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Q.Balance the following Redox reaction by ion-electron method in acidic medium. Cr2O7^2- (aq) + SO2

(g) -> Cr^3+ (aq) + SO4^2- (aq)
Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2018Subjective· 4mImportance★★★★★
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By the ion-electron method in acidic medium: Cr2O7^2- + 3 SO2 + 2 H+ -> 2 Cr^3+ + 3 SO4^2- + H2O.

Step 1 — Identify the two half reactions and oxidation-state changes:

  • Chromium in Cr2O7^2- is +6; in Cr^3+ it is +3 (reduction, gain of electrons).
  • Sulphur in SO2 is +4; in SO4^2- it is +6 (oxidation, loss of electrons).

Step 2 — Reduction half reaction (Cr2O7^2- -> Cr^3+), balance in acidic medium:

Cr2O7^2- -> 2 Cr^3+

Balance O by adding H2O: Cr2O7^2- -> 2 Cr^3+ + 7 H2O

Balance H by adding H+: Cr2O7^2- + 14 H+ -> 2 Cr^3+ + 7 H2O

Balance charge by adding electrons (left charge = -2 + 14 = +12; right = +6; add 6 e- to left):

Cr2O7^2- + 14 H+ + 6 e- -> 2 Cr^3+ + 7 H2O ... (reduction)

Step 3 — Oxidation half reaction (SO2 -> SO4^2-), balance in acidic medium:

SO2 -> SO4^2-

Balance O by adding H2O: SO2 + 2 H2O -> SO4^2-

Balance H by adding H+: SO2 + 2 H2O -> SO4^2- + 4 H+

Balance charge by adding electrons (left = 0; right = -2 + 4 = +2; add 2 e- to right):

SO2 + 2 H2O -> SO4^2- + 4 H+ + 2 e- ... (oxidation)

Step 4 — Equalise electrons:

Reduction gains 6 e-, oxidation loses 2 e-. Multiply the oxidation half reaction by 3:

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

Step 5 — Add the two half reactions and cancel: …

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