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:

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

we can use the relationship between Gibbs free energy change and the standard cell potential (E°) given by the equation:

\[ \Delta G° = -nFE° \]

where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96485 \, C/mol \))
- \( E° \) = standard cell potential in volts

### Step 1: Identify the half-reactions

In the given reaction, we can identify the oxidation and reduction half-reactions:

1. **Reduction half-reaction**:
\[ Ag^+ + e^- \rightarrow Ag \]
(Silver ion is reduced to silver metal)

2. **Oxidation half-reaction**:
\[ 2H^+ + 2e^- \rightarrow H_2 \]
(Hydrogen ions are reduced to hydrogen gas)

### Step 2: Determine the number of electrons transferred (n)

From the half-reactions:
- The reduction of \( 2Ag^+ \) involves 2 electrons (since each \( Ag^+ \) requires 1 electron).
- The oxidation of \( H^+ \) to \( H_2 \) also involves 2 electrons.

Thus, the total number of electrons transferred \( n = 2 \).

### Step 3: Calculate the standard cell potential (E°)

The standard reduction potential for the half-reaction \( Ag^+ + e^- \rightarrow Ag \) is given as:

\[ E°(Ag^+/Ag) = +0.80 \, V \]

For the hydrogen half-reaction, the standard reduction potential is:

\[ E°(H^+/H_2) = 0 \, V \]

To find the standard cell potential \( E° \) for the overall reaction, we use:

\[ E° = E°(cathode) - E°(anode) \]

Here, \( Ag^+ \) is the cathode (reduction) and \( H^+ \) is the anode (oxidation):

\[ E° = E°(Ag^+/Ag) - E°(H^+/H_2) = 0.80 \, V - 0 \, V = 0.80 \, V \]

### Step 4: Substitute values into the Gibbs free energy equation

Now we can substitute the values into the Gibbs free energy equation:

\[ \Delta G° = -nFE° \]

Substituting \( n = 2 \), \( F = 96485 \, C/mol \), and \( E° = 0.80 \, V \):

\[ \Delta G° = -2 \times 96485 \, C/mol \times 0.80 \, V \]

Calculating this gives:

\[ \Delta G° = -2 \times 96485 \times 0.80 = -154776 \, J/mol \]

Converting to kJ:

\[ \Delta G° = -154.776 \, kJ/mol \approx -154.8 \, kJ/mol \]

### Final Answer

The standard free energy change for the reaction is approximately:

\[ \Delta G° \approx -154.8 \, kJ/mol \]