Chemistry Labs

Grade 12

Galvanic cells, electrode potentials, Nernst equation

How a spontaneous redox reaction can drive electrons through a wire, and how the voltage is predicted.

1 Lower secondary2 Upper secondary3 Undergraduate4 Advanced5 Research

IntuitionIntuition: separate the two halves of a redox reaction

Put a zinc strip in a copper(II) solution and zinc dissolves while copper plates out: electrons pass directly from Zn\ce{Zn} to CuX2+\ce{Cu^2+}. If the two metals sit in separate beakers, joined by a wire and a salt bridge, the electrons are forced to travel through the wire, and that current can do work. This is a galvanic cell.

3D galvanic cell with two beakers, electrodes, a wire with moving electrons, a salt bridge and a voltage readout.
Change the electrode pair and read the voltage. Move lg Q and the extent of reaction to see the Nernst equation at work.

SchoolSchool level: electrodes and voltage

In every cell oxidation happens at the anode (the negative pole of a galvanic cell) and reduction at the cathode (the positive pole). For the Daniell cell: Zn→ZnX2++2 eX−\ce{Zn -> Zn^2+ + 2e-} and CuX2++2 eX−→Cu\ce{Cu^2+ + 2e- -> Cu}. The salt bridge lets ions move so that neither beaker builds up charge.

Standard reduction potentials at 25 °C
Half-reactionE° (V)
ZnX2++2 eX−→Zn\ce{Zn^2+ + 2e- -> Zn}−0.76
FeX2++2 eX−→Fe\ce{Fe^2+ + 2e- -> Fe}−0.44
CuX2++2 eX−→Cu\ce{Cu^2+ + 2e- -> Cu}+0.34
AgX++eX−→Ag\ce{Ag+ + e- -> Ag}+0.80
Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

Example: The Daniell cell

Find the standard voltage of the Zn / Cu cell and the standard Gibbs energy of its reaction.

Solution

E∘=0.34−(−0.76)=1.10E^\circ = 0.34 - (-0.76) = 1.10 V. With n=2n = 2: ΔG∘=−nFE∘=−2×96485×1.10≈−212\Delta G^\circ = -nFE^\circ = -2 \times 96485 \times 1.10 \approx -212 kJ/mol, negative, so the reaction is spontaneous.

UndergraduateUndergraduate: the Nernst equation

E=E∘−RTnFln⁡Q  ≈  E∘−0.05916nlg⁡Q(25 ∘C)E = E^\circ - \frac{RT}{nF}\ln Q \;\approx\; E^\circ - \frac{0.05916}{n}\lg Q \quad (25\,^\circ\text{C})

Away from standard concentrations the voltage follows the reaction quotient QQ, here [ZnX2+]/[CuX2+][\ce{Zn^2+}]/[\ce{Cu^2+}]. It falls as the cell discharges (QQ grows) and reaches E=0E = 0 at equilibrium, when Q=KQ = K and ΔG=0\Delta G = 0: a flat battery.

Example: Diluted zinc side

Find EE of the Daniell cell when [ZnX2+]=0.010[\ce{Zn^2+}] = 0.010 mol/L and [CuX2+]=1.0[\ce{Cu^2+}] = 1.0 mol/L.

Solution

Q=0.010Q = 0.010, so lg⁡Q=−2\lg Q = -2 and E=1.10−0.059162(−2)≈1.16E = 1.10 - \frac{0.05916}{2}(-2) \approx 1.16 V. Set lg⁡Q=−2\lg Q = -2 in the simulation to check it.