Calculate standard free energy change for the reaction `2Ag+2H^(+)rarrH_(2)+2Ag^(+)` Given : `E_(Ag^(+)//Ag)^(@)=+0.80V`

Options

A. 308.08 kJ
B. 154.4 kJ
C. 77.2 kJ
D. `-154.4 kJ`

Answer

To calculate the standard free energy change (ΔG°) for the reaction:

\[ 2Ag^+ + 2H^+ \rightarrow H_2 + 2Ag \]

Given the standard reduction potential for the half-reaction:

\[ E^\circ_{Ag^+/Ag} = +0.80 \, V \]

we can follow these steps:

### Step 1: Identify the half-reactions
The overall reaction can be broken down into two half-reactions:

1. **Reduction half-reaction (at the cathode)**:
\[ 2H^+ + 2e^- \rightarrow H_2 \]

2. **Oxidation half-reaction (at the anode)**:
\[ 2Ag \rightarrow 2Ag^+ + 2e^- \]

### Step 2: Determine the standard potentials for the half-reactions
- The reduction potential for the hydrogen half-reaction is:
\[ E^\circ_{H^+/H_2} = 0 \, V \]
- The oxidation potential for the silver half-reaction is the negative of the reduction potential:
\[ E^\circ_{Ag/Ag^+} = -0.80 \, V \]

### Step 3: Calculate the standard cell potential (E°cell)
The standard cell potential is calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Substituting the values:
\[ E^\circ_{cell} = 0 \, V - (-0.80 \, V) = 0.80 \, V \]

### Step 4: Calculate the number of electrons transferred (n)
In this reaction, 2 moles of electrons are transferred (as seen from the half-reactions).

### Step 5: Use the formula for standard free energy change
The relationship between standard free energy change and cell potential is given by:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]

Where:
- \( n = 2 \) (number of moles of electrons)
- \( F = 96500 \, C/mol \) (Faraday's constant)
- \( E^\circ_{cell} = 0.80 \, V \)

### Step 6: Substitute the values into the equation
\[ \Delta G^\circ = -2 \times 96500 \, C/mol \times 0.80 \, V \]

### Step 7: Perform the calculation
Calculating the above expression:
\[ \Delta G^\circ = -2 \times 96500 \times 0.80 = -154400 \, J \]

### Step 8: Convert to kilojoules
To convert joules to kilojoules:
\[ \Delta G^\circ = -154400 \, J \div 1000 = -154.4 \, kJ \]

### Final Answer
Thus, the standard free energy change for the reaction is:
\[ \Delta G^\circ = -154.4 \, kJ \]

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