The standard electrode potentials, `E_(I_(2)//I^(-))^(@), E_(Br^(-)//Br_(2))^(@)`and `E_(Fe//Fe^(2+))^(@)` are respectively `+ 0.54 V, -1.09 V` and `0.44 V` as the basis of given data which of following is/are spontaneous

Options

A. `Br_(2)+2I^(-) rarr 2Br^(-)+I_(2)`
B. `Fe+Br_(2)rarr Fe^(2+)+2Br^(-)`
C. `Fe+I_(2)rarr Fe^(2+)+2I^(-)`
D. `I_(2)2Br^(-)rarr 2I^(-)Br_(2)`

Answer

To determine which of the given reactions are spontaneous based on the standard electrode potentials, we will follow these steps:

### Step 1: Understand the Criteria for Spontaneity
A reaction is spontaneous if the cell potential (E_cell) is greater than 0. The relationship between Gibbs free energy change (ΔG) and cell potential is given by:
\[ \Delta G = -nFE_{cell} \]
Where:
- \( n \) = number of moles of electrons transferred
- \( F \) = Faraday's constant (approximately 96500 C/mol)
- \( E_{cell} \) = cell potential

For spontaneity, we need \( E_{cell} > 0 \).

### Step 2: Calculate E_cell for Each Reaction

#### Reaction A: \( I_2 + 2I^- \rightarrow 2I^- + 0 \)
- Reduction: \( I_2 + 2e^- \rightarrow 2I^- \) (E° = +0.54 V)
- Oxidation: \( I^- \rightarrow I_2 + 2e^- \) (E° = -0.54 V, sign changes)

Using the formula:
\[ E_{cell} = E_{cathode} - E_{anode} \]
\[ E_{cell} = 0.54 - (-1.09) = 0.54 + 1.09 = 1.63 \, V \]

**Conclusion for A:** Since \( E_{cell} = 1.63 \, V > 0 \), Reaction A is spontaneous.

#### Reaction B: \( Fe + Br_2 \rightarrow Fe^{2+} + 2Br^- \)
- Reduction: \( Br_2 + 2e^- \rightarrow 2Br^- \) (E° = -1.09 V, sign changes to +1.09 V)
- Oxidation: \( Fe \rightarrow Fe^{2+} + 2e^- \) (E° = +0.44 V)

Calculating E_cell:
\[ E_{cell} = 1.09 - 0.44 = 0.65 \, V \]

**Conclusion for B:** Since \( E_{cell} = 0.65 \, V > 0 \), Reaction B is spontaneous.

#### Reaction C: \( Fe + I_2 \rightarrow Fe^{2+} + 2I^- \)
- Reduction: \( I_2 + 2e^- \rightarrow 2I^- \) (E° = +0.54 V)
- Oxidation: \( Fe \rightarrow Fe^{2+} + 2e^- \) (E° = +0.44 V)

Calculating E_cell:
\[ E_{cell} = 0.54 - 0.44 = 0.10 \, V \]

**Conclusion for C:** Since \( E_{cell} = 0.10 \, V > 0 \), Reaction C is spontaneous.

#### Reaction D: \( I_2 + 2Br^- \rightarrow 2I^- + Br_2 \)
- Reduction: \( I_2 + 2e^- \rightarrow 2I^- \) (E° = +0.54 V)
- Oxidation: \( Br^- \rightarrow Br_2 + 2e^- \) (E° = -1.09 V, sign changes to +1.09 V)

Calculating E_cell:
\[ E_{cell} = 0.54 - (-1.09) = 0.54 + 1.09 = 1.63 \, V \]

**Conclusion for D:** Since \( E_{cell} = 1.63 \, V > 0 \), Reaction D is spontaneous.

### Final Conclusion
All reactions A, B, C, and D are spontaneous.