the design and operation of galvanic cells, including primary cells, secondary (rechargeable) cells and fuel cells, with reference to the role of anode, cathode, electrolyte, salt bridge and external circuit, and the calculation of cell EMF (E°_cell) from standard electrode potentials at 25°C

What this dot point is asking

VCAA wants you to know the components of a galvanic cell (electrodes, electrolyte, salt bridge, external circuit), to distinguish primary cells, secondary cells and fuel cells, to identify anode, cathode and the direction of electron flow, and to calculate E°_cell from standard electrode potentials at 25°C.

The answer

What a galvanic cell does

A galvanic cell (also called a voltaic cell) converts the chemical energy of a spontaneous redox reaction into electrical energy. The two half-reactions are physically separated into half-cells so that the electrons released by the oxidation half-cell are forced to travel through an external circuit to reach the reduction half-cell. That moving charge is the electric current the cell delivers.

A galvanic cell has five components:

1. Anode: the electrode where oxidation occurs. The species with the more negative E° is oxidised here. The anode is the negative terminal of a galvanic cell.
2. Cathode: the electrode where reduction occurs. The species with the more positive E° is reduced here. The cathode is the positive terminal of a galvanic cell.
3. Electrolyte: the ionic solution in each half-cell that allows ions to flow and supplies the species being oxidised or reduced.
4. Salt bridge: a porous tube or filter paper soaked in inert ionic solution (e.g. KNO3) that lets ions migrate between half-cells to keep both electrolytes electrically neutral. Without it the cell stops working within seconds.
5. External circuit: the wire (often with a voltmeter or load) connecting the two electrodes. Electrons flow from anode to cathode here.

Memory aid: AN OX, RED CAT. The ANode is where OXidation happens; REDuction happens at the CAThode.

Direction of charge

• Electrons flow through the wire from anode to cathode (negative terminal to positive terminal in the external circuit).
• Conventional current is in the opposite direction (cathode to anode externally).
• Cations (positive ions) in the salt bridge flow towards the cathode half-cell.
• Anions (negative ions) in the salt bridge flow towards the anode half-cell.

The ion flow keeps each half-cell electrically neutral. As the anode half-cell builds up Zn^2+, anions move in to balance the charge; as the cathode half-cell consumes Ag^+, cations move in to replace the lost positive charge.

Calculating cell EMF

The EMF (or cell potential) E°_cell is the voltage the cell delivers under standard conditions (1 mol L^-1 solutions, 25°C, 1 atm for gases).

E°_cell = E°(cathode) - E°(anode)

where both E° values come from the data book as reduction potentials.

• If E°_cell > 0, the reaction is spontaneous as written.
• If E°_cell < 0, the reverse reaction is spontaneous; the cell as written will not produce current.

You do not multiply E° by stoichiometric coefficients. E° is a per-electron property that does not scale with the equation.

Primary, secondary and fuel cells

The three cell types in the VCE Study Design:

Primary cells

The redox reaction is one way. Once the reactants are consumed, the cell is dead and discarded. Cheap, simple, used in low-drain applications (remote controls).

Secondary cells

The redox reaction is reversible. Applying an external voltage in reverse drives the reaction backwards, regenerating the original reactants. Higher initial cost, but cheaper per use over the cell's life. The lead acid battery in a petrol car is recharged by the alternator while the engine runs.

Fuel cells

A continuous supply of fuel (e.g. H2) and oxidant (O2) enters the cell. The cell runs as long as the supply continues. Often used where high efficiency, low emissions, or remote operation matter (buses, spacecraft, backup power for hospitals).

The hydrogen oxygen fuel cell

A widely examined example. In an acidic electrolyte:

• Anode: 2H2(g) -> 4H^+(aq) + 4e^-
• Cathode: O2(g) + 4H^+(aq) + 4e^- -> 2H2O(l)
• Overall: 2H2(g) + O2(g) -> 2H2O(l)

The only product is water. E°_cell is about +1.23 V. Advantages: high efficiency, only product is water, modular design. Disadvantages: hydrogen storage is difficult (low energy density as a gas, requires high-pressure tanks or chemical storage), platinum catalyst is expensive, and most hydrogen today is produced from natural gas (releasing CO2 upstream).

Examples in context

Example 1. AGL Hornsdale lithium-ion mega-battery in South Australia. AGL's Hornsdale Power Reserve uses Tesla lithium-ion cells, the world's first $100 \, \text{MW}$ grid-scale battery. The full cell is $\text{LiCoO}_2 / \text{Li}_x \text{C}_6$. Cathode reduction: $\text{Li}^+ + \text{CoO}_2 + e^- \to \text{LiCoO}_2$, $E^{\circ} = +0.60 \, \text{V}$. Anode oxidation: $\text{LiC}_6 \to \text{Li}^+ + \text{C}_6 + e^-$, $E^{\circ} = -3.04 \, \text{V}$ (as $\text{Li}^+ / \text{Li}$). EMF $= 0.60 - (-3.04) = 3.64 \, \text{V}$ per cell, although commercial cells operate around $3.7 \, \text{V}$. The battery delivers $129 \, \text{MWh}$ stored energy in $\sim 30{,}000$ modules. Each charge-discharge cycle involves $\text{Li}^+$ shuttling through a solid polymer electrolyte rather than a liquid salt bridge.

Example 2. AGL Energy Bell Bay hydrogen-fuel-cell pilot. AGL Tasmania trials a $1 \, \text{MW}$ proton-exchange-membrane fuel cell at Bell Bay, powered by green hydrogen from a $5 \, \text{MW}$ electrolyser fed by Hydro Tasmania renewables. Cathode reduction: $\text{O}_2 + 4 \text{H}^+ + 4 e^- \to 2 \text{H}_2 \text{O}$, $E^{\circ} = +1.23 \, \text{V}$. Anode oxidation: $\text{H}_2 \to 2 \text{H}^+ + 2 e^-$, $E^{\circ} = 0 \, \text{V}$. EMF $= 1.23 - 0 = 1.23 \, \text{V}$ per cell, but operating voltage drops to $\sim 0.7 \, \text{V}$ under load. To deliver $1 \, \text{MW}$ at $0.7 \, \text{V}$ and $4$ moles of electrons per mole $\text{O}_2$, the stack consumes $\sim 0.78 \, \text{kg}$ of $\text{H}_2$ per second; only water is the by-product.

Try this

Q1. Sketch and label a galvanic cell using $\text{Zn} | \text{Zn}^{2+}$ and $\text{Cu} | \text{Cu}^{2+}$ electrodes. State the polarity of each electrode. [4 marks]

• Cue. Zn anode (negative, oxidation); Cu cathode (positive, reduction); salt bridge with KCl; voltmeter between; electrons flow Zn to Cu in external wire.

Q2. Calculate the EMF and identify the spontaneous direction for a cell built from $\text{Fe}^{2+} | \text{Fe}$ ($E^{\circ} = -0.44 \, \text{V}$) and $\text{Ag}^+ | \text{Ag}$ ($E^{\circ} = +0.80 \, \text{V}$). [3 marks]

• Cue. Ag higher $E^{\circ}$, is cathode. EMF $= 0.80 - (-0.44) = 1.24 \, \text{V}$. Fe oxidised: $\text{Fe} \to \text{Fe}^{2+} + 2 e^-$; $\text{Ag}^+$ reduced.

Q3. A $\text{Mg} | \text{Mg}^{2+}$ ($E^{\circ} = -2.37 \, \text{V}$) and $\text{Cu} | \text{Cu}^{2+}$ ($E^{\circ} = +0.34 \, \text{V}$) cell operates for $30 \, \text{minutes}$ at $0.20 \, \text{A}$. (a) Calculate the EMF. (b) Calculate moles of $\text{Mg}$ consumed. (c) State the role of the salt bridge. [2+2+2 marks]

• Cue. (a) $0.34 - (-2.37) = 2.71 \, \text{V}$. (b) $Q = 0.20 \times 1800 = 360 \, \text{C}$; $n(e) = 360 / 96{,}500 = 3.73 \times 10^{-3}$; $n(\text{Mg}) = 1.87 \times 10^{-3}$ mol. (c) Allows ion migration to maintain electroneutrality; cations to cathode, anions to anode.