(19) Consider the following redox reaction that occurs in a voltaic cell: Cr2+ (aq) + Cu²+ (aq) → Cr³+ (aq) + Cut (aq) If the cell is operating at 25°C, what is the value of the equilibrium constant (Keq)? (A) 1.41 x 10¹1 1.19 x 1018 5.53 x 105 2.79 x 1033 1.81 x 10-6 (B) (C) (D) (E)

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### Redox Reaction in a Voltaic Cell

**Problem Statement:**

Consider the following redox reaction that occurs in a voltaic cell:
\[ \text{Cr}^{2+} (\text{aq}) + \text{Cu}^{2+} (\text{aq}) \rightarrow \text{Cr}^{3+} (\text{aq}) + \text{Cu}^{+} (\text{aq}) \]

If the cell is operating at 25°C, what is the value of the equilibrium constant (K_eq)?

**Options:**

(A) \( 1.41 \times 10^{11} \)

(B) \( 1.19 \times 10^{18} \)

(C) \( 5.53 \times 10^{5} \)

(D) \( 2.79 \times 10^{33} \)

(E) \( 1.81 \times 10^{6} \)

---

**Explanation:**

To determine the equilibrium constant (\(K_{eq}\)) for the given redox reaction at 25°C, it's essential to understand the relationship between the cell potential and the equilibrium constant. At standard conditions (25°C or 298 K):

\[ \Delta G^\circ = -RT \ln K_{eq} \]

For a voltaic cell:

\[ \Delta G^\circ = -nFE^\circ_{cell} \]

Where:
- \( \Delta G^\circ \) is the standard Gibbs free energy change.
- \( R \) is the universal gas constant (8.314 J/mol·K).
- \( T \) is the temperature in Kelvin (25°C = 298 K).
- \( n \) is the number of moles of electrons transferred in the balanced redox equation.
- \( F \) is the Faraday constant (96485 C/mol).
- \( E^\circ_{cell} \) is the standard cell potential.

By equating the equations for \( \Delta G^\circ \):

\[ -nFE^\circ_{cell} = -RT \ln K_{eq} \]

Simplifying and solving for \( K_{eq} \):

\[ \ln K_{eq} = \frac{nFE^\circ_{cell}}{RT} \]

\[ K_{eq} = e^{\left( \frac{nFE^\circ_{cell}}{RT} \right)} \]

Given these relationships, you
Transcribed Image Text:### Redox Reaction in a Voltaic Cell **Problem Statement:** Consider the following redox reaction that occurs in a voltaic cell: \[ \text{Cr}^{2+} (\text{aq}) + \text{Cu}^{2+} (\text{aq}) \rightarrow \text{Cr}^{3+} (\text{aq}) + \text{Cu}^{+} (\text{aq}) \] If the cell is operating at 25°C, what is the value of the equilibrium constant (K_eq)? **Options:** (A) \( 1.41 \times 10^{11} \) (B) \( 1.19 \times 10^{18} \) (C) \( 5.53 \times 10^{5} \) (D) \( 2.79 \times 10^{33} \) (E) \( 1.81 \times 10^{6} \) --- **Explanation:** To determine the equilibrium constant (\(K_{eq}\)) for the given redox reaction at 25°C, it's essential to understand the relationship between the cell potential and the equilibrium constant. At standard conditions (25°C or 298 K): \[ \Delta G^\circ = -RT \ln K_{eq} \] For a voltaic cell: \[ \Delta G^\circ = -nFE^\circ_{cell} \] Where: - \( \Delta G^\circ \) is the standard Gibbs free energy change. - \( R \) is the universal gas constant (8.314 J/mol·K). - \( T \) is the temperature in Kelvin (25°C = 298 K). - \( n \) is the number of moles of electrons transferred in the balanced redox equation. - \( F \) is the Faraday constant (96485 C/mol). - \( E^\circ_{cell} \) is the standard cell potential. By equating the equations for \( \Delta G^\circ \): \[ -nFE^\circ_{cell} = -RT \ln K_{eq} \] Simplifying and solving for \( K_{eq} \): \[ \ln K_{eq} = \frac{nFE^\circ_{cell}}{RT} \] \[ K_{eq} = e^{\left( \frac{nFE^\circ_{cell}}{RT} \right)} \] Given these relationships, you
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