Question

Arrange Ag, Cr, and Hg metals in the increasing order of reducing power.

Given:

$$\begin{gathered} {{\mathrm{E}}^0}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}=+0.80\mathrm{V} \\ {{\mathrm{E}}^0}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}=-0.74\mathrm{V} \\ {{\mathrm{E}}^0}_{{\mathrm{Hg}}^{2+}/\mathrm{Hg}}=+0.79\mathrm{V} \end{gathered}$$

• A. $\mathrm{Cr}>\mathrm{Hg}>\mathrm{Ag}$
• B. $\mathrm{Hg}>\mathrm{Ag}>\mathrm{Cr}$
• C. $\mathrm{Ag}>\mathrm{Cr}>\mathrm{Hg}$
• D. $\mathrm{Ag}>\mathrm{Hg}>\mathrm{Cr}$

Answer 1

The correct option is A

$\mathrm{Cr}>\mathrm{Hg}>\mathrm{Ag}$

Explanation for correct option:

A. $\mathrm{Cr}>\mathrm{Hg}>\mathrm{Ag}$

Electrochemical series:

• “It is a list that describes the arrangement of elements in order of their increasing electrode potential values.”
• In electrochemical series as the reduction potential increases, the tendency to get reduced increases, and hence oxidizing power will increase.
• The values of reduction potential are given below:

$$\begin{gathered} {{\mathrm{E}}^0}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}=+0.80\mathrm{V} \\ {{\mathrm{E}}^0}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}=-0.74\mathrm{V} \\ {{\mathrm{E}}^0}_{{\mathrm{Hg}}^{2+}/\mathrm{Hg}}=+0.79\mathrm{V} \end{gathered}$$

• The higher the negative ${\mathrm{E}}^0$ value, the more the position of equilibrium will lie to the left, that is the more readily the metal will lose its electrons.
• Hence, the more negative the value of redox potential, the stronger the reducing agent, and the metal will have high reducing power.
• Chromium has the most negative ${\mathrm{E}}^0$ value.
• The trend for increasing order of reducing power will be:

$\mathrm{Cr}>\mathrm{Hg}>\mathrm{Ag}$

Explanation for incorrect options:

As the order of increasing reducing power is $\mathrm{Cr}>\mathrm{Hg}>\mathrm{Ag}$. Thus, options B ,C, and D are incorrect.

Hence, the correct option is A i.e. $\mathrm{Cr}>\mathrm{Hg}>\mathrm{Ag}$.