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?
