QuestionJuly 2, 2025

Given: Cr_(2)O_(7)^2-(aq)+6Fe^2+(aq)+14H^+(aq)arrow 2Cr^3+(aq)+6Fe^3+(aq)+7H_(2)O(l) for this reaction varepsilon _(cell)^circ =0.51V Calculate the equilibrium constant for this reaction at standard conditions. Express your answer in scientific notations using 5.12^ast 10^wedge -6 format to TWO decimal places.

Given: Cr_(2)O_(7)^2-(aq)+6Fe^2+(aq)+14H^+(aq)arrow 2Cr^3+(aq)+6Fe^3+(aq)+7H_(2)O(l) for this reaction varepsilon _(cell)^circ =0.51V Calculate the equilibrium constant for this reaction at standard conditions. Express your answer in scientific notations using 5.12^ast 10^wedge -6 format to TWO decimal places.
Given:
Cr_(2)O_(7)^2-(aq)+6Fe^2+(aq)+14H^+(aq)arrow 2Cr^3+(aq)+6Fe^3+(aq)+7H_(2)O(l) for this reaction varepsilon _(cell)^circ =0.51V
Calculate the equilibrium constant for this reaction at standard conditions.
Express your answer in scientific notations using 5.12^ast 10^wedge -6 format to TWO decimal places.

Solution
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Answer

5.12 \times 10^{51} Explanation 1. Use the Nernst Equation The relationship between the standard cell potential and the equilibrium constant is given by **\varepsilon_{cell}^{\circ} = \frac{RT}{nF} \ln K**. At standard conditions (298 K), this simplifies to **\varepsilon_{cell}^{\circ} = \frac{0.0257}{n} \ln K**. 2. Determine the Number of Electrons Transferred In the balanced reaction, 6 electrons are transferred from Fe^{2+} to Cr_{2}O_{7}^{2-}. 3. Solve for the Equilibrium Constant K Rearrange the equation to solve for K: \[ \ln K = \frac{n \cdot \varepsilon_{cell}^{\circ}}{0.0257} \] Substitute the values: \[ \ln K = \frac{6 \times 0.51}{0.0257} \approx 119.07 \] Convert \ln K to K: \[ K = e^{119.07} \approx 5.12 \times 10^{51} \]

Explanation

1. Use the Nernst Equation<br /> The relationship between the standard cell potential and the equilibrium constant is given by **$\varepsilon_{cell}^{\circ} = \frac{RT}{nF} \ln K$**. At standard conditions (298 K), this simplifies to **$\varepsilon_{cell}^{\circ} = \frac{0.0257}{n} \ln K$**.<br />2. Determine the Number of Electrons Transferred<br /> In the balanced reaction, 6 electrons are transferred from $Fe^{2+}$ to $Cr_{2}O_{7}^{2-}$.<br />3. Solve for the Equilibrium Constant $K$<br /> Rearrange the equation to solve for $K$: <br />\[ \ln K = \frac{n \cdot \varepsilon_{cell}^{\circ}}{0.0257} \]<br /> Substitute the values: <br />\[ \ln K = \frac{6 \times 0.51}{0.0257} \approx 119.07 \]<br /> Convert $\ln K$ to $K$: <br />\[ K = e^{119.07} \approx 5.12 \times 10^{51} \]
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