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?

  • $\begingroup$ 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". $\endgroup$
    – user55119
    May 27, 2019 at 1:51
  • $\begingroup$ 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. $\endgroup$
    – M.A.R.
    May 27, 2019 at 7:28

1 Answer 1


In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.


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