Q.Balance the following equations by the oxidation number method.
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Start your 14-day free trial to unlock the full solution →Step 1 (i). Cr: +6→+3 (gain 3e⁻ per Cr, 6e⁻ per Cr2O7²⁻); I: −1→0 (lose 1e⁻ per I, so 6 KI needed for 6e⁻). This gives 1 K2Cr2O7 : 6 KI, and the atom/charge balance (verified by counting K, S, O, H) requires 7 H2SO4 and 7 H2O, 4 K2SO4, 3 I2: K2Cr2O7 + 6KI + 7H2SO4 → 4K2SO4 + Cr2(SO4)3 + 3I2 + 7H2O.
Step 2 (ii). Mn: +7→+4 (gain 3e⁻); S (in SO3²⁻): +4→+6 (lose 2e⁻). LCM(3,2)=6 needs 2 KMnO4 and 3 Na2SO3. Balancing O and H (this is a neutral/basic-medium reaction) needs 1 H2O on the reactant side and gives 2 KOH: 2KMnO4 + 3Na2SO3 + H2O → 2MnO2 + 3Na2SO4 + 2KOH.
Step 3 (iii). Cu: 0→+2 (lose 2e⁻); N (in the reduced product NO2): +5→+4 (gain 1e⁻ each), so 2 HNO3 are reduced per Cu. A further 2 HNO3 are needed as the (unreduced) nitrate ions in Cu(NO3)2, for 4 HNO3 total: Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O. …
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