Q.Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

The key idea is that a substance can oxidise $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ if its standard reduction potential is greater than $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\ \text{V}$.

1. The half-reaction for oxidation is: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (reverse of the reduction).
2. Any oxidising agent with $E^\circ > +0.77\ \text{V}$ will spontaneously accept electrons from $\text{Fe}^{2+}$.
3. From the standard potential series, three common examples are:
   • $\text{Cl}_2$: $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$, $E^\circ = +1.36\ \text{V}$
   • $\text{Br}_2$: $\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-$, $E^\circ = +1.09\ \text{V}$ …

Why Standard Electrode Potentials Decide the Reaction

Oxidation is the loss of electrons. For $\ce{Fe^{2+}}$ to be oxidised to $\ce{Fe^{3+}}$, it must give away an electron. That electron must be accepted by some other species — the oxidising agent. The tendency of a species to accept electrons is measured by its standard reduction potential ($E^\circ$). The more positive the $E^\circ$, the stronger the oxidising agent.

The half-reaction for ferrous oxidation is:

This value tells us that $\ce{Fe^{3+}}$ has a moderate tendency to get reduced back to $\ce{Fe^{2+}}$. To push the reaction in the opposite direction — to force $\ce{Fe^{2+}}$ to lose an electron — we need an oxidising agent that is even more eager to gain electrons. In other words, the oxidising agent's reduction potential must be greater than $+0.77\ \text{V}$.

Step-by-step selection

1. Recall the standard reduction potentials of common oxidising agents. From the standard table (at $25^\circ\text{C}$, $1\ \text{atm}$, $1\ \text{M}$), we have:

   • $\ce{F2 + 2e- -> 2F-}$: $E^\circ = +2.87\ \text{V}$
   • $\ce{O3 + 2H+ + 2e- -> O2 + H2O}$: $E^\circ = +2.07\ \text{V}$
   • $\ce{Cl2 + 2e- -> 2Cl-}$: $E^\circ = +1.36\ \text{V}$
   • $\ce{MnO4- + 8H+ + 5e- -> Mn^{2+} + 4H2O}$: $E^\circ = +1.51\ \text{V}$
   • $\ce{Cr2O7^{2-} + 14H+ + 6e- -> 2Cr^{3+} + 7H2O}$: $E^\circ = +1.33\ \text{V}$
   • $\ce{Br2 + 2e- -> 2Br-}$: $E^\circ = +1.09\ \text{V}$
   • $\ce{I2 + 2e- -> 2I-}$: $E^\circ = +0.54\ \text{V}$

2. Compare each with $+0.77\ \text{V}$. Any species with $E^\circ > 0.77\ \text{V}$ can, in principle, oxidise $\ce{Fe^{2+}}$. Those with $E^\circ < 0.77\ \text{V}$ (like $\ce{I2}$) cannot.

3. Select three distinct substances that are commonly available and clearly satisfy the condition. Good choices are:

   • Fluorine gas ($\ce{F2}$): $E^\circ = +2.87\ \text{V}$ — the strongest oxidising agent known.
   • Chlorine gas ($\ce{Cl2}$): $E^\circ = +1.36\ \text{V}$ — a classic laboratory oxidiser.
   • Ozone ($\ce{O3}$): $E^\circ = +2.07\ \text{V}$ — a powerful oxidant used in water treatment.

   Other valid options include $\ce{KMnO4}$ (acidified), $\ce{K2Cr2O7}$ (acidified), or $\ce{Br2}$, but the question asks for three substances, and the above are unambiguous. …

Method: Using Standard Electrode Potentials to Predict Redox Feasibility

Method name: The $E^\circ_{\text{cell}}$ Sign Rule (or Spontaneity Rule)

Concept behind the method

A substance can oxidise ferrous ions ($\text{Fe}^{2+}$) if it can accept electrons from $\text{Fe}^{2+}$, converting $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$.

The half-reaction for oxidation of ferrous ions is:

For a reaction to be spontaneous (feasible), the overall cell potential must be positive:

Here, the anode (oxidation) is $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ with $E^\circ = +0.77\ \text{V}$.

The cathode (reduction) is the substance that will accept electrons — its $E^\circ$ must be greater than $+0.77\ \text{V}$.

Steps

1. Identify the half-reaction for the substance to be oxidised

   • Here: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$, $E^\circ = +0.77\ \text{V}$

2. Recall the condition for a substance to act as an oxidising agent

   • The oxidising agent (cathode) must have a higher reduction potential than the substance being oxidised (anode).

3. Consult the standard electrode potential table

   • Look for half-reactions with $E^\circ > +0.77\ \text{V}$.

4. Select three substances whose reduction potentials are greater than $+0.77\ \text{V}$.

Three substances that can oxidise $\text{Fe}^{2+}$

| Substance | Reduction half-reaction | $E^\circ$ (V) | …

Here are the common mistakes students make when tackling this exact type of question from Standard Electrode Potentials, along with clear strategies to avoid each.

Mistake 1: Confusing “Oxidise” with “Reduce”

• The error: Students think any substance with a higher reduction potential than $\text{Fe}^{2+}/\text{Fe}$ will oxidise $\text{Fe}^{2+}$. They often pick metals like Zinc ($E^\circ = -0.76\ \text{V}$) because “it’s more reactive.”
• Why it’s wrong: To oxidise $\text{Fe}^{2+}$ (turn it into $\text{Fe}^{3+}$), the oxidising agent must itself be reduced. This means the oxidising agent must have a higher reduction potential than the $\text{Fe}^{3+}/\text{Fe}^{2+}$ couple ($+0.77\ \text{V}$). A lower potential means the substance is a reducing agent, not an oxidising agent.
• How to avoid: Always write the half-reaction for the species being oxidised:

Then, the oxidising agent must have a more positive $E^\circ$ than $+0.77\ \text{V}$.

Mistake 2: Forgetting the Correct Half-Cell for Iron

• The error: Using the $\text{Fe}^{2+}/\text{Fe}$ couple ($-0.44\ \text{V}$) instead of the $\text{Fe}^{3+}/\text{Fe}^{2+}$ couple ($+0.77\ \text{V}$).
• Why it’s wrong: The question asks to oxidise ferrous ions ($\text{Fe}^{2+}$) to ferric ions ($\text{Fe}^{3+}$). The relevant half-reaction is:

The $\text{Fe}^{2+}/\text{Fe}$ couple is for reduction to metallic iron, which is a different process.

• How to avoid: Memorise the key couples for iron:
  • $\text{Fe}^{3+}/\text{Fe}^{2+}$: $+0.77\ \text{V}$ (for oxidation of $\text{Fe}^{2+}$)
  • $\text{Fe}^{2+}/\text{Fe}$: $-0.44\ \text{V}$ (for reduction to metal)

Mistake 3: Picking Substances with $E^\circ$ Exactly Equal to $+0.77\ \text{V}$

• The error: Choosing $\text{I}_2/\text{I}^-$ ($+0.54\ \text{V}$) or $\text{Cu}^{2+}/\text{Cu}$ ($+0.34\ \text{V}$) because they are “close” to $+0.77\ \text{V}$.
• Why it’s wrong: For a spontaneous reaction, the oxidising agent must have $E^\circ > +0.77\ \text{V}$. If $E^\circ$ is lower, the reaction is non-spontaneous under standard conditions. The reaction will not proceed.
• How to avoid: Use the spontaneity rule:

For a spontaneous reaction, $E^\circ_{\text{cell}} > 0$. Here:

So you need $E^\circ_{\text{oxidising agent}} > 0.77\ \text{V}$.

Mistake 4: Not Checking the Table Carefully

• The error: Listing substances like $\text{Cl}_2$ ($+1.36\ \text{V}$) but forgetting that $\text{Cl}_2$ is a gas and may not be “suitable” under all conditions (e.g., in solution, it can also oxidise water).
• Why it’s wrong: The question says “under suitable conditions.” While $\text{Cl}_2$ works in principle, students often miss that $\text{MnO}_4^-$ ($+1.51\ \text{V}$) or $\text{Cr}_2\text{O}_7^{2-}$ ($+1.33\ \text{V}$) are more common lab oxidising agents for $\text{Fe}^{2+}$.
• How to avoid: List three distinct substances from the table with $E^\circ > +0.77\ \text{V}$. Good choices:
  • $\text{Cl}_2$ ($+1.36\ \text{V}$)
  • $\text{MnO}_4^-$ in acid ($+1.51\ \text{V}$)
  • $\text{Cr}_2\text{O}_7^{2-}$ in acid ($+1.33\ \text{V}$) …