4. Electrochemistry & Redox Cells
Electrochemistry converts between chemical and electrical energy. Every cell splits into two half-reactions: oxidation (loss of electrons, at the anode) and reduction (gain of electrons, at the cathode) — remember "an ox, red cat."
Voltaic (Galvanic) Cells
A spontaneous redox reaction drives electrons through an external wire while ions cross a salt bridge to balance charge.
Example (Daniell cell): $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$
Anode (oxidation): $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$. Cathode (reduction): $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$.
Anode (oxidation): $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$. Cathode (reduction): $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$.
Standard Cell Potentials
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$
Example: $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V, $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76$ V.
$$E^\circ_{\text{cell}} = 0.34 - (-0.76) = 1.10\ \text{V}$$
Free Energy and the Nernst Equation
$$\Delta G^\circ = -nFE^\circ \qquad E = E^\circ - \frac{0.0592}{n}\log Q \;\; (25\,^\circ\text{C})$$
Example: For the Daniell cell ($n = 2$) with $[\text{Zn}^{2+}] = 1.0$ M and $[\text{Cu}^{2+}] = 0.010$ M:
$$E = 1.10 - \frac{0.0592}{2}\log\frac{1.0}{0.010} = 1.10 - 0.0592 = 1.04\ \text{V}$$
Practice Problems
- Identify the anode and cathode in: Mg(s) + 2 Ag+(aq) → Mg2+(aq) + 2 Ag(s).
- Calculate E°cell for the Mg/Ag cell using E°(Ag+/Ag) = +0.80 V and E°(Mg2+/Mg) = −2.37 V.
- Is the Mg/Ag cell reaction spontaneous under standard conditions? How do you know?
- An electrolytic cell vs a voltaic cell: which one needs an external power source?
Answers: 1. Anode: Mg (oxidation); cathode: Ag+ (reduction) 2. E°cell = 0.80 − (−2.37) = 3.17 V 3. Yes — E°cell > 0 means ΔG° < 0 4. The electrolytic cell