`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)toCr^(+ + +)+7H_(2)O,E^(@)=1.33V,3xx[2I^(-)toI_(2)+2e^(-)],E^(@)=-0.54V`
Find out the value of the equilibrium constant and Gibbs free energy change in the reaction given above.

Explanation

To solve the problem, we need to find the equilibrium constant (K) and the Gibbs free energy change (ΔG) for the given redox reaction. Let's break down the solution step by step.

### Step 1: Identify the half-reactions and their standard electrode potentials
We have two half-reactions:
1. Reduction of dichromate ion:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}, \quad E^\circ = 1.33 \, \text{V} \]
2. Oxidation of iodide ion:
\[ 2\text{I}^- \rightarrow \text{I}_2 + 2e^-, \quad E^\circ = -0.54 \, \text{V} \]

### Step 2: Convert oxidation potential to reduction potential
To find the standard reduction potential for the iodide half-reaction, we take the negative of the given oxidation potential:
\[ E^\circ_{\text{I}_2/\text{I}^-} = -(-0.54) = 0.54 \, \text{V} \]

### Step 3: Calculate the standard cell potential (E°cell)
The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, the cathode is the reduction half-reaction (dichromate) and the anode is the oxidation half-reaction (iodide):
\[ E^\circ_{\text{cell}} = 1.33 \, \text{V} - 0.54 \, \text{V} = 0.79 \, \text{V} \]

### Step 4: Calculate the Gibbs free energy change (ΔG)
The Gibbs free energy change can be calculated using the formula:
\[ \Delta G = -nFE^\circ_{\text{cell}} \]
Where:
- \( n \) = number of moles of electrons transferred (6 from the reduction half-reaction)
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))
- \( E^\circ_{\text{cell}} = 0.79 \, \text{V} \)

Substituting the values:
\[ \Delta G = -6 \times 96500 \, \text{C/mol} \times 0.79 \, \text{V} \]
Calculating this gives:
\[ \Delta G = -457410 \, \text{J} \quad \text{or} \quad -457.41 \, \text{kJ} \]

### Step 5: Calculate the equilibrium constant (K)
Using the relationship between ΔG and K:
\[ \Delta G = -RT \ln K \]
We can rearrange this to solve for K:
\[ K = e^{-\Delta G / RT} \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \)

Substituting the values:
\[ K = e^{-(-457410) / (8.314 \times 298)} \]
Calculating the exponent:
\[ K = e^{\frac{457410}{2477.572}} \approx e^{184.47} \]
Calculating \( K \):
\[ K \approx 1.511 \times 10^{80} \]

### Final Results
- Gibbs free energy change (ΔG): \( -457.41 \, \text{kJ} \)
- Equilibrium constant (K): \( 1.511 \times 10^{80} \)