Q.Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

Concept: Standard Electrode Potentials — A reaction is feasible if the overall cell potential $E^\circ_{\text{cell}} > 0$, i.e., the species with the higher reduction potential gets reduced, and the other gets oxidised.

(i) $Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq)$; $E^\circ = +0.77\ \text{V}$

$I_2(s) + 2e^- \rightarrow 2I^-(aq)$; $E^\circ = +0.54\ \text{V}$

$Fe^{3+}$ has higher $E^\circ$, so it is reduced; $I^-$ is oxidised.

$E^\circ_{\text{cell}} = 0.77 - 0.54 = +0.23\ \text{V} > 0$ → feasible.

(ii) $Ag^+(aq) + e^- \rightarrow Ag(s)$; $E^\circ = +0.80\ \text{V}$

$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$; $E^\circ = +0.34\ \text{V}$

$Ag^+$ is reduced, $Cu$ is oxidised.

$E^\circ_{\text{cell}} = 0.80 - 0.34 = +0.46\ \text{V} > 0$ → feasible.

(iii) $Fe^{3+}/Fe^{2+} = +0.77\ \text{V}$; $Br_2(l) + 2e^- \rightarrow 2Br^-$; $E^\circ = +1.09\ \text{V}$

$Br_2$ has higher $E^\circ$, so $Br^-$ cannot reduce $Fe^{3+}$; the reverse would occur.

$E^\circ_{\text{cell}} = 0.77 - 1.09 = -0.32\ \text{V} < 0$ → not feasible. …

The key idea is simple: for any redox reaction, we need to identify which species gets reduced (gains electrons) and which gets oxidised (loses electrons). The standard electrode potential $E^\circ$ tells us the tendency of a species to get reduced — a higher (more positive) $E^\circ$ means a stronger oxidising agent. So the species with the higher $E^\circ$ will be reduced, and the one with the lower $E^\circ$ will be oxidised. The reaction is spontaneous (feasible) when the cell potential $E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} > 0$.

Let's recall the relevant standard reduction potentials from the NCERT table (Table 2.1, at 298 K):

Now let's go through each case.

1. Fe³⁺(aq) and I⁻(aq)

   Fe³⁺ can be reduced to Fe²⁺ ($E^\circ = +0.77$ V). I⁻ can be oxidised to I₂ ($E^\circ = +0.54$ V for the reverse of reduction). Since Fe³⁺ has a higher reduction potential, it acts as the oxidising agent (gets reduced), and I⁻ acts as the reducing agent (gets oxidised).

   $E^\circ_{\text{cell}} = 0.77 - 0.54 = +0.23$ V > 0.

   Feasible. The reaction is: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$.

2. Ag⁺(aq) and Cu(s)

   Ag⁺ can be reduced to Ag ($E^\circ = +0.80$ V). Cu can be oxidised to Cu²⁺ ($E^\circ = +0.34$ V for reduction of Cu²⁺). Ag⁺ has the higher reduction potential, so it gets reduced; Cu gets oxidised.

   $E^\circ_{\text{cell}} = 0.80 - 0.34 = +0.46$ V > 0.

   Feasible. Reaction: $2\text{Ag}^+ + \text{Cu} \rightarrow 2\text{Ag} + \text{Cu}^{2+}$.

3. Fe³⁺(aq) and Br⁻(aq)

   Fe³⁺ reduction: $E^\circ = +0.77$ V. Br₂/Br⁻ has $E^\circ = +1.09$ V, which is higher — meaning Br₂ is a stronger oxidising agent than Fe³⁺. So Fe³⁺ cannot oxidise Br⁻ to Br₂; if anything, the reverse happens (Br₂ oxidising Fe²⁺).

   For the reaction Fe³⁺ + Br⁻ → Fe²⁺ + ½Br₂, the cell potential is $0.77 - 1.09 = -0.32$ V < 0.

   Not feasible. …

Method: Predicting Feasibility Using Standard Electrode Potentials ($E^\circ$)

We use the Gibbs free energy relation:

A reaction is feasible (spontaneous) if $\Delta_r G^\circ < 0$, which means:

Steps

1. Identify the half-reactions — one is oxidation (anode), the other is reduction (cathode).
2. Look up standard reduction potentials ($E^\circ_{\text{red}}$) from the NCERT table.
3. Calculate cell potential:

where cathode = reduction half-cell, anode = oxidation half-cell.

4. Check sign:

• If $E^\circ_{\text{cell}} > 0$ → feasible
• If $E^\circ_{\text{cell}} < 0$ → not feasible

(i) $Fe^{3+}(aq)$ and $I^-(aq)$

• Reduction: $Fe^{3+} + e^- \rightarrow Fe^{2+}$ $E^\circ = +0.77\ \text{V}$ (cathode)
• Oxidation: $2I^- \rightarrow I_2 + 2e^-$ $E^\circ_{\text{red}}(I_2/I^-) = +0.54\ \text{V}$ → $E^\circ_{\text{anode}} = +0.54\ \text{V}$

Feasible ✓

(ii) $Ag^+(aq)$ and $Cu(s)$

• Reduction: $Ag^+ + e^- \rightarrow Ag$ $E^\circ = +0.80\ \text{V}$ (cathode)
• Oxidation: $Cu \rightarrow Cu^{2+} + 2e^-$ $E^\circ_{\text{red}}(Cu^{2+}/Cu) = +0.34\ \text{V}$ → $E^\circ_{\text{anode}} = +0.34\ \text{V}$

Feasible ✓

(iii) $Fe^{3+}(aq)$ and $Br^-(aq)$

• Reduction: $Fe^{3+} + e^- \rightarrow Fe^{2+}$ $E^\circ = +0.77\ \text{V}$ (cathode)
• Oxidation: $2Br^- \rightarrow Br_2 + 2e^-$ $E^\circ_{\text{red}}(Br_2/Br^-) = +1.09\ \text{V}$ → $E^\circ_{\text{anode}} = +1.09\ \text{V}$

Not feasible ✗

(iv) $Ag(s)$ and $Fe^{3+}(aq)$

• Reduction: $Fe^{3+} + e^- \rightarrow Fe^{2+}$ $E^\circ = +0.77\ \text{V}$ (cathode) …

Here are the common mistakes students make when predicting the feasibility of redox reactions using standard electrode potentials, along with how to avoid each.

Mistake 1: Confusing which half-reaction gets reversed

The error:

Students often reverse the wrong half-reaction when writing the overall cell reaction. They forget that the more negative (or less positive) $E^\circ$ gets reversed (oxidation), while the more positive $E^\circ$ stays as reduction.

How to avoid:

Always follow this rule:

• Higher $E^\circ$ → Reduction (remains as written)
• Lower $E^\circ$ → Oxidation (reverse the reaction)

Example (i):

$Fe^{3+}/Fe^{2+}$ has $E^\circ = +0.77\ \text{V}$

$I_2/I^-$ has $E^\circ = +0.54\ \text{V}$

Since $0.77 > 0.54$, $Fe^{3+}$ is reduced and $I^-$ is oxidised.

Correct: $2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$

Mistake 2: Forgetting to multiply $E^\circ$ when balancing electrons

The error:

Students think $E^\circ$ changes when they multiply a half-reaction by a coefficient. For example, they might double $E^\circ$ for $2I^- \rightarrow I_2 + 2e^-$.

How to avoid:

$E^\circ$ is an intensive property — it does not depend on the number of electrons. Never multiply $E^\circ$ by a coefficient.

Correct approach:

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

(using the values as given in the table, no scaling)

Mistake 3: Using $E^\circ$ of the wrong species

The error:

For a reaction like $Ag(s) + Fe^{3+}(aq)$, students might look up $E^\circ$ for $Ag^+/Ag$ instead of $Ag/Ag^+$ or confuse $Fe^{3+}/Fe^{2+}$ with $Fe^{2+}/Fe$.

How to avoid:

• Always identify which species is being reduced and which is being oxidised.
• Use the correct half-cell from the table:
  • $Fe^{3+} + e^- \rightarrow Fe^{2+}$ → $E^\circ = +0.77\ \text{V}$
  • $Ag^+ + e^- \rightarrow Ag$ → $E^\circ = +0.80\ \text{V}$
  • $Br_2 + 2e^- \rightarrow 2Br^-$ → $E^\circ = +1.09\ \text{V}$

Mistake 4: Thinking a positive $E^\circ_{\text{cell}}$ always means “fast reaction”

The error:

Students assume that if $E^\circ_{\text{cell}} > 0$, the reaction will occur instantly or at a measurable rate.

How to avoid:

$E^\circ_{\text{cell}} > 0$ only tells you the reaction is thermodynamically feasible (spontaneous). It says nothing about kinetics. Some reactions with positive $E^\circ_{\text{cell}}$ are very slow (e.g., due to high activation energy).

Example: $Fe^{3+}$ and $Br^-$ have $E^\circ_{\text{cell}} = 0.77 - 1.09 = -0.32\ \text{V}$ (not feasible). But even if it were positive, it might still be slow.

Mistake 5: Ignoring the sign convention for $E^\circ_{\text{cell}}$

The error:

Students sometimes compute $E^\circ_{\text{cell}}$ as $E^\circ_{\text{ox}} + E^\circ_{\text{red}}$ and get the wrong sign.

How to avoid:

Use the standard formula:

Where:

• Cathode = reduction half-cell (higher $E^\circ$)
• Anode = oxidation half-cell (lower $E^\circ$)

Example (ii):

$Ag^+$ reduction: $+0.80\ \text{V}$

$Cu$ oxidation: $+0.34\ \text{V}$

$E^\circ_{\text{cell}} = 0.80 - 0.34 = +0.46\ \text{V}$ → feasible

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