Balance the following equations by the oxidation number method. (i) Fe2+ + H+ + Cr2O72- →Cr3+ + Fe3+ + H2O … S4O62- (iv) MnO, +…
Answers:
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– + 4H2O
To equalize O.N. multiply CO2 by 2. –
MnO2 + C2O2-4 → Mn2+ + 2CO2
Balance H and O by adding 2H2O on right side, and 4H+ on left side of equation.
MnO2 + C2O2-4 + 4H+ → Mn2+ + 2CO2 + 2H2O .