`Ag(s)+Fe^(3+)(aq) to Ag^(+) (aq)+ Fe^(2+)(aq)`
Given standard electrode potentials -
` E_(Fe^(3+)//Fe^(2+))^(@)=+0.77V ` and `E_(Ag^(+)//Ag(s)))^(@)=+0.80V`
If the reaction is feasible , enter 1.00 as answer elewise enter 0.00.

Answer

To determine whether the reaction \( \text{Ag}(s) + \text{Fe}^{3+}(aq) \rightarrow \text{Ag}^+(aq) + \text{Fe}^{2+}(aq) \) is feasible, we will follow these steps:

### Step 1: Identify Oxidation and Reduction
- **Oxidation**: Silver (Ag) is oxidized from 0 to +1. Thus, \( \text{Ag}(s) \rightarrow \text{Ag}^+(aq) + e^- \).
- **Reduction**: Iron (Fe) is reduced from +3 to +2. Thus, \( \text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \).

### Step 2: Write the Standard Electrode Potentials
- Given:
- \( E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77 \, \text{V} \)
- \( E^\circ(\text{Ag}^+/Ag) = +0.80 \, \text{V} \)

### Step 3: Determine the Anode and Cathode
- **Anode**: The electrode where oxidation occurs (Ag).
- **Cathode**: The electrode where reduction occurs (Fe).

### Step 4: Calculate the Standard Cell Potential \( E^\circ_{\text{cell}} \)
- The formula for the standard cell potential is:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\circ(\text{Ag}^+/Ag) = 0.77 \, \text{V} - 0.80 \, \text{V} = -0.03 \, \text{V} \]

### Step 5: Calculate \( \Delta G \)
- The relationship between Gibbs free energy and cell potential is given by:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
where \( n \) is the number of moles of electrons transferred (1 in this case), and \( F \) is Faraday's constant (\( 96485 \, \text{C/mol} \)).
- Substituting the values:
\[ \Delta G^\circ = -1 \times 96485 \, \text{C/mol} \times (-0.03 \, \text{V}) = 2894.55 \, \text{J/mol} \approx 2.89 \, \text{kJ/mol} \]

### Step 6: Determine Feasibility
- Since \( \Delta G^\circ \) is positive, the reaction is not feasible.

### Final Answer
- The answer to whether the reaction is feasible is **0.00**.

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