Question 11.7: Description of a Concentration Cell Determine the direction ...
Description of a Concentration Cell
Determine the direction of electron flow, designate the anode and cathode, and calculate the potential at 25°C for the cell represented in Fig. 11.12
The concentrations of Fe^{2+} ion in the compartments can eventually be equalized by transferring electrons from the left compartment to the right. This will cause Fe^{2+} to be formed in the left compartment, and iron metal will be deposited on the right electrode, thus consuming Fe^{2+} ions. Therefore, electron flow is from left to right, oxidation occurs in the left compartment (the anode), and reduction occurs in the right compartment (the cathode).
To calculate the cell potential, we use the Nernst equation in the form
ξ = ξ° – \frac{0.0591}{n} \log (Q)
where n = 2 because the cell half-reaction is Fe^{2+} + 2 e^{-} → Fe or its opposite. Also, ξ° = 0 (as always) for a concentration cell, and Q = 0.01/0.10 because Fe^{2+} is being formed in the compartment with the lower concentration. Thus
ξ = 0 – \frac{0.0591}{2} \log (\frac{0.01}{0.10} ) = – 0.0296 (- 1.00 ) = 0.0296 V