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Q.Balance the given equation in acidic medium using Half reaction method:
Fe2+(aq) + Cr2O7^2-(aq) → Fe3+(aq) + Cr3+(aq)

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2019Subjective· 3mImportance★★★★★
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Splitting the reaction into oxidation and reduction half-reactions, balancing each for atoms/charge, and combining with equal electrons gives the fully balanced equation.

Step 1 — write the two half-reactions:

Oxidation (Fe²⁺ → Fe³⁺): Fe2+→Fe3++e−\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-

Reduction (Cr₂O₇²⁻ → Cr³⁺): Cr2O72−→2Cr3+\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}

Step 2 — balance atoms other than O and H in the reduction half (Cr already balanced at 2 on each side).

Step 3 — balance O by adding H₂O, then H by adding H⁺ (acidic medium):

Cr2O72−+14H+→2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

Step 4 — balance charge by adding electrons:

Left side charge: −2+14=+12-2 + 14 = +12. Right side charge: 2(+3)=+62(+3) = +6. Add 6 electrons to the left to balance:

Cr2O72−+14H++6e−→2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

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