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Q.Balance the following equation by the ion-electron method: Cr2O7^2-(aq) + SO2(g) -> Cr^3+(aq) + SO4^2-(aq) (in acidic medium).

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2024Subjective· 4mImportance★★★★★
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Split into a reduction half-reaction (Cr2O7^2- to Cr3+) and an oxidation half-reaction (SO2 to SO4^2-), balance each for atoms/charge, equalise electrons, and add.

Step 1: Write the two half-reactions (skeletal, unbalanced).

Reduction: Cr2O7^2- -> Cr3+

Oxidation: SO2 -> SO4^2-

Step 2: Balance atoms other than O and H first.

Reduction: Cr2O7^2- -> 2Cr3+ (Cr balanced)

Oxidation: SO2 -> SO4^2- (S balanced)

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

Reduction: Cr2O7^2- + 14H+ -> 2Cr3+ + 7H2O

Oxidation: SO2 + 2H2O -> SO4^2- + 4H+

Step 4: Balance charge by adding electrons.

Reduction: left charge = -2 + 14 = +12; right charge = +6; add 6e- to the left:

Cr2O7^2- + 14H+ + 6e- -> 2Cr3+ + 7H2O

Oxidation: left charge = 0; right charge = -2 + 4 = +2; add 2e- to the right:

SO2 + 2H2O -> SO4^2- + 4H+ + 2e-

Step 5: Equalise electrons lost and gained. Reduction needs 6e-, oxidation releases 2e-, so multiply the oxidation half-reaction by 3:

3SO2 + 6H2O -> 3SO4^2- + 12H+ + 6e-

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