AP Chemistry · Lesson 2

Galvanic cells and cell potential

Know Know it

A galvanic cell separates a favorable redox reaction into two half-cells joined by a wire and a salt bridge. Oxidation (loss of electrons) happens at the anode; reduction (gain of electrons) happens at the cathode. Electrons flow through the wire from the anode to the cathode.

In the salt bridge, anions drift toward the anode and cations drift toward the cathode, which keeps both solutions electrically neutral. Without it, charge would build up and the current would stop. Electrons never travel through the salt bridge.

Tables list standard reduction potentials. The half-reaction with the more positive potential is reduced (the cathode). Then Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}, using both values as reductions. A galvanic cell has a positive Ecell∘E^\circ_{\text{cell}}. Multiplying a half-reaction to balance electrons does not change its potential.

Standard reduction potentials at 25 °C
Half-reactionE∘E^\circ (V)
Ag+(aq)+e−→Ag(s)\text{Ag}^{+}(aq) + e^{-} \rightarrow \text{Ag}(s)+0.80
Cu2+(aq)+2e−→Cu(s)\text{Cu}^{2+}(aq) + 2\text{e}^{-} \rightarrow \text{Cu}(s)+0.34
Pb2+(aq)+2e−→Pb(s)\text{Pb}^{2+}(aq) + 2\text{e}^{-} \rightarrow \text{Pb}(s)-0.13
Ni2+(aq)+2e−→Ni(s)\text{Ni}^{2+}(aq) + 2\text{e}^{-} \rightarrow \text{Ni}(s)-0.25
Zn2+(aq)+2e−→Zn(s)\text{Zn}^{2+}(aq) + 2\text{e}^{-} \rightarrow \text{Zn}(s)-0.76
Al3+(aq)+3e−→Al(s)\text{Al}^{3+}(aq) + 3\text{e}^{-} \rightarrow \text{Al}(s)-1.66

Worked example Easy

A galvanic cell is built from a nickel electrode in 1.0 M Ni2+\text{Ni}^{2+} and a silver electrode in 1.0 M Ag+\text{Ag}^{+}. The standard reduction potentials are +0.80 V for Ag+(aq)+e−→Ag(s)\text{Ag}^{+}(aq) + e^{-} \rightarrow \text{Ag}(s) and -0.25 V for Ni2+(aq)+2e−→Ni(s)\text{Ni}^{2+}(aq) + 2\text{e}^{-} \rightarrow \text{Ni}(s). What is Ecell∘E^\circ_{\text{cell}}?

  1. +0.55 V
  2. +1.85 V
  3. +1.05 VAnswer
  4. -1.05 V

How to solve it

  1. Cathode: silver, +0.80 V (the more positive reduction potential). Anode: nickel, -0.25 V.
  2. Ecell∘=0.80−(−0.25)=+1.05E^\circ_{\text{cell}} = 0.80 - (-0.25) = +1.05 V.
  3. The silver half-reaction is doubled to balance two electrons, but its potential stays +0.80 V.

Why each choice is right or wrong

  • A. Incorrect. +0.55 V adds the two table values, 0.80 + (-0.25), instead of subtracting the anode potential from the cathode potential.
  • B. Incorrect. +1.85 V doubles the silver potential to match the electrons, but multiplying a half-reaction does not change its potential.
  • C. Correct. Silver has the higher reduction potential, so it is the cathode: Ecell∘=0.80−(−0.25)=+1.05E^\circ_{\text{cell}} = 0.80 - (-0.25) = +1.05 V.
  • D. Incorrect. -1.05 V puts nickel at the cathode; a galvanic cell always runs with a positive cell potential.

Trap Do not multiply the voltage

To balance electrons you may multiply a half-reaction by 2 or 3, but its potential stays the same: potential is energy per unit of charge, not a total. The coefficients matter for n in ΔG∘=−nFEcell∘\Delta G^\circ = -nFE^\circ_{\text{cell}}, not for Ecell∘E^\circ_{\text{cell}}.

Explain Explain it

Task verbs: calculate, explain

Cell potential and free energy are linked by ΔG∘=−nFEcell∘\Delta G^\circ = -nFE^\circ_{\text{cell}}, where n is the moles of electrons transferred in the balanced equation and F = 96,485 C/mol. With Ecell∘E^\circ_{\text{cell}} in volts, the result is in joules (1 V · C = 1 J). A positive Ecell∘E^\circ_{\text{cell}} gives a negative ΔG∘\Delta G^\circ and K greater than 1.

Under nonstandard conditions, the cell potential depends on Q. If Q is less than 1 (few products), the potential is above Ecell∘E^\circ_{\text{cell}}; if Q is greater than 1, it is below. As the cell runs, Q rises toward K and the potential falls, reaching zero at equilibrium: a dead battery.

Worked example Medium

For the cell reaction Pb(s)+Cu2+(aq)→Pb2+(aq)+Cu(s)\text{Pb}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Pb}^{2+}(aq) + \text{Cu}(s), Ecell∘=+0.47E^\circ_{\text{cell}} = +0.47 V. What is ΔG∘\Delta G^\circ for the reaction? (F = 96,485 C/mol)

  1. -45.3 kJ/mol
  2. -90.7 kJ/molAnswer
  3. +90.7 kJ/mol
  4. -181 kJ/mol

How to solve it

  1. Electrons transferred: Pb gives 2 electrons and Cu2+\text{Cu}^{2+} takes 2, so n = 2.
  2. ΔG∘=−nFEcell∘=−(2)(96,485)(0.47)=−9.07×104\Delta G^\circ = -nFE^\circ_{\text{cell}} = -(2)(96{,}485)(0.47) = -9.07 \times 10^{4} J.
  3. In kilojoules: -90.7 kJ per mole of reaction, negative because the cell potential is positive.

Why each choice is right or wrong

  • A. Incorrect. -45.3 kJ/mol uses n = 1, but each lead atom gives two electrons to each copper(II) ion, so n = 2.
  • B. Correct. ΔG∘=−(2)(96,485)(0.47)=−90,700\Delta G^\circ = -(2)(96{,}485)(0.47) = -90{,}700 J, or -90.7 kJ per mole of reaction.
  • C. Incorrect. +90.7 kJ/mol has the right size but the wrong sign; a positive cell potential means a negative ΔG∘\Delta G^\circ.
  • D. Incorrect. -181 kJ/mol uses n = 4, counting the two electrons once for each half-reaction; only 2 mol of electrons move per mole of reaction.

Explain Your turn

In a galvanic cell, a zinc electrode sits in zinc nitrate solution and a copper electrode sits in copper(II) nitrate solution. The two beakers are joined by a wire and by a salt bridge filled with potassium nitrate solution. Explain what the salt bridge does, including which way the nitrate ions in it move.

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Compare with a model answer

A full answer does this

  • Says the salt bridge keeps each half-cell electrically neutral so the current can keep flowing
  • Says the nitrate ions move toward the zinc (anode) half-cell

Model answer

As the cell runs, zinc is oxidized at the anode, adding Zn2+\text{Zn}^{2+} ions to the zinc beaker, while copper(II) ions are reduced at the cathode, leaving extra nitrate in the copper beaker. The salt bridge lets ions move to balance these charges, keeping both solutions neutral so that electrons keep flowing through the wire. The nitrate ions in the bridge move toward the zinc (anode) beaker, and the potassium ions move toward the copper (cathode) beaker.

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