1.3 - Voltaic Cell

Structure and Energy Conversion

• A voltaic cell or galvanic cell has two different electrodes connected through an external circuit and electrolytes that permit ion movement.

• A spontaneous redox reaction converts chemical energy to electrical energy.

• Oxidation occurs at the anode (negative terminal); reduction occurs at the cathode (positive terminal). Electrons flow through the wire from anode to cathode.

• Two half-cells are joined by a salt bridge or separated by a porous pot so ions can move and electrical neutrality is maintained.

• In the simple Mg/Cu cell with copper(II) chloride solution, magnesium is the anode: $\mathrm{Mg}\to \mathrm{M}{\mathrm{g}}^{2+}+2{\mathrm{e}}^{-}$; copper is the cathode: $\mathrm{C}{\mathrm{u}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$.

• Conventional current flows through the external circuit from the positive terminal to the negative terminal, opposite to electron flow.

Daniell Cell and Cell Notation

• In a Daniell cell:
  • anode: $\mathrm{Zn}(\mathrm{s})\to \mathrm{Z}{\mathrm{n}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}$;
  • cathode: $\mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{Cu}(\mathrm{s})$;
  • overall: $\mathrm{Zn}(\mathrm{s})+\mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})\to \mathrm{Z}{\mathrm{n}}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$.
• Cell notation places the anode on the left and cathode on the right. A single line separates phases; a double line represents the salt bridge or porous partition: $\mathrm{Zn}(\mathrm{s})\mid \mathrm{Z}{\mathrm{n}}^{2+}(\mathrm{aq})\parallel \mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})\mid \mathrm{Cu}(\mathrm{s})$
• With $E^0(\mathrm{C}{\mathrm{u}}^{2+}/Cu)=+0.34\mathrm{V}$ and $E^0(\mathrm{Z}{\mathrm{n}}^{2+}/Zn)=-0.76\mathrm{V}$: $E_{\text{cell}}^0=(+0.34)-(-0.76)=+1.10\mathrm{V}$
• Use platinum as an inert electrode when a half-cell contains only ions or gases.

Experiment 1A — Voltage of a Voltaic Cell

• Aim: Determine the voltage of a voltaic cell using different pairs of metals.

• Problem statement: How do different metal pairs dipped in electrolytes affect cell voltage?

• Hypothesis: The greater the difference between the metals’ $E^0$ values, the greater the voltage.

• Variables:

  • manipulated: pair of metals;
  • responding: voltage reading;
  • fixed: volume and concentration of electrolytes.

• Operational definition of a chemical cell: The voltmeter gives a reading when two different metals are dipped into electrolytes.

• Materials: Different metal plates and suitable metal-salt solutions of $1.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$.

• Apparatus: Sandpaper, voltmeter and connecting wires with crocodile clips, with apparatus suitable for a simple cell or two half-cells.

• Measurements: Cell voltage, negative terminal (anode) and positive terminal (cathode) for each metal pair in a simple cell or two connected half-cells.

• Relationship: The larger the difference in electrode potentials, the larger the cell voltage; $E_{\text{cell}}^0=E_{\text{cathode}}^0-E_{\text{anode}}^0$.