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

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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.

Explanation:

Oxidation:
Elemental iodine is oxidized to $I O_{3}^{-}$ $IO_3^-$ , iodate.

$\frac{1}{2} I_{2} + 3 H_{2} O \to I O_{3}^{-} + 5 e^{-} + 6 H^{+}$ $1/2I_2 +3H_2O rarr IO_3^(-) +5e^(-) + 6H^+$ $(i)$ $(i)$

Reduction:

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

$H N O_{3} + H^{+} + e^{-} \to N O_{2} + H_{2} O$ $HNO_3 +H^+ +e^(-)rarr NO_2 +H_2O$ $(i i)$ $(ii)$

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

$\frac{1}{2} I_{2} + 5 H N O_{3} \to I O_{3}^{-} + 5 N O_{2} + 2 H_{2} O + H^{+}$ $1/2I_2 + 5HNO_3 rarr IO_3^(-) +5NO_2 + 2H_2O +H^+$

Alternatively, I could write:

$\frac{1}{2} I_{2} + 5 H N O_{3} \to H I O_{3} + 5 N O_{2} + 2 H_{2} O$ $1/2I_2 + 5HNO_3 rarr HIO_3 +5NO_2 + 2H_2O$

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 $I_{2}$ $I_2$ ?

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

1 Answer

Explanation:

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