Question

Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).

Hint:

Use the Nernst equation to adjust the standard potential based on the concentration of the species involved in the reaction.

Answer 1

To solve the problem, we need to calculate the potential for a half-cell containing the following concentrations: 0.01 M K₂Cr₂O₇ (aq), 0.01 M Cr³⁺ (aq), and 1.0 × 10⁻⁴ M H⁺ (aq).

1. Understanding the Reaction and Nernst Equation:
We are given a half-cell containing a redox couple involving Cr₂O₇²⁻ (dichromate) and Cr³⁺ (chromium ion). The half-reaction for this system is: \[ \text{Cr}_2\text{O}_7^{2-}(aq) + 14H^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7H_2O(l) \] To calculate the potential, we can use the Nernst equation: \[ E = E^\circ - \frac{0.0591}{n} \log Q \] where: - \( E^\circ \) is the standard electrode potential for the half-reaction, - \( n \) is the number of electrons transferred, - \( Q \) is the reaction quotient, which is the ratio of concentrations of products to reactants.

2. Standard Electrode Potential:
The standard electrode potential for the half-reaction is given as: \[ E^\circ = +1.33 \, \text{V} \]

3. Reaction Quotient (Q):
The reaction quotient \( Q \) is given by the expression: \[ Q = \frac{[\text{Cr}^{3+}]^2 [\text{H}_2\text{O}]^7}{[\text{Cr}_2\text{O}_7^{2-}] [\text{H}^+]^{14}} \] However, since water (\( \text{H}_2\text{O} \)) is a pure liquid, its concentration is considered constant and does not appear in the expression for \( Q \). Therefore, we have: \[ Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}] [\text{H}^+]^{14}} \]

4. Substituting the Concentrations:
We are given: - \( [\text{Cr}_2\text{O}_7^{2-}] = 0.01 \, \text{M} \) - \( [\text{Cr}^{3+}] = 0.01 \, \text{M} \) - \( [\text{H}^+] = 1.0 \times 10^{-4} \, \text{M} \) Substituting these values into the expression for \( Q \), we get: \[ Q = \frac{(0.01)^2}{(0.01)(1.0 \times 10^{-4})^{14}} \]

5. Nernst Equation Calculation:
Now, we substitute the values into the Nernst equation: \[ E = 1.33 - \frac{0.0591}{6} \log Q \] Calculating the value of \( Q \) and then applying the Nernst equation will give us the potential for the half-cell under the given conditions.

Final Answer:
We can calculate \( E \) using these equations and constants to find the potential of the half-cell.