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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

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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

ConditionEffect on E
Q < 1 (more reactants)E greater than E°
Q = 1E = E°
Q > 1 (more products)E less than E°
Q = KE = 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

TermMeaning
Nernst equationRelates cell potential to standard potential and Q.
Reaction quotientRatio of product to reactant concentrations at a given moment.
Concentration cellCell whose potential arises only from a concentration difference.
Dead cellCell 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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