Question 21.8: Calculating the Potential of a Concentration Cell Problem A ......
Calculating the Potential of a Concentration Cell
Problem A voltaic cell consists of two Ag/Ag^+ half-cells. In half-cell A, the electrolyte is 0.010 M AgNO_3; in half-cell B, it is 4.0×10^{−4} M AgNO_3. What is E_{\text{cell}} at 298.15 K?
Plan The standard half-cell reactions are identical, so E^\circ _{\text{cell}} is zero, and we find E_{\text{cell}} from the Nernst equation. Half-cell A has a higher [Ag^+], so Ag^+ ions are reduced and plate out on electrode A. In half-cell B, Ag atoms of the electrode are oxidized and Ag^+ ions enter the solution. As in all voltaic cells, reduction occurs at the cathode, so it is positive.
Solution Writing the spontaneous reaction: The [Ag^+] decreases in half-cell A and increases in half-cell B, so the spontaneous reaction is
Ag^+(aq; 0.010 M) [\text{half-cell A}] ⟶ Ag^+(aq; 4.0×10^{−4} M) [\text{half-cell B}]
Calculating E_{\text{cell}}, with n = 1:
E_{\text{cell}} = E^\circ _{\text{cell}} − \frac{0.0592 V}{1} \log \frac{[Ag^+]_{\text{dil}}}{[Ag^+]_{\text{conc}}} = 0 V − \left( 0.0592 V \log \frac{4.0×10^{−4}}{0.010} \right)
= 0.083 V