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Question #d0f6f

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Answer 1 · 2017-08-30T13:39:05

This is a redox reaction (and we will see why!), and we separate the reaction into individual oxidation reduction couples.......

Explanation:

$Oxidation reaction:$ $"Oxidation reaction:"$ $F e (I I)$ $Fe(II)$ is OXIDIZED to $F e (I I I)$ $Fe(III)$

$F e I_{2} \to F e^{3 +} + 2 I^{-} + e^{-}$ $FeI_2 rarrFe^(3+) +2I^(-) +e^(-)$ $(i)$ $(i)$

$Reduction reaction:$ $"Reduction reaction:"$ $Iodate ion$ $"Iodate ion"$ is REDUCED to $I^{+ I}$ $I^(+I)$

$V + I O_{3}^{-} + 6 H^{+} + 4 e^{-} \to I^{+} + 3 H_{2} O$ $stackrel(V+)IO_3^(-) + 6H^+ + 4e^(-) rarrI^(+) +3H_2O$ $(i i)$ $(ii)$

And for each half equation, charge and mass are balanced ABSOLUTELY, as indeed they must be if we purport to represent chemical reality. I add them in such a way as to eliminate the electrons from the final equation, i.e. $4 \times (i) + (i i)$ $4xx(i) + (ii)$ :

$4 F e I_{2} + I O_{3}^{-} + 6 H^{+} \to 4 F e^{3 +} + 8 I^{-} + I^{+} + 3 H_{2} O$ $4FeI_2 +IO_3^(-) + 6H^+ rarr4Fe^(3+) +8I^(-) +I^(+) +3H_2O$

And this is in fact balanced with respect to mass and charge. But we could add $1 \times C l^{-}$ $1xxCl^-$ to each side of the chemical equation....

$4 F e I_{2} + I O_{3}^{-} + 6 H^{+} + C l^{-} \to 4 F e^{3 +} + 8 I^{-} + I C l + 3 H_{2} O$ $4FeI_2 +IO_3^(-) + 6H^+ +Cl^(-)rarr4Fe^(3+) +8I^(-) +ICl +3H_2O$

And to make it even simpler, I could remove $8 \times I^{-}$ $8xxI^-$ FROM EACH SIDE of the EQUATION:

$4 F e^{2 +} + I O_{3}^{-} + 6 H^{+} + C l^{-} \to 4 F e^{3 +} + I C l + 3 H_{2} O$ $4Fe^(2+) +IO_3^(-) + 6H^+ +Cl^(-)rarr4Fe^(3+) +ICl +3H_2O$ ,

i.e. a four electron reduction with respect to $I O_{3}^{-}$ $IO_3^-$ , and a four electron oxidation with respect to 4 equiv of ferrous ion.

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Question #d0f6f

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