(i) What is the oxidation number of Mn in K₂MnO₄?
(ii) Balance the following chemical equation by ion-electron method: Cr₂O₇²⁻ + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O [1 + 2]

**OR**

(i) Balance the following chemical equation by oxidation number method: NaNO₃ + Zn + NaOH → NH₃ + Na₂ZnO

Question

(i) What is the oxidation number of Mn in K₂MnO₄?
(ii) Balance the following chemical equation by ion-electron method: Cr₂O₇²⁻ + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O [1 + 2]

**OR**

(i) Balance the following chemical equation by oxidation number method: NaNO₃ + Zn + NaOH → NH₃ + Na₂ZnO₂ + H₂O
(ii) What is the oxidation number of S in S₈? [2 + 1]

Accepted Answer

>

> [!TLDR]
> Use the sum-of-oxidation-numbers-equals-charge rule for Mn, then balance the dichromate/Fe²⁺ redox couple using separate reduction and oxidation half-reactions.
>
> **(i) Oxidation number of Mn in $K_2MnO_4$:**
> Let the oxidation number of Mn be $x$. K is $+1$ (×2), O is $-2$ (×4), and the overall compound is neutral:
> $$2(+1) + x + 4(-2) = 0 \Rightarrow 2 + x - 8 = 0 \Rightarrow x = +6$$
>
> **(ii) Balancing $Cr_2O_7^{2-} + Fe^{2+} + H^+ \rightarrow Cr^{3+} + Fe^{3+} + H_2O$ by the ion–electron (half-reaction) method:**
>
> **Reduction half-reaction** (Cr goes from $+6$ to $+3$, a gain of 3 electrons per Cr, ×2 Cr atoms = 6 electrons):
> $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$
>
> **Oxidation half-reaction** (Fe goes from $+2$ to $+3$, loss of 1 electron):
> $$Fe^{2+} \rightarrow Fe^{3+} + e^-$$
> …

Q.(i) What is the oxidation number of Mn in K₂MnO₄?

Concept understanding — Balancing Redox Reactions

Q.(i) What is the oxidation number of Mn in K₂MnO₄?

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