Chemistry Clarity · chemistryclarity.com

Galvanic Cells and Electrode Potentials

How does a galvanic cell work and how do you calculate its voltage?

IntermediateRedox & ElectrochemistryLast reviewed 4 October 2026
Download lesson (PDF)

What is it?

A galvanic (or voltaic) cell uses a spontaneous redox reaction to produce electricity. The oxidation and the reduction happen in separate half-cells, so the electrons must travel through an external wire, where they can do useful work.

In the Daniell cell, a zinc electrode in ZnX2+\ce{Zn^2+} solution is connected to a copper electrode in CuX2+\ce{Cu^2+} solution:

  • Anode (zinc): oxidation, Zn→ZnX2++2 eX−\ce{Zn -> Zn^2+ + 2e-}. It is the negative electrode.
  • Cathode (copper): reduction, CuX2++2 eX−→Cu\ce{Cu^2+ + 2e- -> Cu}. It is the positive electrode.
  • A salt bridge lets ions move between the solutions to keep them electrically neutral.

Key idea

Electrons flow through the wire from the anode (oxidation) to the cathode (reduction). The cell voltage is E°cell=E°cathode−E°anodeE°_\text{cell} = E°_\text{cathode} - E°_\text{anode}; a positive E°cellE°_\text{cell} means the reaction is spontaneous.

Why does it matter?

  • Batteries. Every battery, from AA cells to lithium-ion packs, is a galvanic cell (or several in series).
  • Corrosion. Rusting is an unwanted galvanic process; zinc coatings protect steel because zinc is oxidized first.
  • Industry. Electrolysis (the reverse process, driven by electricity) produces aluminium, chlorine and copper, and electroplates metals.

How does it work?

1. Standard reduction potentials

Each half-cell has a standard reduction potential, E°E°, measured against the standard hydrogen electrode (defined as 0.00 V) at 25 °C, 1 M and 1 bar. A more positive E°E° means a stronger oxidizing agent (more easily reduced).

Half-reaction (reduction)E°E° (V)
FX2+2 eX−→2 FX−\ce{F2 + 2e- -> 2F-}+2.87
MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}+1.51
ClX2+2 eX−→2 ClX−\ce{Cl2 + 2e- -> 2Cl-}+1.36
AgX++eX−→Ag\ce{Ag+ + e- -> Ag}+0.80
FeX3++eX−→FeX2+\ce{Fe^3+ + e- -> Fe^2+}+0.77
CuX2++2 eX−→Cu\ce{Cu^2+ + 2e- -> Cu}+0.34
2 HX++2 eX−→HX2\ce{2H+ + 2e- -> H2}0.00
NiX2++2 eX−→Ni\ce{Ni^2+ + 2e- -> Ni}−0.25
ZnX2++2 eX−→Zn\ce{Zn^2+ + 2e- -> Zn}−0.76
AlX3++3 eX−→Al\ce{Al^3+ + 3e- -> Al}−1.66
MgX2++2 eX−→Mg\ce{Mg^2+ + 2e- -> Mg}−2.37

2. Cell voltage

E°cell=E°cathode−E°anodeE°_\text{cell} = E°_\text{cathode} - E°_\text{anode}

Use both values as reduction potentials from the table, and do not multiply E°E° by the coefficients: a potential does not depend on how much substance reacts.

3. Cell notation

The anode is written on the left and the cathode on the right; a single line | separates phases and a double line || shows the salt bridge:

Zn(s) ∣ ZnX2+(aq) ∣∣ CuX2+(aq) ∣ Cu(s)\ce{Zn(s) | Zn^2+(aq) || Cu^2+(aq) | Cu(s)}

4. Free energy and spontaneity

The cell voltage is linked to the Gibbs free energy change:

ΔG°=−nFE°cell\Delta G° = -nFE°_\text{cell}

where nn is the moles of electrons transferred and F=96 485F = 96\,485 C/mol (the Faraday constant, the charge on one mole of electrons). Positive E°cellE°_\text{cell} gives negative ΔG°\Delta G°: a spontaneous reaction.

5. Electrolysis

In electrolysis, an external power supply forces a non-spontaneous redox reaction. The amount of product follows from the charge passed (Faraday’s law):

Q=Itn(e−)=QFQ = It \qquad n(e^-) = \frac{Q}{F}

where II is the current in amperes (A = C/s) and tt is the time in seconds.

Think of it like this

A galvanic cell is like water flowing downhill through a water wheel: the reaction “wants” to happen, and the cell makes it do work on the way. Electrolysis is like pumping water back uphill: it needs an external energy source.

More precisely

Standard potentials apply at 1 M concentrations and 25 °C. At other concentrations, the Nernst equation, E=E°−RTnFln⁡QE = E° - \dfrac{RT}{nF}\ln Q, gives the actual cell voltage; it falls as the cell runs and Q increases, reaching zero when the reaction is at equilibrium (a “flat” battery).

Visualise it

The Daniell cell. On the left, a zinc electrode in Zn2+ solution is the anode (negative), where oxidation happens: Zn → Zn2+ + 2e−. On the right, a copper electrode in Cu2+ solution is the cathode (positive), where reduction happens: Cu2+ + 2e− → Cu. A wire through a voltmeter reading 1.10 V joins the electrodes, and electrons flow from zinc to copper. A KNO3 salt bridge connects the two solutions. Cell notation: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s).
Oxidation at the anode, reduction at the cathode, electrons through the wire, ions through the salt bridge.

Worked example

Worked example: Cell voltage and ΔG°

Question: A cell is made from Ag⁺/Ag and Cu²⁺/Cu half-cells. Find E°cellE°_\text{cell}, write the overall reaction, and calculate ΔG°\Delta G°.

  1. The more positive E°E° (Ag⁺/Ag, +0.80 V) is reduced at the cathode; copper (+0.34 V) is oxidized at the anode.

  2. E°cell=+0.80 V−(+0.34 V)=+0.46 VE°_\text{cell} = +0.80\ \text{V} - (+0.34\ \text{V}) = +0.46\ \text{V}

  3. Overall: Cu+2 AgX+→CuX2++2 Ag\ce{Cu + 2Ag+ -> Cu^2+ + 2Ag}, so n=2n = 2 mol of electrons.

  4. Substitute (1 C × 1 V = 1 J):

    ΔG°=−(2 mol)×(96 485 Cmol)×(0.46 V)=−8.9×104 J=−89 kJ\begin{aligned} &\Delta G° = -(2\ \text{mol}) \\[4pt] &\quad \times \left(96\,485\ \tfrac{\text{C}}{\text{mol}}\right) \\[4pt] &\quad \times (0.46\ \text{V}) \\[4pt] &= -8.9 \times 10^{4}\ \text{J} \\[4pt] &= -89\ \text{kJ} \end{aligned}

  5. E°cellE°_\text{cell} is positive and ΔG°\Delta G° negative: the reaction is spontaneous.

Worked example: Electroplating copper

Question: A current of 2.00 A passes through copper(II) sulfate solution for 30.0 min. What mass of copper is deposited? (CuX2++2 eX−→Cu\ce{Cu^2+ + 2e- -> Cu}; Cu = 63.55 g/mol)

  1. t=30.0 min×60 s1 min=1800 st = 30.0\ \text{min} \times \dfrac{60\ \text{s}}{1\ \text{min}} = 1800\ \text{s}
  2. Q=It=2.00 C/s×1800 s=3600 CQ = It = 2.00\ \text{C/s} \times 1800\ \text{s} = 3600\ \text{C}
  3. n(e−)=3600 C96 485 C/mol=0.03731 moln(e^-) = \dfrac{3600\ \text{C}}{96\,485\ \text{C/mol}} = 0.03731\ \text{mol}
  4. n(Cu)=0.03731 mol e−×1 mol Cu2 mol e−=0.01866 moln(\ce{Cu}) = 0.03731\ \text{mol}\ e^- \times \dfrac{1\ \text{mol Cu}}{2\ \text{mol}\ e^-} = 0.01866\ \text{mol}
  5. m=0.01866 mol×63.55 g/mol=m = 0.01866\ \text{mol} \times 63.55\ \text{g/mol} = 1.19 g

Common mistake

Common mistake: Multiplying E° by the coefficients

In Cu+2 AgX+→CuX2++2 Ag\ce{Cu + 2Ag+ -> Cu^2+ + 2Ag}, use E°=+0.80E° = +0.80 V for silver, not 2 × 0.80 V. Potentials are intensive: they don’t depend on amount.

Common mistake: Flipping signs twice

Use E°cell=E°cathode−E°anodeE°_\text{cell} = E°_\text{cathode} - E°_\text{anode} with both values copied straight from the reduction table. Don’t also change the sign of the anode value, or you will subtract twice.

Common mistake: Mixing up anode and cathode

In any cell, oxidation happens at the anode and reduction at the cathode (“an ox, red cat”). In a galvanic cell the anode is negative; in electrolysis it is positive.

Notation note

  • 1 V = 1 J/C, so C × V = J; 1 A = 1 C/s.
  • FF = 96 485 C/mol is sometimes rounded to 96 500 C/mol.
  • “SHE” = standard hydrogen electrode, the zero of the potential scale.

Remember this

Remember this

  • Galvanic cell: spontaneous redox makes electricity; anode = oxidation (−), cathode = reduction (+).
  • E°cell=E°cathode−E°anodeE°_\text{cell} = E°_\text{cathode} - E°_\text{anode}; positive means spontaneous. Don’t multiply E° by coefficients.
  • ΔG°=−nFE°cell\Delta G° = -nFE°_\text{cell} with F=96 485F = 96\,485 C/mol.
  • Electrolysis: Q=ItQ = It, n(e−)=Q/Fn(e^-) = Q/F, then use the half-equation’s mole ratio.

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    What is a galvanic cell?
    Answer

    A cell that uses a spontaneous redox reaction to produce electricity.

  2. Question
    What happens at the anode and the cathode?
    Answer

    Anode: oxidation. Cathode: reduction. ("An ox, red cat")

  3. Question
    Which way do electrons flow in the external wire?
    Answer

    From the anode to the cathode.

  4. Question
    What does the salt bridge do?
    Answer

    Lets ions move between the half-cells to keep each solution electrically neutral.

  5. Question
    Formula for E°cell?
    Answer

    E°cell=E°cathode−E°anodeE°_\text{cell} = E°_\text{cathode} - E°_\text{anode} (both as reduction potentials)

  6. Question
    E°cell for the Daniell cell (Cu²⁺/Cu +0.34 V, Zn²⁺/Zn −0.76 V)?
    Answer

    +0.34 V − (−0.76 V) = +1.10 V

  7. Question
    Write the cell notation for the Daniell cell.
    Answer

    Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

  8. Question
    How are ΔG° and E°cell related?
    Answer

    ΔG°=−nFE°cell\Delta G° = -nFE°_\text{cell}, with F = 96 485 C/mol; positive E° means negative ΔG° (spontaneous).

  9. Question
    Should E° be multiplied by the coefficients in the equation?
    Answer

    No. Potentials do not depend on the amount of substance.

  10. Question
    In electrolysis, how do you find the moles of electrons?
    Answer

    Q=ItQ = It (charge in C), then n(e−)=Q/Fn(e^-) = Q/F.

Quiz

7 questions

  1. Question 1EasyIn a galvanic cell, where does oxidation take place?
    Show answer

    Answer: At the anode

    Oxidation always occurs at the anode and reduction at the cathode.

  2. Question 2MediumUsing E°(Ag⁺/Ag) = +0.80 V and E°(Zn²⁺/Zn) = −0.76 V, what is E°cell for a zinc–silver cell?
    Show answer

    Answer: +1.56 V

    Silver is reduced (cathode): E°cell = +0.80 V − (−0.76 V) = +1.56 V. Doubling the silver value to 1.60 V would be wrong; the answer 2.36 V does that.

  3. Question 3MediumWhich species is the strongest oxidizing agent: Ag⁺ (+0.80 V), Cu²⁺ (+0.34 V), Zn²⁺ (−0.76 V) or Mg²⁺ (−2.37 V)?
    Show answer

    Answer: Ag⁺

    The most positive reduction potential means the species is most easily reduced, so it is the strongest oxidizing agent.

  4. Question 4HardWill copper metal react with 1 M Zn²⁺ solution? (Cu²⁺/Cu +0.34 V, Zn²⁺/Zn −0.76 V)
    Show answer

    Answer: No, E°cell = −1.10 V

    For Cu + Zn²⁺ → Cu²⁺ + Zn, Zn²⁺ would be the cathode: E°cell = −0.76 V − (+0.34 V) = −1.10 V. Negative means not spontaneous.

  5. Question 5MediumWhat is ΔG° for the Daniell cell (E°cell = +1.10 V, n = 2)?
    Show answer

    Answer: −212 kJ

    ΔG° = −(2 mol)(96 485 C/mol)(1.10 V) = −2.12 × 10⁵ J = −212 kJ (C × V = J). The answer −106 kJ uses n = 1.

  6. Question 6EasyWhat is the purpose of the salt bridge?
    Show answer

    Answer: To let ions flow and keep the solutions neutral

    Electrons travel through the wire; ions move through the salt bridge. Without it, charge would build up and the current would stop.

  7. Question 7MediumA current of 2.00 A flows for 30.0 min. How much charge passes?
    Show answer

    Answer: 3600 C

    Q = It = 2.00 C/s × (30.0 min × 60 s/min) = 2.00 C/s × 1800 s = 3600 C. The answer 60.0 C forgets to convert minutes to seconds.

Notes and downloads

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/galvanic-cells/

Spotted a mistake? Let us know and we'll fix it.