Why a Silver Spoon Is Safe in Copper Sulphate but an Iron Nail Is Not

Drop zinc into copper sulphate solution and the zinc becomes coated with copper while the solution warms slightly — electrons pass straight from zinc to copper ions, and the energy is lost as heat. Separate the two halves and connect them with a wire, and the same electrons flow through the wire as a useful current. That arrangement is a galvanic cell.

This lesson covers the construction, working and notation of galvanic cells, and standard electrode potentials with the electrochemical series.

A galvanic cell converts the chemical energy of a spontaneous redox reaction into electrical energy by keeping oxidation and reduction in separate half-cells joined by a wire and a salt bridge, and its notation lists the anode on the left and the cathode on the right.

The Daniell cell:

- Anode (negative) — a zinc rod in zinc sulphate solution, where oxidation occurs: $\mathrm{Zn}\to \mathrm{Z}{\mathrm{n}}^{2+}+2{\mathrm{e}}^{-}$
- Cathode (positive) — a copper rod in copper sulphate solution, where reduction occurs: $\mathrm{C}{\mathrm{u}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$
- Overall reaction: $\mathrm{Zn}+\mathrm{C}{\mathrm{u}}^{2+}\to \mathrm{Z}{\mathrm{n}}^{2+}+\mathrm{Cu}$
- Electrons flow through the external wire from zinc to copper; conventional current flows the opposite way

The salt bridge:

- A U-tube holding an inert electrolyte such as KCl or $\mathrm{KN}{\mathrm{O}}_3$ set in agar gel
- Completes the circuit by letting ions move between the half-cells
- Keeps each solution electrically neutral: anions drift towards the anode, where $\mathrm{Z}{\mathrm{n}}^{2+}$ builds up, and cations drift towards the cathode, where $\mathrm{C}{\mathrm{u}}^{2+}$ is used up
- Without it, charge builds up and the current stops almost at once

Cell notation:

$$\mathrm{Zn}(\mathrm{s})|\mathrm{Z}{\mathrm{n}}^{2+}(\mathrm{aq})||\mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})|\mathrm{Cu}(\mathrm{s})$$

- The anode is written on the left and the cathode on the right
- A single vertical line marks a phase boundary, and a double line marks the salt bridge

Worked example. For $\mathrm{Mg}+2A{\mathrm{g}}^{+}\to \mathrm{M}{\mathrm{g}}^{2+}+2Ag$, magnesium is oxidised, so it is the anode:

$$\mathrm{Mg}(\mathrm{s})|\mathrm{M}{\mathrm{g}}^{2+}(\mathrm{aq})||\mathrm{A}{\mathrm{g}}^{+}(\mathrm{aq})|\mathrm{Ag}(\mathrm{s})$$

An everyday example. The dry cell in a TV remote is a galvanic cell: zinc is oxidised at the casing while manganese dioxide is reduced around the carbon rod, driving current through the remote's circuit.

The substance. The anode is negative in a galvanic cell but positive in an electrolytic cell — what defines the anode is that oxidation happens there, whatever its sign.

**A standard electrode potential is the reduction potential of a half-cell measured against the standard hydrogen electrode, a cell's standard emf is $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$, and a positive value means the reaction is feasible.

Standard hydrogen electrode:**

- Platinum coated with platinum black, dipped in 1 M ${\mathrm{H}}^{+}$ with hydrogen gas bubbling at 1 bar and 298 K
- Its potential is taken as exactly 0 V for $2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2$
- Any half-cell connected to it gives an emf equal to that half-cell's standard reduction potential

The electrochemical series arranges standard reduction potentials in order:

- $\mathrm{L}{\mathrm{i}}^{+}/Li$ $-3.04$ V and ${\mathrm{K}}^{+}/K$ $-2.93$ V — the most negative values, the strongest reducing agents
- $\mathrm{Z}{\mathrm{n}}^{2+}/Zn$ $-0.76$ V and $\mathrm{F}{\mathrm{e}}^{2+}/Fe$ $-0.44$ V
- ${\mathrm{H}}^{+}/{\mathrm{H}}_2$ $0.00$ V
- $\mathrm{C}{\mathrm{u}}^{2+}/Cu$ $+0.34$ V and $\mathrm{A}{\mathrm{g}}^{+}/Ag$ $+0.80$ V
- ${\mathrm{F}}_2/{\mathrm{F}}^{-}$ $+2.87$ V — the most positive value, the strongest oxidising agent

What the series tells you:

- A more negative $E^{\circ }$ means a stronger reducing agent, more easily oxidised
- A metal with a negative $E^{\circ }$ can release hydrogen from dilute acids
- A metal displaces from solution the ions of any metal with a more positive $E^{\circ }$

Worked example 1 — the Daniell cell.

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

Worked example 2 — feasibility. Can silver displace copper? For $2Ag+\mathrm{C}{\mathrm{u}}^{2+}\to 2A{\mathrm{g}}^{+}+\mathrm{Cu}$, silver would be the anode:

$$E_{\text{cell}}^{\circ }=0.34-0.80=-0.46\text{ V}$$

The negative value means the reaction is not feasible, so a silver spoon stays unchanged in copper sulphate. For iron, $E_{\text{cell}}^{\circ }=0.34-(-0.44)=+0.78$ V, so an iron nail does displace copper.

An everyday example. Galvanised iron buckets and roofing sheets are coated with zinc, whose electrode potential is more negative than iron's, so the zinc corrodes first and protects the iron beneath.

The substance. Multiplying a half-reaction does not change its electrode potential — $E^{\circ }$ is an intensive property, so $2A{\mathrm{g}}^{+}+2{\mathrm{e}}^{-}\to 2Ag$ still has $E^{\circ }=+0.80$ V.