7.2 Electrochemical cells and standard reduction potentials

Electrochemical cells convert chemical energy into electrical energy by separating a redox reaction into two half-cells. Understanding how these cells work and how to calculate their potentials lets you predict whether a reaction is spontaneous, which is the foundation for understanding batteries, corrosion, and electrolysis.

Electrochemical Cells

Components of galvanic cells

A galvanic (voltaic) cell uses a spontaneous redox reaction to generate electrical current. The reaction is split into two half-cells, each containing an electrode dipped in a solution of its ions.

• Anode half-cell: where oxidation occurs. The electrode loses electrons. In a Zn-Cu cell, this is the Zn electrode ($Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$).
• Cathode half-cell: where reduction occurs. The electrode gains electrons. In a Zn-Cu cell, this is the Cu electrode ($Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$).

Electrons flow from the anode to the cathode through an external circuit (a wire connected to a light bulb, voltmeter, etc.). That flow of electrons is the electric current you can measure or use.

A salt bridge connects the two solutions and maintains electrical neutrality. Without it, positive charge would build up in the cathode compartment and negative charge in the anode compartment, and the reaction would stop almost immediately. The salt bridge contains an inert electrolyte like $KCl$ or $NaNO_3$ that lets ions migrate between compartments to balance the charge.

Calculation of standard cell potentials

The standard cell potential ($E°_{cell}$) is the voltage a cell produces when all species are at standard conditions: 1 M concentration, 1 atm pressure, and 25°C.

You calculate it from the standard reduction potentials ($E°_{red}$) listed in a reference table:

$E°_{cell} = E°_{cathode} - E°_{anode}$

Both values you plug in are the reduction potentials from the table. You don't flip the sign of the anode value yourself; the subtraction takes care of it.

Step-by-step for a Zn-Cu cell:

1. Look up the standard reduction potentials:

   • $Cu^{2+} + 2e^- \rightarrow Cu(s)$, $E°_{red} = +0.34 \text{ V}$
   • $Zn^{2+} + 2e^- \rightarrow Zn(s)$, $E°_{red} = -0.76 \text{ V}$

2. Identify which half-reaction is the cathode and which is the anode. The species with the more positive $E°_{red}$ gets reduced (cathode). The species with the more negative $E°_{red}$ gets oxidized (anode).

   • Cathode: Cu (higher $E°_{red}$)
   • Anode: Zn (lower $E°_{red}$)

3. Apply the formula:

   • $E°_{cell} = (+0.34) - (-0.76) = +1.10 \text{ V}$

A positive $E°_{cell}$ confirms the reaction is spontaneous under standard conditions.

One common mistake: do not multiply $E°$ values by the number of electrons transferred. Standard reduction potentials are intensive properties (they don't change when you scale the reaction).

Spontaneity in redox reactions

The sign of $E°_{cell}$ tells you directly about spontaneity:

• $E°_{cell} > 0$: spontaneous (galvanic cell, produces electricity)
• $E°_{cell} < 0$: non-spontaneous (requires energy input, like an electrolytic cell)
• $E°_{cell} = 0$: system is at equilibrium

You can connect cell potential to thermodynamics through Gibbs free energy:

$\Delta G° = -nFE°_{cell}$

• $n$ = number of moles of electrons transferred in the balanced equation
• $F$ = Faraday's constant = 96,485 C/mol

This equation shows why the signs work out: a positive $E°_{cell}$ gives a negative $\Delta G°$, which is the thermodynamic condition for spontaneity. For the Zn-Cu cell with $n = 2$:

$\Delta G° = -(2)(96{,}485)(1.10) = -212{,}267 \text{ J} \approx -212 \text{ kJ}$

That large negative value confirms a strongly spontaneous reaction.

Notation for electrochemical cells

Cell diagrams follow a standard convention so anyone can read them at a glance:

• The anode is drawn on the left; the cathode is on the right.
• Electrodes are labeled with chemical species and physical states.
• A double vertical line (‖) represents the salt bridge.
• An arrow shows electron flow from anode to cathode through the external wire.

Shorthand (line) notation compresses all of this into one line:

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

How to read it:

• A single vertical line ( | ) represents a phase boundary (solid electrode touching its aqueous ion solution).
• The double line ( ‖ ) represents the salt bridge.
• Read left to right: oxidation happens on the left side, reduction on the right.

So for this cell: zinc metal is oxidized to $Zn^{2+}$ ions at the anode, and $Cu^{2+}$ ions are reduced to copper metal at the cathode.

If a half-cell involves species that are both in solution (like $Fe^{3+}$ being reduced to $Fe^{2+}$), you'd use an inert electrode such as platinum, and separate the aqueous species with a comma:

$Zn(s) \mid Zn^{2+}(aq) \| Fe^{3+}(aq), Fe^{2+}(aq) \mid Pt(s)$