Question

Beakers A, B and C contain zinc sulphate, silver nitrate and iron (II) sulphate solutions respectively. Copper pieces are added to each beaker. The blue colour will appear in case of

• A. beaker A
• B. beaker B
• C. beaker C
• D. all the beakers

Answer 1

The correct option is B

beaker B

The explanation for the correct option B:

• Arrangement of the metals on the basis of their reactivity towards other elements is termed as activity series. The activity series in decreasing order is:

$\mathrm{K}>\mathrm{Na}>\mathrm{Li}>\mathrm{Ba}>\mathrm{Sr}>\mathrm{Ca}>\mathrm{Mg}>\mathrm{Al}>\mathrm{Mn}>\mathrm{Zn}>\mathrm{Cr}>\mathrm{Fe}>\mathrm{Cd}>\mathrm{Co}>\mathrm{Ni}>\mathrm{Sn}>\mathrm{Pb}>\mathrm{H}>\mathrm{Sb}>\mathrm{Bi}>\mathrm{Cu}>\mathrm{Hg}>\mathrm{Ag}>\mathrm{Au}>\mathrm{Pt}$

• Elements at the starting of this series are comparatively more reactive than elements present after them in the series.
• In a single displacement reaction, an element from its compound is substituted by another element to form two product molecules. Elements that are more reactive according to the activity series can displace the elements less reactive than them from their compounds.
• Since Copper $\left(\mathrm{Cu}\right)$ is more reactive than Silver $\left(\mathrm{Ag}\right)$ in the activity series, Copper can displace Silver from Silver sulfate $\left({\mathrm{Ag}}_2{\mathrm{SO}}_4\right)$ to form Copper sulfate $\left({\mathrm{CuSO}}_4\right)$ and Silver $\left(\mathrm{Ag}\right)$.

$\mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{Ag}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)\to {\mathrm{CuSO}}_4\left(\mathrm{s}\right)+2\mathrm{Ag}\left(\mathrm{s}\right)$

• Product Copper sulfate $\left({\mathrm{CuSO}}_4\right)$ is blue in color.

The explanation for incorrect options:

Since Copper$\left(\mathrm{Cu}\right)$ is less reactive than Zinc $\left(\mathrm{Zn}\right)$, and Iron $\left(\mathrm{Fe}\right)$, Copper can not displace these elements from their solution. Hence, no reaction occurs in the case of Zinc sulfate $\left({\mathrm{ZnSO}}_4\right)$ (in beaker A) and iron(II) sulfate $\left({\mathrm{FeSO}}_4\right)$ (in beaker B).

Final Answer: Hence, option B is correct. The blue color will be observed in beaker B.