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I found out the reductant, oxidant, oxidation half reaction, reduction half reaction; balanced ions (adding $\ce{H2O},$ $\ce{H+},$ electrons) and charges:

$$\ce{K\overset{-1}{I} + \overset{+2}{Cu}SO4 -> K2SO4 + \overset{+1}{Cu}_2I2 + \overset{0}{I}_2}$$

$$ \begin{align} \ce{KI &-> K+ + I + e-} \tag{ox} \\ \ce{CuSO4 + 8 H+ + e- &-> Cu+ + S^6+ + 4 H2O} \tag{red} \\ \hline \ce{KI + CuSO4 + 8 H+ &-> K+ + I + Cu+ + S^6+ + 4 H2O} \tag{redox} \end{align} $$

Products should come across to $\ce{K2SO4 + Cu2I2 + I2}.$ But adding the half reactions I get $\ce{I + K+ + Cu+ + 4H2O + S^6+}.$ Water isn't present in the actual products.

Also, what do I do with the $\ce{H+}$ ions in the left? Where did it go wrong and how to proceed further?

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    $\begingroup$ I think you're trying to balance the equation following some recipe too closely, and you're throwing away the actual chemistry in the process, which is making the problem way harder than it should be. Both potassium and sulfate ions are spectators in the reaction, so you might as well just disregard them, and that would simplify things a lot - for one, it wouldn't have caused you to make the mistake of adding $\ce{H2O}$ as a product, or add $\ce{H+}$ to compensate for that. "$\ce{S^{6+}}$" also doesn't exist as an ion. $\endgroup$
    – Nicolau Saker Neto
    Commented Oct 9, 2022 at 8:39
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    $\begingroup$ Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in CH SE titles. $\endgroup$
    – Poutnik
    Commented Oct 9, 2022 at 9:38
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    $\begingroup$ Many mistakes. First, it is not $\ce{CuSO4}$ that reacts, because in solution $\ce{CuSO4}$ does not exist. It is transformed into $\ce{Cu^{2+}}$ and $\ce{SO4^{2-}}$. Then, $\ce{Cu^{2+}}$ is reduced into $\ce{Cu^+}$ without any other change. Second, the ion $\ce{SO4^{2-}}$ is not modified at all, although you changed it into a non-existant $\ce{S^{6+}}$ ion. Third, in solution the substance $\ce{KI}$ is transformed into two ions : the ion $\ce{K^+}$ that does not react, and the ion $\ce{I^-}$ which is transformed into $\ce{I2}$ (and not $\ce{I}$) in the reaction. So start again $\endgroup$
    – Maurice
    Commented Oct 9, 2022 at 9:56

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First issue is the incorrect formula for copper(I) iodide. $\ce{Cu2I2}$ is a common moiety in dicopper(I) iodide complexes (C.N. 4) with various ligands, but the correct formula unit for copper(I) iodide is $\ce{CuI}.$ Unlike mercury(I) iodide with $\ce{Hg2I2}$ formula unit depicting $\ce{Hg2^2+}$ moiety, there are no $\ce{Cu-Cu}$ linkages in copper(I) iodide.

Second, assigning states of aggregation would help to identify spectator ions and write net ionic equation \eqref{a-rxn-r2}. For simplicity, copper(I) iodide assumed to exist as precipitate due to its low solubility, and iodine as soluble compound due to formation of triiodide ion with potassium iodide $(\ce{KI3(aq)})$:

$$ \begin{align} \ce{\overset{+2}{Cu}SO4(aq) + K\overset{-1}{I}(aq) &-> K2SO4(aq) + \overset{+1}{Cu}I(s) + \overset{0}{I}_2(aq)}\label{a-rxn-r1} \tag{R1} \\ \ce{\overset{+2}{Cu}^2+(aq) + \overset{-1}{I}^-(aq) &-> \overset{+1}{Cu}I(s) + \overset{0}{I}_2(aq)}\label{a-rxn-r2} \tag{R2} \end{align} $$

Third problem is improper half-reactions which contain non-existing chemicals in solution such as $\ce{I}$ or $\ce{S^6+}$. It also looks like you are mixing up dissociation and redox reactions, which is unnecessary at this point once you've came up with the net ionic equation. Correct half-reactions and balanced net ionic equations are as follows (states of aggregation which have already served their purpose are omitted for clarity):

$$ \begin{align} \ce{\overset{+2}{Cu}^2+ + I- + e- &-> \overset{+1}{Cu}I} &|\times 2 \tag{red} \\ \ce{2\overset{-1}{I}^- &-> \overset{0}{I}_2 + 2 e-} & \tag{ox} \\ \hline \ce{2 Cu^2+ + 4 I- &-> 2 CuI + I2} \tag{redox} \end{align} $$

To complete the full balanced reaction you need to add spectator ions — potassium and sulfate — with corresponding coefficients:

$$\ce{2 CuSO4(aq) + 4 KI(aq) -> 2 K2SO4(aq) + 2 CuI(s) + I2(aq)}\label{a-rxn-r3}\tag{R3}$$

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  • $\begingroup$ Shouldn't $\ce{KI3}$ be a separate entity in the final equation? See Jan's answer to What happens when potassium iodide is added to a copper (II) salt?. $\endgroup$
    – Nilay Ghosh
    Commented Nov 5, 2022 at 11:08
  • $\begingroup$ @NilayGhosh $\ce{KI3}$ is implied as $\ce{I2(aq)}$ for simplicity since it's not the major point of the question and it exists among the reactants and the products, too. $\endgroup$
    – andselisk ♦
    Commented Nov 5, 2022 at 13:15

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