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Q.Balance the following equation in acid medium by Ion-electron method:
Fe^2+(aq) + Cr2O7^2-(aq) -> Fe^3+(aq) + Cr^3+(aq)

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2018Subjective· 4mImportance★★★★★
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Balancing by the ion-electron (half-reaction) method: write, balance, and add the oxidation and reduction half-reactions separately, then combine so electrons cancel.

Reaction: Fe2+(aq)+Cr2O72−(aq)→Fe3+(aq)+Cr3+(aq)\text{Fe}^{2+}(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Cr}^{3+}(aq) (acidic medium)

Step 1 — Split into two half-reactions:

Oxidation: Fe2+→Fe3+\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}

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

Step 2 — Balance atoms other than O and H, then O (with H2O\text{H}_2\text{O}), then H (with H+\text{H}^+, acidic medium):

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

Balance the 7 oxygens with 7 H2O\text{H}_2\text{O}: Cr2O72−→2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

Balance the 14 hydrogens (from the 7 H2O\text{H}_2\text{O}) with 14 H+\text{H}^+ on the left: 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 3 — Balance charge with electrons:

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

Reduction half: left side charge =−2+14=+12= -2 + 14 = +12; right side charge =2(+3)=+6= 2(+3) = +6; add 6 electrons to the left to balance charge:

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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