄ contains manganese in the +7 oxidation state. The medium (acidic, neutral, or alkaline) determines how far Mn(VII) is reduced. In strong acid, it goes all the way to Mn²⁺ (+2), a 5‑electron change. In strong alkali, it stops at manganate ion MnO₄²⁻ (+6), a 1‑electron change. So the same compound has very different equivalent weights in different conditions.

Step‑by‑step reasoning

1. Molar mass of KMnO₄

   K = 39, Mn = 55, O₄ = 64 → total = 158 g/mol.

2. Reaction in acidic medium

   Half‑reaction:

Oxidation state of Mn changes from +7 to +2 → gain of 5 electrons.

n‑factor = 5.

Equivalent weight = $\frac{158}{5} = 31.6$.

3. Reaction in strongly alkaline medium Half‑reaction: $\text{MnO}_4^- + e^- \rightarrow \text{MnO}_4^{2-}$ …