E°cell = +1.10 V — spontaneous as written (a galvanic cell)
Overall: Cu²⁺ + Zn → Cu + Zn²⁺
- Electrons (n)
- 2
- ΔG°
- -212.3 kJ/mol
- K
- 1.54 × 10³⁷
- E (Nernst)
- 1.1000 V
- E°cell = E°cathode − E°anode = 0.34 − (-0.76) =
1.10 V - ΔG° = −nFE° = −2 × 96,485 × 1.10 ÷ 1000 =
-212.3 kJ/mol - E = E° − (RT ÷ nF) ln Q at 298.15 K
How it works
E°cell = E°cathode − E°anode, using standard reduction potentials for both. The half-reactions are multiplied so that the electrons lost at the anode equal those gained at the cathode, giving n and the overall equation. Then ΔG° = −nFE° and K = e^(nFE°/RT).
The Nernst equation, E = E° − (RT ÷ nF) ln Q, gives the potential when concentrations aren't 1 M. At 25 °C, RT ÷ F = 0.025693 V. Potentials in the table are standard values at 25 °C versus the standard hydrogen electrode.
Frequently asked questions
- What is the voltage of a Daniell cell?
- Cu²⁺/Cu (+0.34 V) as cathode and Zn²⁺/Zn (−0.76 V) as anode: E° = 0.34 − (−0.76) = +1.10 V.
- How do I know which electrode is the cathode?
- In a galvanic cell, the half-reaction with the more positive reduction potential is the cathode. If E°cell comes out negative, swap them.
- Do I multiply E° when I multiply a half-reaction?
- No. Electrode potentials are intensive: they don't change when a half-equation is multiplied. Only n and ΔG° scale.
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