Redox refresher. Electrochemistry studies the link between chemical change and electricity. Every cell reaction is a redox reaction: one species is oxidised (loses electrons, oxidation number rises) while another is reduced (gains electrons, oxidation number falls). The two are inseparable because electrons released in oxidation must be taken up in reduction. The trick of electrochemistry is to physically separate these two half-reactions so the electrons are forced to travel through an external wire, giving us usable current.
Galvanic (voltaic) cell. A galvanic cell converts the energy of a spontaneous redox reaction into electrical energy. The classic example is the Daniell cell: a zinc rod in $\text{ZnSO}_4$ solution and a copper rod in $\text{CuSO}_4$ solution, joined by a salt bridge. Zinc, being more reactive, is oxidised: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ (this electrode is the anode, marked negative). The electrons travel through the wire to the copper electrode, where reduction occurs: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (the cathode, marked positive). The net reaction is $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$.
The salt bridge (a U-tube of agar gel saturated with $\text{KCl}$ or $\text{KNO}_3$) completes the circuit and keeps each half-cell electrically neutral by allowing ions to migrate, without letting the two solutions mix.
Electrode potential. At each electrode a potential difference develops between the metal and its ion in solution. This is the electrode potential. We cannot measure a single electrode in isolation, so we measure it against a reference. The agreed reference is the standard hydrogen electrode (SHE) — platinum coated with platinum black, with $\text{H}_2$ gas at $1\,\text{bar}$ bubbled over it in $1\,\text{M}\;\text{H}^+$ at $298\,\text{K}$ — whose potential is fixed at exactly zero volts.
Electrochemical series. Arranging electrodes by their standard reduction potentials $E^0$ gives the electrochemical series. A more negative $E^0$ means a stronger tendency to be oxidised (a better reducing agent); a more positive $E^0$ means a stronger tendency to be reduced. For the Daniell cell, $E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34\,\text{V}$ and $E^0_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}$.
Cell EMF. The electromotive force of a cell is the potential difference when no current flows: $E_{cell} = E_{cathode} - E_{anode}$ (both as reduction potentials). For the Daniell cell, $E^0_{cell} = 0.34 - (-0.76) = 1.10\,\text{V}$. A positive EMF means the reaction is spontaneous as written.
Nernst equation. EMF depends on concentration. The Nernst equation at $298\,\text{K}$ is $E = E^0 - \frac{0.059}{n}\log Q$, where $n$ is the number of electrons transferred and $Q$ is the reaction quotient. At equilibrium $E = 0$ and $Q = K_c$, linking EMF to the equilibrium constant.
EMF and Gibbs energy. The maximum electrical work a cell can do equals the fall in Gibbs energy: $\Delta G = -nFE_{cell}$, where $F = 96500\,\text{C mol}^{-1}$ is the Faraday constant. A positive $E_{cell}$ gives a negative $\Delta G$ — confirming a spontaneous, energy-releasing process.
Write the cell reaction and calculate the standard EMF of the Daniell cell given $E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34\,\text{V}$ and $E^0_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}$.
Solution- Step 1: Zinc has the more negative $E^0$, so it is oxidised (anode); copper is reduced (cathode).
- Step 2: Anode: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$; Cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$.
- Step 3: Net: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$.
- Step 4: $E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.34 - (-0.76)$.
- Step 5: $E^0_{cell} = 1.10\,\text{V}$.
Answer: Cell reaction $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$; $E^0_{cell} = 1.10\,\text{V}$.
Calculate the standard Gibbs energy change for the Daniell cell ($E^0_{cell} = 1.10\,\text{V}$, $n = 2$, $F = 96500\,\text{C mol}^{-1}$).
Solution- Step 1: Use $\Delta G^0 = -nFE^0_{cell}$.
- Step 2: Substitute $n = 2$, $F = 96500$, $E^0_{cell} = 1.10$.
- Step 3: $\Delta G^0 = -2 \times 96500 \times 1.10$.
- Step 4: $\Delta G^0 = -212300\,\text{J} = -212.3\,\text{kJ mol}^{-1}$.
Answer: $\Delta G^0 = -212.3\,\text{kJ mol}^{-1}$, so the reaction is spontaneous.
For the cell $\text{Zn} \mid \text{Zn}^{2+}(0.1\,\text{M}) \parallel \text{Cu}^{2+}(1\,\text{M}) \mid \text{Cu}$, find the EMF at $298\,\text{K}$. Take $E^0_{cell} = 1.10\,\text{V}$.
Solution- Step 1: Nernst: $E = E^0 - \dfrac{0.059}{n}\log\dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$ with $n = 2$.
- Step 2: $Q = \dfrac{0.1}{1} = 0.1$, so $\log Q = -1$.
- Step 3: $E = 1.10 - \dfrac{0.059}{2}\times(-1) = 1.10 + 0.0295$.
- Step 4: $E = 1.1295\,\text{V} \approx 1.13\,\text{V}$.
Answer: The cell EMF is about $1.13\,\text{V}$.
Using the electrochemical series, predict whether zinc can displace copper from copper sulphate solution. ($E^0_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}$, $E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34\,\text{V}$)
Solution- Step 1: A metal displaces another from solution if the combined cell EMF is positive.
- Step 2: Here Zn is oxidised and $\text{Cu}^{2+}$ reduced: $E^0_{cell} = 0.34 - (-0.76) = +1.10\,\text{V}$.
- Step 3: Positive EMF means $\Delta G < 0$, so the reaction is spontaneous.
- Step 4: Zinc, being more reactive (more negative $E^0$), reduces $\text{Cu}^{2+}$ to copper.
Answer: Yes — zinc displaces copper because $E^0_{cell} = +1.10\,\text{V}$ (spontaneous).
The EMF of a cell in which the reaction $\text{Ni} + 2\text{Ag}^+ \rightarrow \text{Ni}^{2+} + 2\text{Ag}$ occurs is $1.05\,\text{V}$. Calculate $\Delta G$ ($n = 2$, $F = 96500\,\text{C mol}^{-1}$).
Solution- Step 1: $\Delta G = -nFE_{cell}$.
- Step 2: $\Delta G = -2 \times 96500 \times 1.05$.
- Step 3: $\Delta G = -202650\,\text{J}$.
- Step 4: $\Delta G = -202.65\,\text{kJ mol}^{-1}$.
Answer: $\Delta G = -202.65\,\text{kJ mol}^{-1}$.
Why is a salt bridge used in a galvanic cell? State two functions.
Solution- Step 1: As the cell runs, $\text{Zn}^{2+}$ ions build up in the anode beaker and $\text{Cu}^{2+}$ ions deplete in the cathode beaker.
- Step 2: This charge imbalance would stop the reaction; the salt bridge lets ions migrate to restore neutrality.
- Step 3: Function 1 — it completes the internal circuit, allowing current to flow.
- Step 4: Function 2 — it maintains electrical neutrality of the two half-cells without letting the solutions mix.
Answer: The salt bridge completes the circuit and keeps each half-cell electrically neutral.