Question

On addition of which metal, copper sulphate solution (blue colour) will be changed to colourless solution?
A. ${\text{Fe}}$
B. ${\text{Ag}}$
C. ${\text{Zn}}$
D. ${\text{Hg}}$

Answer 1

Hint: To answer this question, you should have an idea of the standard reduction potentials of various elements. Lower is the standard reduction potential value of the element, higher is its tendency to form positive ions. When Cu will be reduced from the solution, it will turn colorless.

Complete step by step answer:
We know that lower the standard reduction potential of a metal, higher is its reactivity. In the given question we observe that copper in copper sulphate is reduced to copper. In order to reduce the copper ion $\left( {{\text{C}}{{\text{u}}^{{\text{2 + }}}}} \right)$ from its sulphate salt, the element must have a standard reduction potential lower than that of ${\text{Cu}}$.
We know that the standard reduction potential of Copper is:
${{\text{E}}^{\text{o}}}{\text{(C}}{{\text{u}}^{{\text{2 + }}}}{\text{/Cu)}} = 0.34{\text{V}}$
Also, the standard reduction potentials of the given metals are:
${{\text{E}}^{\text{o}}}{\text{(F}}{{\text{e}}^{{\text{2 + }}}}{\text{/Fe)}} = - 0.44{\text{V}}$
${{\text{E}}^{\text{o}}}{\text{(A}}{{\text{g}}^{\text{ + }}}{\text{/Ag)}} = 0.80{\text{V}}$
${{\text{E}}^{\text{o}}}{\text{(Z}}{{\text{n}}^{{\text{2 + }}}}{\text{/Zn)}} = - 0.76{\text{V}}$
${{\text{E}}^{\text{o}}}{\text{(H}}{{\text{g}}^{{\text{2 + }}}}{\text{/Hg}}_2^{2 + }{\text{)}} = 0.92{\text{V}}$
From the values of the standard reduction potentials of the metals, you can clearly see that Iron and zinc are stronger reducing agents than copper while mercury and silver are weaker reducing agents than copper. Thus, iron and zinc can displace copper from copper sulphate solution.
Iron will form ${\text{FeS}}{{\text{O}}_{\text{4}}}$ and zinc will form ${\text{ZnS}}{{\text{O}}_{\text{4}}}$
Ferrous sulphate is a coloured crystal and thus forms a green colour solution, while zinc sulphate forms a colourless solution.

Thus, the correct answer is C.

Note:
A problem that you may face during the question is determining whether Fe is oxidised to $ + 3$or $ + 2$ oxidation state. You can predict this using the various standard reduction potentials of Fe.
${{\text{E}}^{\text{o}}}{\text{(F}}{{\text{e}}^{{\text{3 + }}}}{\text{/Fe)}} = - 0.04{\text{V,}}{{\text{E}}^{\text{o}}}{\text{(F}}{{\text{e}}^{{\text{2 + }}}}{\text{/Fe)}} = - 0.44{\text{V ,}}{{\text{E}}^{\text{o}}}\left( {{\text{F}}{{\text{e}}^{{\text{3 + }}}}{\text{/F}}{{\text{e}}^{{\text{2 + }}}}} \right) = 0.77{\text{V}}$
Since the reduction potential of ferric ion to iron has a smaller value than that of cupric ion to copper, ferric ion should have been formed and ferric sulphate should be obtained as a product. But ferric ions are reduced by iron atoms into ferrous ions. Thus, the final product formed is Ferrous sulphate.