${\rm{Cu}}\left( {{\rm{II}}} \right)$ sulphate solution is treated separately with ${\rm{KCl}}$ and KI. In which case, ${\rm{C}}{{\rm{u}}^{2 + }}$ be reduced to ${\rm{C}}{{\rm{u}}^ + }?$

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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Analyzing +4 | -1

\mathrm{Cu}\left( {\mathrm{II}} \right)

sulphate solution is treated separately with

\mathrm{KCl}

and KI. In which case,

\mathrm{C}{\mathrm{u}^{2 + }}

be reduced to

\mathrm{C}{\mathrm{u}^ + }?

View answer and explanation

Correct option: B

With KI

Option B (With KI) is correct because I- is a stronger reducing agent than Cl-, and Cu2+ is reduced to Cu+ by I-.

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\mathrm{A}{\mathrm{l}^{3 + }}