Question

balance the given equation by oxidation method I2 + HNO3 ------> HIO3 + NO2 + H2O PLS no unnecessary answers !!

Answer 1

. Consider the case of Nitrogen-

In reactant side the oxidation number of nitrogen is +5, while the oxidation number of nitrogen in product side is +4, so the reduction reaction is being carried out in a manner:-

N(+5) + e(-) == N(+4) ———————- (1)

2. Consider the case of Iodine-

In reactant side the oxidation number of iodine is 0 and in product side the oxidation number of iodine is +5, so the oxidation reaction reaction can be written as:-

I2(0) == 2I(+5) + 10e(-) ———————— (2)

Now, balancing reaction (1) and (2) gives us the following equation-

10N(+5) + I2(0) == 2I(+5) + 10N(+4)

Thus, in original form the reaction can be written as-

10HNO3 + I2 == 2HIO3 + 10NO2 + 4H2O