Question

(b.)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

The half-cell reaction is: \[ \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \] The standard electrode potential for this half-cell reaction is \( E^\circ = 1.33 \, \text{V} \). Using the Nernst equation: \[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \] Substituting values: \[ E = 1.33 - \frac{0.0591}{6} \log \left( \frac{[\text{Cr}^{3+}]^2 \times [\text{H}_2\text{O}]^7}{[\text{Cr}_2\text{O}_7^{2-}] \times [\text{H}^+]^{14}} \right) \] \[ E = 1.33 - \frac{0.0591}{6} \log \left( \frac{(0.01)^2 \times 1}{(0.01) \times (1.0 \times 10^{-4})^{14}} \right) \] After simplification, calculate the final \(E\).