Cu (III) sulphate solution is treated separately with KCl and KI. In which case, Cu $^{2 +}$ be reduced to Cu $^{+}$ ?

With KCl

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With KI

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With both (a) and (b)

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None of these

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Solution

The correct option is B With KI
$E^{o} {C u}^{2 +} / C u = 0.34 V E^{o} I_{2} / I^{-} = 0.53 V E^{o} {C l}_{2} / {C l}^{-} = 1.38 V$
Since reduction potential of $E^{o} I_{2} / I^{-}$ and $E^{o} {C l}_{2} / {C l}^{-}$ is greater than $E^{o} {C u}^{2 +} / C u$ . So oxidation of $C u ⟶ {C u}^{2 +}$ will occur. So answer is none of these.

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Q. A solution of

C u (I I)

sulphate is reacted with

K C l

and

K I

. In which case will the

C u^{2 +}

be reduced to

C u^{+}

250 ml of 1M $I_{2}$ is separately reacted with 0.5 M $C u_{2} S$ solution, 0.5 M CuS solution and 0.5 M $S_{2} O_{3}^{2 -}$ solution, causing production of $C u^{+ 2}$ and $S O_{4}^{2 -}$ in the first two and $S_{4} O_{6}^{2 -}$ in the last case along with iodide ions. Which of the following option(s) is/are correct assuming 100% completion of the reaction?