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Answer the following (group 2) · Q5

Q.Balance the following equation

(i) KMnO4+H2C2O4+H2SO4 -> MnSO4 + K2SO4 + H2O + O2
(ii) K2Cr2O7 + KI + H2SO4 -> K2SO4 + Cr2(SO4)3 + 7H2O + 3I2
Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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(i) The permanganate–oxalic acid redox in acid medium:

2KMnO4+5H2C2O4+3H2SO4⟶2MnSO4+K2SO4+10CO2+8H2O\mathrm{2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \longrightarrow 2MnSO_4 + K_2SO_4 + 10CO_2 + 8H_2O}

Mn goes from +7 to +2 (gain of 5 e⁻ each) while each oxalate carbon goes from +3 to +4 (loss of 2 e⁻ per H2C2O4\mathrm{H_2C_2O_4}): 2×5=5×22 \times 5 = 5 \times 2 electrons balance, and the atom balance follows.

Note

The book prints the product of this equation as "O₂"; the oxidation of oxalic acid gives CO₂ (as the chapter's own section 8.7.2 a iv shows). We balance with CO₂.

(ii) The dichromate–iodide redox:

K2Cr2O7+6KI+7H2SO4⟶4K2SO4+Cr2(SO4)3+7H2O+3I2\mathrm{K_2Cr_2O_7 + 6KI + 7H_2SO_4 \longrightarrow 4K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3I_2}

Cr goes from +6 to +3 (2 Cr gain 6 e⁻); six I⁻ lose 6 e⁻ to give 3 I23\,\mathrm{I_2}. …

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