Question

Q24. Balance the following equations by the oxidation number method. (i) Fe2+ + H+ + Cr2072- →Cr3+ + Fe3+ + H20 (ii) I2 + N0–3→ N02 +I03 (iii) I2 + S2032- →I– + S4062- ‘ (iv) MnO, + C2042-→ Mn2+ + CO2​

Answer 1

Explanation:

Total increase in O.N. = 5×2= 10

Total decrease in O.N. = 1

To equalize O.N. multiply NO3–, by 10

I2 + 10no3–→ 10no2 + IO3–

Balancing atoms other than O and H

I2 + 10no3–→ 10NO2 + 2 IO3–

Balancing O and H

I2 + lO no3– + 8H+→ 10NO2 + 2 IO3– + 4H20

Answer 2

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