# Reaction between copper(II) iodate and iodide

I was attempting the following question from IChO 1989, which involved balancing of redox equations. One may just expect this to be a typical iodate-iodide redox reaction, usually employed in iodometry. However, the presence of $$\ce{Cu(II)}$$ makes things a bit more complicated since it can also be reduced to give $$\ce{CuI}$$.

To determine the solubility product of copper(II) iodate, $$\ce{Cu(IO3)2}$$, by iodometric titration in an acidic solution ($$\pu{25 ^\circ C}$$), $$\pu{30.00 cm3}$$ of a $$0.100$$ molar sodium thiosulfate solution are needed to titrate $$\pu{20.00 cm3}$$ of a saturated aqueous solution of $$\ce{Cu(IO3)2}$$.

1.1 Write the sequence of balanced equations for the above described reactions. [...]

After realising that $$\ce {Cu^2+}$$ is reduced as well, I wrote the following overall equation :

$$\ce{2Cu^2+ + 2IO3^- + 12H^+ + 14I^- -> 7I2 + 2 CuI + 6H2O} \tag{1}$$

which is based on

\begin{align} \ce{2Cu^2+ + 4I^- &-> 2CuI + I2} \tag{2} \label{eq:cuii} \\ \ce{2IO3^- + 12H^+ + 10I^- &-> 6I2 + 6H2O} \tag{3} \label{eq:iodate} \end{align}

However, it turns out that the equation given in the answer scheme seems to have multiplied eq. $$(\ref{eq:iodate})$$ by $$2$$ and the overall equation given is:

$$\ce{2Cu^2+ + 4IO3^- + 24H^+ + 24I^- -> 13I2 + 2CuI + 12H2O} \tag{4}$$

Usually, this is a trivial thing and does not matter. But in this case, it does since the reaction stiochiometry is completely different based on the equation given. Have I gone wrong somewhere?

• Well, I get what you got. Iodate to iodide is +6e balanced with protons and water. Cupric to cuprous is + 1e. A total of +7e for reduction. Two iodide to iodine is -2e oxidation. Multiply reduction by 2 and the oxidation by 7 and it works out. Can't speak for the "answer scheme". May 27 '19 at 1:51
• This question is clearly not closeable as homework because there is effort provided. If you want to close it, please describe your reasoning in a comment. May 27 '19 at 7:28