Question

Cu (III) sulphate solution is treated separately with $KCl$ and $KI$. In which case $C{{u}^{2+}}$be reduced to $C{{u}^{+}}$?
A. With $KCl$
B. With $KI$
C. With both (A) and (B)
D. None of these

Answer 1

Hint: If you have a little idea about the Electrode reduction potential ${{E}^{\circ }}$of $C{{u}^{2+}}/Cu,{{I}_{2}}/{{I}^{-}},C{{l}_{2}}/C{{l}^{-}}$, then you can easily solve the problem. The element which has the higher reduction electrode potential will replace the element with lower potential from its salt solution.

Complete step-by-step answer:
Let us first understand what electrode potential is. The tendency of an electrode to lose or gain electrons is called electrode potential. In the given question Copper sulphate (II) is being converted to copper (I), it means that it is losing one electron, therefore there should be a species which would accept this electron also, so that the solution remains neutral. That means the reaction should involve gain and loss of electrons and we call such reactions as redox reactions.

We know that a redox reaction is one which involves oxidation and reduction of a species simultaneously. And we should remember the fact that., A redox reaction is feasible only if the species having higher reduction potential is gaining electrons and the species to release electrons must have lower reduction potential.

Now, we know that reduction potential are as follows:
$
E_{C{{u}^{2+}}/Cu}^{\circ }=0.153 \\
E_{C{{l}_{2}}/C{{l}^{-}}}^{\circ }=1.36 \\
E_{{{I}_{2}}/{{I}^{-}}}^{\circ }=0.54 \\
$

Now, if we take the copper (II) and chlorine. In the question, it is given to us that copper is losing one electron, so for this reaction to occur chlorine should accept the electron. But we see that Reduction potential of chlorine is greater than copper. Also, we have studies in the above paragraph that species with high reduction potential should lose electrons.

Therefore, as per this rule chlorine instead of accepting should lose the electrons. This cannot be possible as both species that are copper and chlorine cannot lose electrons. This reaction is wrong.
Similarly. Reduction potential of iodine is greater than copper (II). Therefore, copper also cannot gain electrons and will have the tendency to only lose electrons. Hence, this reaction will also not be spontaneous.

So, we see that neither iodide ion nor the chlorine ion can replace Copper (II) to copper (I). or in other words we can say that neither $KCl$ nor $KI$ will reduce the $C{{u}^{2+}}\to C{{u}^{+}}$.

Hence, the correct answer is option ‘D’.

Note: It becomes very tedious to learn the electrode potential. So, one can go through the above method if the potential is mentioned in the question. I do not, then just remember the electrochemical series. In electrochemical series (starting from fluorine) the potential is arranged in descending form. So, you can just get the idea that if chlorine and iodine lie above copper, then they can only release the electrons. The series should be remembered in sequence.