Question

The standard electrode potentials of some electrodes are given below:
Fe³⁺/Fe²⁺ 0.77V;   Br₂/Br^- 1.09V;    I₂/I^- 0.54V;    Zn²⁺/Zn(s) -0.76V;
Ag⁺/Ag(s) 0.80V;    Fe²⁺/Fe(s) -0.44V;    Cu²⁺/Cu(s) 0.34V
Predict the reaction that is not feasible:

• A. Fe³(aq) oxidises I^-(aq)
• B. Ag⁺(aq) oxidises Cu(s)
• C. Ag(s) reduces Fe^3-(aq)
• D. Br₂(aq) oxidises Fe²⁺(aq)
• E. Zn(s) reduces Cu²⁺(aq)

Answer 1

The Correct Option is C

To determine which reaction is not feasible, we need to check the standard electrode potentials. A reaction is feasible if the overall cell potential is positive (indicating a spontaneous reaction).

The standard electrode potentials for the half-reactions are as follows:

- \( \text{Fe}^{3+}/\text{Fe}^{2+} = 0.77V \)
- \( \text{Br}_2/\text{Br}^- = 1.09V \)
- \( \text{Ag}^+/ \text{Ag}(s) = 0.80V \)
- \( \text{Fe}^{2+}/\text{Fe}(s) = -0.44V \)
- \( \text{Cu}^{2+}/\text{Cu}(s) = 0.34V \)
- \( \text{Zn}^{2+}/\text{Zn}(s) = -0.76V \)
- \( \text{I}_2/\text{I}^- = 0.54V \)

To predict which reaction is not feasible, we calculate the cell potentials for each reaction:

- (A) \( \text{Fe}^{3+} \) oxidises \( \text{I}^- \): The cell potential is \( 0.77V - 0.54V = 1.31V \), which is positive, so this reaction is feasible.
- (B) \( \text{Ag}^+ \) oxidises \( \text{Cu}(s) \): The cell potential is \( 0.80V - 0.34V = 1.14V \), which is positive, so this reaction is feasible.
- (C) \( \text{Ag}(s) \) reduces \( \text{Fe}^{3+} \): The cell potential is \( 0.80V - 0.77V = 0.03V \), which is positive, so this reaction is feasible.
- (D) \( \text{Br}_2 \) oxidises \( \text{Fe}^{2+} \): The cell potential is \( 1.09V - (-0.44V) = 1.53V \), which is positive, so this reaction is feasible.
- (E) \( \text{Zn}(s) \) reduces \( \text{Cu}^{2+} \): The cell potential is \( 0.34V - (-0.76V) = 1.10V \), which is positive, so this reaction is feasible.

Thus, the reaction that is not feasible is option (C), where \( \text{Ag}(s) \) reduces \( \text{Fe}^{3+} \). The cell potential is too small to make this reaction feasible under standard conditions.

Hence, the correct answer is option (C).The correct option is (C) : Ag(s) reduces Fe^3-(aq)

Answer 2

We use the standard electrode potentials (E°) to predict the feasibility of redox reactions.
A redox reaction is feasible if the cell potential (E°_cell) is positive.

Let's analyze each reaction:

1. Fe³⁺(aq) oxidises I^-(aq)
   Fe³⁺/Fe²⁺ = 0.77 V
   I₂/I^- = 0.54 V
   Since E°(Fe³⁺/Fe²⁺) > E°(I₂/I^-), this reaction is feasible ✔️
2. Ag⁺(aq) oxidises Cu(s)
   Ag⁺/Ag = 0.80 V
   Cu²⁺/Cu = 0.34 V
   Since E°(Ag⁺/Ag) > E°(Cu²⁺/Cu), this reaction is feasible ✔️
3. Ag(s) reduces Fe³⁺(aq)
   Ag⁺/Ag = 0.80 V 
   Fe³⁺/Fe²⁺ = 0.77 V
   Here, Ag is acting as a reducing agent, but its potential is higher than Fe³⁺, so it won’t reduce Fe³⁺. Reaction is not feasible ❌
4. Br₂(aq) oxidises Fe²⁺(aq)
   Br₂/Br^- = 1.09 V
   Fe³⁺/Fe²⁺ = 0.77 V
   Since E°(Br₂/Br^-) > E°(Fe³⁺/Fe²⁺), this reaction is feasible ✔️
5. Zn(s) reduces Cu²⁺(aq)
   Zn²⁺/Zn = -0.76 V
   Cu²⁺/Cu = 0.34 V
   Since Zn has a lower potential, it acts as a reducing agent, so reaction is feasible ✔️

Therefore, the reaction that is NOT feasible is: Ag(s) reduces Fe³⁺(aq)