Q.Use the data given in Q.8 ($E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V$; $E^\circ_{Cl_2/Cl^-} = 1.36\ V$; $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51\ V$; $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$) and find out which of the following is the strongest oxidising agent.

Concept: Standard Electrode Potentials — the more positive the $E^\circ$ value, the stronger the oxidising agent (greater tendency to get reduced).

Reasoning:

1. Oxidising agents are themselves reduced. Their strength is directly given by their reduction potential.
2. From the data:
   • $E^\circ_{Cl_2/Cl^-} = 1.36\ \text{V}$ (oxidiser: $Cl_2$, not $Cl^-$)
   • $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51\ \text{V}$
   • $E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ \text{V}$
   • $E^\circ_{Cr^{3+}/Cr} = -0.74\ \text{V}$ (oxidiser: $Cr^{3+}$ is very weak) …

The question asks us to identify the strongest oxidising agent from a list of four species. An oxidising agent is a substance that gains electrons (gets reduced) in a redox reaction. The tendency of a species to get reduced is measured by its standard reduction potential ($E^\circ$). The more positive the $E^\circ$ value, the stronger the oxidising agent.

We are given the standard reduction potentials for four half-reactions. But note: the options include both the oxidised and reduced forms of some couples. Let’s map each option to the relevant half-reaction and its $E^\circ$ value.

1. Option (i): $Cl^-$ This is the reduced form of the $Cl_2/Cl^-$ couple. The half-reaction is:

$Cl^-$ itself is already reduced — it would need to lose electrons to act as an oxidising agent, which is not favourable. So $Cl^-$ is a very weak oxidising agent (in fact, it is a reducing agent). Its oxidising strength is negligible here.

2. Option (ii): $Mn^{2+}$ This is the reduced form of the $MnO_4^-/Mn^{2+}$ couple. The half-reaction is:

$Mn^{2+}$ is the product after reduction — to act as an oxidising agent, it would need to be further reduced (to Mn metal or lower oxidation state), but that is not the relevant couple here. The $Mn^{2+}$ ion is a very poor oxidising agent; its standard reduction potential (for $Mn^{2+}/Mn$) is $-1.18\ V$, which is highly negative. So $Mn^{2+}$ is a weak oxidiser.

3. Option (iii): $MnO_4^-$

   This is the oxidised form of the $MnO_4^-/Mn^{2+}$ couple. Its reduction potential is $+1.51\ V$, which is the highest among all the given $E^\circ$ values. A high positive $E^\circ$ means it has a strong tendency to gain electrons and get reduced. Therefore, $MnO_4^-$ is a very strong oxidising agent.

4. Option (iv): $Cr^{3+}$

   This is the oxidised form of the $Cr^{3+}/Cr$ couple:

   $Cr^{3+} + 3e^- \rightarrow Cr \quad E^\circ = -0.74\ V$ …

Method: Comparison of Standard Electrode Potentials ($E^\circ$)

This method uses the principle that higher the standard reduction potential ($E^\circ$), stronger the oxidising agent.

Steps

1. Identify the oxidising agents from the options — these are the species that get reduced (gain electrons).

   • (i) $Cl^-$ → already reduced form, not an oxidising agent.
   • (ii) $Mn^{2+}$ → already reduced form, not an oxidising agent.
   • (iii) $MnO_4^-$ → can be reduced to $Mn^{2+}$ ($E^\circ = 1.51\ V$).
   • (iv) $Cr^{3+}$ → can be reduced to $Cr$ ($E^\circ = -0.74\ V$).

2. List the $E^\circ$ values for the reduction half-reactions of the possible oxidising agents: …

Common Mistakes Students Make on Standard Electrode Potential Questions

Mistake 1: Confusing Oxidising Agent with Reducing Agent

The error: Students often pick the species with the most negative $E^\circ$ value, thinking "more negative = more reactive." This is wrong for oxidising agents.

Why it happens: You may be mixing up the trend for reducing agents (which are stronger when $E^\circ$ is more negative) with oxidising agents.

How to avoid: Remember this rule:

• Strongest oxidising agent → highest (most positive) $E^\circ$ value
• Strongest reducing agent → lowest (most negative) $E^\circ$ value

Key insight: An oxidising agent gains electrons (gets reduced). A higher $E^\circ$ means a greater tendency to gain electrons.

Mistake 2: Picking the Reduced Form Instead of the Oxidised Form

The error: Given the half-reaction $MnO_4^-/Mn^{2+}$, students sometimes pick $Mn^{2+}$ as the oxidising agent because it appears in the data.

Why it happens: You may look at the list of options and see $Mn^{2+}$ and $MnO_4^-$ — and forget that the oxidised form (the one on the left of the slash) is the oxidising agent.

How to avoid: For any half-cell written as:

The oxidised form is the oxidising agent. In this question:

• $Cr_2O_7^{2-}$ (not $Cr^{3+}$) is the oxidising agent in that couple
• $MnO_4^-$ (not $Mn^{2+}$) is the oxidising agent
• $Cl_2$ (not $Cl^-$) is the oxidising agent

So from the options:

• (i) $Cl^-$ → reduced form → not an oxidising agent
• (ii) $Mn^{2+}$ → reduced form → not an oxidising agent
• (iii) $MnO_4^-$ → oxidised form → candidate
• (iv) $Cr^{3+}$ → reduced form (in $Cr_2O_7^{2-}/Cr^{3+}$) → not an oxidising agent

Mistake 3: Comparing All Given $E^\circ$ Values Without Checking Relevance

The error: Students compare $E^\circ$ of $Cr^{3+}/Cr = -0.74\ V$ with others and get confused.

Why it happens: You see four numbers and try to rank all of them, including the $Cr^{3+}/Cr$ couple which is irrelevant to the oxidising agents in the options.

How to avoid: Only compare $E^\circ$ values for the oxidised forms that appear in the options. Here: …