What is the correct order of increasing strength of the following ion…

The correct answer is option 2. \(Cr_2O_7^{2-} < Cr^{3+} < Cr^{2+}\).

Let us delve into the concept of reducing agents and standard reduction potentials to clarify the order of increasing strength of the given ions as reducing agents:

Standard Reduction Potentials

The standard reduction potential (\(E^\circ\)) of a half-reaction indicates the tendency of a species to gain electrons and be reduced. The more positive the reduction potential, the greater the species' affinity for electrons and the stronger it is as an oxidizing agent. Conversely, a species with a more negative reduction potential is more likely to donate electrons and act as a reducing agent.

Given Ions and Their Reduction Potentials

1. \(\text{Cr}^{3+}\) to \(\text{Cr}\)

\(\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \quad E^\circ = -0.74 \, \text{V}\)

This reduction potential indicates that \(\text{Cr}^{3+}\) is less likely to gain electrons compared to species with more positive potentials, making it a weaker reducing agent.

2. \(\text{Cr}^{2+}\) to \(\text{Cr}\)

\(\text{Cr}^{2+} + 2e^- \rightarrow \text{Cr} \quad E^\circ = -0.91 \, \text{V}\)

This more negative reduction potential indicates that \(\text{Cr}^{2+}\) is more likely to donate electrons compared to

\(\text{Cr}^{3+}\), making it a stronger reducing agent.

3. \(Cr_2O_7^{2-}\) to \(\text{Cr}^{3+}\)

\(\text{Cr}_2\text{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7H_2O \quad E^\circ = +1.33 \, \text{V}\)

This very positive reduction potential means that \(\text{Cr}_2\text{O}_7^{2-}\) strongly prefers to gain electrons, making it a very poor reducing agent and a strong oxidizing agent.

Reducing Agent Strength

A reducing agent donates electrons to another species. The strength of a reducing agent is inversely related to its reduction potential: the more negative the reduction potential, the stronger the reducing agent.

Order of Reducing Strength

Given the reduction potentials:

\(\text{Cr}_2\text{O}_7^{2-}\) with a very positive potential is a very weak reducing agent (strong oxidizing agent)

\(\text{Cr}^{3+}\) with a moderately negative potential is a moderate reducing agent.

\(\text{Cr}^{2+}\) with a more negative potential is the strongest reducing agent among the three.Thus, the order of increasing strength of reducing agents (from weakest to strongest) is:
\(\text{Cr}_2\text{O}_7^{2-} < \text{Cr}^{3+} < \text{Cr}^{2+}\)

So, the correct option is: \(Cr_2O_7^{2-} < Cr^{3+} < Cr^{2+}\).