9.10Thermodynamics and Electrochemistry
Cell Potential Under Nonstandard Conditions — AP CHEMISTRY Revision Notes & Topical Questions
5 min read 9 parts Lesson 14 of 15 1 quiz
Lesson summary
9.10 Cell Potential Under Nonstandard Conditions — AP CHEMISTRY notes: Use Q and the Nernst equation to find how concentrations change the cell potential.Standard cell — all concentrations 1 M (Q = 1), so E = E°Nonstandard cell — concentrations differ, so E ≠ E°Le Châtelier — qua…
- [Zn2+] = 5.00 M and [Cu2+] = 1.00 M
- [Zn2+] = 1.00 M and [Cu2+] = 5.00 M
- Definition
- Fe3+/Fe2+ with 0.034 M and 0.64 M
- Cu2+/Cu with 0.0010 M
- Reaction quotient
Section 1 of 9Standard and nonstandard cells
Standard cell
- Standard cell all concentrations 1 M (Q = 1), so E = E°
Nonstandard cell
- Nonstandard cell concentrations differ, so E ≠ E°
Le Châtelier
- Le Châtelier qualitative equilibrium arguments are applied through Q and E, not by shifting the cell
Section 2 of 9Nernst equation
- Equation
- E = E° − (RT/nF) ln Q
- At 298 K this becomes E = E° − (0.0592/n) log Q.
- n is the number of electrons transferred and F = 96 485 C mol-1.
- Q is built from aqueous and gaseous species only; solids are omitted.
Section 3 of 9Effect of Q on E
| Condition | Effect on E |
|---|---|
| Q < 1 (more reactants) | E greater than E° |
| Q = 1 | E = E° |
| Q > 1 (more products) | E less than E° |
| Q = K | E = 0; the cell is dead |
Section 4 of 9Pattern: Zn/Cu cell (E° = 1.10 V)
[Zn2+]
5.00 M and [Cu2+] = 1.00 M
- Q = 5, E = 1.10 − (0.0592/2) log 5 = 1.08 V
[Zn2+]
1.00 M and [Cu2+] = 5.00 M
- Q = 0.2, E = 1.12 V
Section 5 of 9Concentration cells
Definition
- Definition both half-cells contain the same species at different concentrations, so E° = 0
Direction
- Direction oxidation at the more dilute half-cell and reduction at the more concentrated half-cell
Pattern
- 0.50 M and 1.00 M Zn2+: Q = 0.50 and E = −(0.0592/2) log 0.50 = +0.0089 V
Over time
- Over time concentrations approach each other, Q approaches 1 and E falls to zero
Section 6 of 9Single half-cell potential
1Fe3+/Fe2+ with 0.034 M and 0.64 M
- Fe3+/Fe2+ with 0.034 M and 0.64 M for reduction Q = [Fe2+]/[Fe3+], so E = 0.77 − 0.0592 log(0.64/0.034) = +0.69 V
2Cu2+/Cu with 0.0010 M
- Q = 1/[Cu2+], so E = 0.34 − (0.0592/2) log(1/0.0010) = +0.25 V
Section 7 of 9More from the specification
More from the specification
Electrochemical oxidation at the anode
- Electrochemical oxidation at the anode metal atoms lose electrons at the sacrificial anode, releasing hydrated cations into the surrounding aqueous solution.
Electrolysis
- The decomposition of a molten or aqueous ionic compound by an electric current.
Charge
- Q = It, measured in coulombs.What the symbols mean: Q = charge (C) · I = current (A) · t = time (s)
Electrolytic cell
- A cell using electricity to drive a non-spontaneous reaction.
Reaction quotient
- The ratio of product to reactant concentrations at any moment.
Electrochemical cell
- A device linking redox reactions and electricity.
Faraday constant
- The charge of one mole of electrons, 96 500 C mol-1.
Voltaic cell
- A cell that generates electricity from a spontaneous redox reaction.
Cathode
- The negative electrode, where reduction occurs.
Section 8 of 9Key terms
| Term | Meaning |
|---|---|
| Nernst equation | Relates cell potential to standard potential and Q. |
| Reaction quotient | Ratio of product to reactant concentrations at a given moment. |
| Concentration cell | Cell whose potential arises only from a concentration difference. |
| Dead cell | Cell at equilibrium with E = 0. |
Section 9 of 9Exam technique
1
Including solids in Q.
2
Using log in place of ln without changing the constant.
3
Using Q upside down — Q is products over reactants for the written cell reaction.
4
Thinking the cell potential is zero at standard conditions.
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