Chapter 19: Problem 73

Calculate the emf of the following concentration cell at $25^{\circ} \mathrm{C}$ : $$ \mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(0.080 \mathrm{M}) \| \mathrm{Cu}^{2+}(1.2 \mathrm{M})\right| \mathrm{Cu}(s) $$

Short Answer

Expert verified

The emf for the given concentration cell at 25°C is -0.09 V.

Step by step solution

– Recall the Nernst equation

The Nernst equation can be expressed as:\n\n\[ E = E_0 - \frac{0.0592}{n} \log \frac{{[Cu^{2+}]}_{cathode}}{{[Cu^{2+}]}_{anode}} \]\n\nwhere n is the number of electron transfers in the cell reaction (for Cu -> Cu^{2+} its value is 2), [Cu^{2+}]_{cathode} and [Cu^{2+}]_{anode} are the molar concentrations of copper ions at the cathode and anode respectively, E is the cell potential we are solving for, and E_0 is the standard cell potential. Because both the reduction and the oxidation occur in the same type of ion, the standard cell potential (E_0) equals 0.

– Substitute in values

Substitute the given values into the equation. The concentration of Cu^{2+} ions at the cathode is given as 1.2 M and at the anode is given as 0.08 M.\n\n\[ E = 0 - \frac{0.0592}{2} \log \frac{1.2}{0.08} \]