How would you balance the following equation: I2+HNO3-->HIO3+NO2+H2O?

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Answer 1 · 2015-11-13T21:38:17

This is a redox reaction, which may be approached by oxidation and reduction half equations.

Oxidation:
Elemental iodine is oxidized to $IO_3^-$ , iodate.

$1/2I_2 +3H_2O rarr IO_3^(-) +5e^(-) + 6H^+$ $(i)$

Reduction:

Nitric acid is reduced to nitric oxide $N^(V)rarrN^(IV)$ :

$HNO_3 +H^+ +e^(-)rarr NO_2 +H_2O$ $(ii)$

The overall redox reaction is $(i) + 5xx(ii)$ $=$ the overall redox equation:

$1/2I_2 + 5HNO_3 rarr IO_3^(-) +5NO_2 + 2H_2O +H^+$

Alternatively, I could write:

$1/2I_2 + 5HNO_3 rarr HIO_3 +5NO_2 + 2H_2O$

Are the final equations balanced with respect to mass and to charge? Don't trust my arithmetic. How would I remove the half coefficient on $I_2$ ?