Why Zinc Gives Up Its Electrons to Copper in a Cell

Drop a zinc strip into blue copper sulphate solution and it becomes coated with copper as the colour fades — electrons pass straight from zinc to copper ions. Separate the two halves of that reaction and the electrons are forced through a wire instead, where they can light a bulb.

This part covers galvanic and electrolytic cells, standard electrode potentials, the Nernst equation, and the link between emf, Gibbs energy and equilibrium.

A galvanic cell turns the chemical energy of a spontaneous redox reaction into electrical energy, while an electrolytic cell uses electrical energy to drive a non-spontaneous reaction; in cell notation the anode is written on the left and the cathode on the right, with a double line for the salt bridge.

Galvanic versus electrolytic:

- Galvanic — spontaneous reaction; anode negative, cathode positive
- Electrolytic — an external source drives the reaction; anode positive, cathode negative
- In both, oxidation happens at the anode and reduction at the cathode

Daniell cell:

$$\text{Zn}(s)|{\text{Zn}}^{2+}(aq)||{\text{Cu}}^{2+}(aq)|\text{Cu}(s)$$

- Anode: $\text{Zn}\to {\text{Zn}}^{2+}+2e^{-}$
- Cathode: ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$

Salt bridge. An inverted tube of KCl or KNO${}_3$ set in agar jelly completes the circuit and keeps both solutions electrically neutral.

Worked example. If $0.1$ mol of zinc dissolves, $0.1\times 2=0.2$ mol of electrons flow through the wire, and $0.1\times 63.5=6.35$ g of copper is deposited.

An everyday example. A torch cell works as a galvanic cell, while charging a phone's battery runs it as an electrolytic cell.

The substance. The signs of the electrodes swap between the two cell types, but oxidation is always at the anode.

**The standard electrode potential $E^{\circ }$ is the reduction potential of an electrode at $1$ M concentration, $1$ bar and $298$ K, measured against the standard hydrogen electrode taken as $0.00$ V; the cell emf is $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$, and a positive value means the reaction is feasible.

Standard hydrogen electrode.** Platinum coated with platinum black dips into $1$ M H${}^{+}$ with hydrogen gas bubbled at $1$ bar:

$$2{\text{H}}^{+}(aq)+2e^{-}\to {\text{H}}_2(g),E^{\circ }=0.00\text{ V}$$

Electrochemical series:

- **More positive $E^{\circ }$** — stronger oxidising agent, such as F${}_2$ at $+2.87$ V
- **More negative $E^{\circ }$** — stronger reducing agent, such as Li at $-3.05$ V

Worked example. For the Daniell cell, $E^{\circ }$ is $+0.34$ V for copper and $-0.76$ V for zinc:

$$E_{cell}^{\circ }=0.34-(-0.76)=1.10\text{ V}$$

Positive, so zinc reduces copper ions spontaneously. With silver at $+0.80$ V, a copper–silver cell gives $0.80-0.34=0.46$ V, so copper displaces silver.

An everyday example. An iron nail left in copper sulphate solution turns copper-coloured, because iron, at $-0.44$ V, lies below copper in the series.

The substance. Electrode potentials do not scale with the equation — doubling a half-reaction leaves $E^{\circ }$ unchanged.

**The Nernst equation corrects a standard potential for real concentrations: for a cell at $298$ K, $E_{cell}=E_{cell}^{\circ }-\frac{0.059}{n}\log Q$, where $n$ is the number of electrons transferred and $Q$ is the reaction quotient.

For an electrode** $M^{n+}+n{e}^{-}\to M$:

$$E=E^{\circ }-\frac{2.303RT}{nF}\log \frac{1}{[M^{n+}]}=E^{\circ }-\frac{0.059}{n}\log \frac{1}{[M^{n+}]}$$

For the Daniell cell:

$$E_{cell}=E_{cell}^{\circ }-\frac{0.059}{2}\log \frac{[{\text{Zn}}^{2+}]}{[{\text{Cu}}^{2+}]}$$

Worked example. A Daniell cell has $[{\text{Zn}}^{2+}]=0.1$ M and $[{\text{Cu}}^{2+}]=0.001$ M:

$$E_{cell}=1.10-\frac{0.059}{2}\log 100=1.10-0.059=1.041\text{ V}$$

With the two concentrations swapped, $E_{cell}=1.10+0.059=1.159$ V.

**Where $0.059$ comes from.** At $298$ K, $\frac{2.303RT}{F}=\frac{2.303\times 8.314\times 298}{96487}\approx 0.0591$ V.

An everyday example. A cell's voltage slowly drops as it is used, because the concentrations inside change — exactly what the Nernst equation predicts.

The substance. Two identical electrodes at different concentrations still give an emf — a concentration cell — even though its $E_{cell}^{\circ }$ is zero.

**The electrical work a cell can do equals the fall in Gibbs energy, so ${\Delta }_r{G}^{\circ }=-nF{E}_{cell}^{\circ }$, and since ${\Delta }_r{G}^{\circ }=-2.303RT\log K$, the equilibrium constant follows from $\log K=\frac{n{E}_{cell}^{\circ }}{0.059}$ at $298$ K.

Key relations:**

$${\Delta }_r{G}^{\circ }=-nF{E}_{cell}^{\circ },{\Delta }_r{G}^{\circ }=-2.303RT\log K$$

- **$E_{cell}^{\circ }>0$** — ${\Delta }_r{G}^{\circ }<0$, spontaneous, $K>1$
- **$E_{cell}^{\circ }<0$ — not spontaneous as written
- At equilibrium**, $E_{cell}=0$

Worked example. For the Daniell cell, $n=2$ and $E_{cell}^{\circ }=1.10$ V, with $F=96487$ C mol${}^{-1}$:

$${\Delta }_r{G}^{\circ }=-2\times 96487\times 1.10\approx -212{\text{ kJ mol}}^{-1}$$

$$\log K=\frac{2\times 1.10}{0.059}\approx 37.3\Rightarrow K\approx 2\times 10^{37}$$

Such a large $K$ means the reaction goes essentially to completion.

An everyday example. A completely flat cell has reached equilibrium — its $E_{cell}$ has fallen to zero, so it can do no more work.

The substance. **$E^{\circ }$ does not depend on how much reaction occurs, but ${\Delta }_r{G}^{\circ }$ does** — the factor $n$ carries that dependence.