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AP Chemistry Flashcards: Cell Potential Under Nonstandard Conditions

Study Cell Potential Under Nonstandard Conditions in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Cell Potential Under Nonstandard Conditions, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Chemistry Flashcards: Cell Potential Under Nonstandard Conditions

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QUESTION

Calculate ΔG\Delta GΔG if n=2n=2n=2 and E=0.25 VE=0.25\ \text{V}E=0.25 V (use F=96485 C mol−1F=96485\ \text{C mol}^{-1}F=96485 C mol−1).

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ANSWER

ΔG=−4.82×104 J mol−1\Delta G=-4.82\times 10^4\ \text{J mol}^{-1}ΔG=−4.82×104 J mol−1. ΔG=−(2)(96485)(0.25)=−48242.5\Delta G=-(2)(96485)(0.25)=-48242.5ΔG=−(2)(96485)(0.25)=−48242.5 J/mol.

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Flashcard 1: Calculate ΔG\Delta GΔG if n=2n=2n=2 and E=0.25 VE=0.25\ \text{V}E=0.25 V (use F=96485 C mol−1F=96485\ \text{C mol}^{-1}F=96485 C mol−1).

Answer: ΔG=−4.82×104 J mol−1\Delta G=-4.82\times 10^4\ \text{J mol}^{-1}ΔG=−4.82×104 J mol−1. ΔG=−(2)(96485)(0.25)=−48242.5\Delta G=-(2)(96485)(0.25)=-48242.5ΔG=−(2)(96485)(0.25)=−48242.5 J/mol.

Flashcard 2: Calculate E∘E^\circE∘ at 25∘C25^\circ\text{C}25∘C if K=1.0×106K=1.0\times 10^6K=1.0×106 and n=2n=2n=2.

Answer: E∘=0.178 VE^\circ=0.178\ \text{V}E∘=0.178 V. E∘=0.05922log⁡(106)=0.0296×6E^\circ=\frac{0.0592}{2}\log(10^6)=0.0296\times 6E∘=20.0592​log(106)=0.0296×6.

Flashcard 3: Calculate KKK at 25∘C25^\circ\text{C}25∘C if E∘=0.30 VE^\circ=0.30\ \text{V}E∘=0.30 V and n=2n=2n=2.

Answer: K=1.4×1010K=1.4\times 10^{10}K=1.4×1010. 0.30=0.05922log⁡K0.30=\frac{0.0592}{2}\log K0.30=20.0592​logK; log⁡K=10.14\log K=10.14logK=10.14.

Flashcard 4: Find QQQ at 25∘C25^\circ\text{C}25∘C if E∘=0.50 VE^\circ=0.50\ \text{V}E∘=0.50 V, E=0.44 VE=0.44\ \text{V}E=0.44 V, and n=2n=2n=2.

Answer: Q=4.0×102Q=4.0\times 10^2Q=4.0×102. 0.44=0.50−0.05922log⁡Q0.44=0.50-\frac{0.0592}{2}\log Q0.44=0.50−20.0592​logQ; solve for QQQ.

Flashcard 5: Calculate EEE at 25∘C25^\circ\text{C}25∘C if E∘=0.80 VE^\circ=0.80\ \text{V}E∘=0.80 V, n=1n=1n=1, and Q=102Q=10^2Q=102.

Answer: E=0.68 VE=0.68\ \text{V}E=0.68 V. E=0.80−0.0592log⁡(100)=0.80−0.118E=0.80-0.0592\log(100)=0.80-0.118E=0.80−0.0592log(100)=0.80−0.118.

Flashcard 6: Calculate EEE at 25∘C25^\circ\text{C}25∘C if E∘=1.10 VE^\circ=1.10\ \text{V}E∘=1.10 V, n=2n=2n=2, and Q=10−4Q=10^{-4}Q=10−4.

Answer: E=1.22 VE=1.22\ \text{V}E=1.22 V. E=1.10−0.05922log⁡(10−4)=1.10+0.118E=1.10-\frac{0.0592}{2}\log(10^{-4})=1.10+0.118E=1.10−20.0592​log(10−4)=1.10+0.118.

Flashcard 7: What is nnn in the Nernst equation, and how is it determined from the reaction?

Answer: nnn is moles of electrons transferred in the balanced redox reaction. Count electrons in half-reactions after balancing.

Flashcard 8: What is the relationship between EEE and E∘E^\circE∘ when Q=1Q=1Q=1?

Answer: E=E∘E=E^\circE=E∘. Standard conditions: all activities equal 1.

Flashcard 9: What is the sign of the Nernst correction when Q<1Q<1Q<1 (relative to E∘E^\circE∘)?

Answer: E>E∘E>E^\circE>E∘. Reactants favored, so potential increases from standard.

Flashcard 10: What is the sign of the Nernst correction when Q>1Q>1Q>1 (relative to E∘E^\circE∘)?

Answer: E<E∘E<E^\circE<E∘. Products favored, so potential decreases from standard.

Flashcard 11: State the relationship between cell potential and Gibbs free energy under any conditions.

Answer: ΔG=−nFE\Delta G=-nFEΔG=−nFE. Negative sign shows spontaneous reactions have E>0E>0E>0.

Flashcard 12: Identify which species are omitted from QQQ: pure solids, pure liquids, aqueous ions, or gases.

Answer: Pure solids and pure liquids are omitted from QQQ. Only dissolved species and gases affect cell potential.

Flashcard 13: What does QQQ represent in the Nernst equation for an electrochemical cell?

Answer: Reaction quotient from activities in the balanced net ionic equation. Ratio of products to reactants raised to stoichiometric powers.

Flashcard 14: State the relationship between standard cell potential and standard Gibbs free energy.

Answer: ΔG∘=−nFE∘\Delta G^\circ=-nFE^\circΔG∘=−nFE∘. Links thermodynamics to electrochemistry at standard state.

Flashcard 15: State the relationship between E∘E^\circE∘ and the equilibrium constant KKK at 25∘C25^\circ\text{C}25∘C.

Answer: E∘=0.0592nlog⁡KE^\circ=\frac{0.0592}{n}\log KE∘=n0.0592​logK. Derived from ΔG∘=−RTln⁡K=−nFE∘\Delta G^\circ=-RT\ln K=-nFE^\circΔG∘=−RTlnK=−nFE∘.

Flashcard 16: What is the cell potential at equilibrium, and what is the corresponding value of QQQ?

Answer: E=0E=0E=0 and Q=KQ=KQ=K. No driving force at equilibrium; ΔG=0\Delta G=0ΔG=0.

Flashcard 17: Identify the direction of spontaneity when E>0E>0E>0 for the cell reaction as written.

Answer: Spontaneous as written. Positive EEE means ΔG<0\Delta G<0ΔG<0.

Flashcard 18: Identify the direction of spontaneity when E<0E<0E<0 for the cell reaction as written.

Answer: Nonspontaneous as written; spontaneous in reverse. Negative EEE means ΔG>0\Delta G>0ΔG>0 forward.

Flashcard 19: State the Nernst equation at 25∘C25^\circ\text{C}25∘C using base-10 logarithms.

Answer: E=E∘−0.0592nlog⁡QE=E^\circ-\frac{0.0592}{n}\log QE=E∘−n0.0592​logQ. Simplified form at 298 K using log⁡\loglog instead of ln⁡\lnln.

Flashcard 20: State the formula for the cell potential under nonstandard conditions (Nernst equation).

Answer: E=E∘−RTnFln⁡QE=E^\circ-\frac{RT}{nF}\ln QE=E∘−nFRT​lnQ. Modifies standard potential by concentration effects via ln⁡Q\ln QlnQ term.

Flashcard 21: Identify the condition on QQQ when E=0E=0E=0 for a cell at equilibrium.

Answer: Q=KQ=KQ=K. At equilibrium, E=0E=0E=0 and reaction quotient equals equilibrium constant.

Flashcard 22: State the relationship between E∘E^\circE∘ and the equilibrium constant KKK at 25∘C25^\circ\text{C}25∘C.

Answer: E∘=0.0592nlog⁡KE^\circ=\frac{0.0592}{n}\log KE∘=n0.0592​logK. Derived from E=0E=0E=0 when Q=KQ=KQ=K in Nernst equation.

Flashcard 23: State the general relationship between ΔG\Delta GΔG and cell potential EEE.

Answer: ΔG=−nFE\Delta G=-nFEΔG=−nFE. Relates electrical work to thermodynamic spontaneity.

Flashcard 24: State the relationship between ΔG∘\Delta G^\circΔG∘ and the equilibrium constant KKK.

Answer: ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln KΔG∘=−RTlnK. Fundamental thermodynamic relationship at equilibrium.

Flashcard 25: Identify how EEE changes when QQQ increases (with E∘E^\circE∘, TTT, and nnn constant).

Answer: EEE decreases. Larger QQQ means more products, driving reverse reaction.

Flashcard 26: Calculate log⁡K\log KlogK at 25∘C25^\circ\text{C}25∘C if E∘=0.30 VE^\circ=0.30\ \text{V}E∘=0.30 V and n=3n=3n=3.

Answer: log⁡K≈15.2\log K\approx15.2logK≈15.2. 0.30=0.05923log⁡K0.30=\frac{0.0592}{3}\log K0.30=30.0592​logK; log⁡K=0.900.0592≈15.2\log K=\frac{0.90}{0.0592}\approx15.2logK=0.05920.90​≈15.2.

Flashcard 27: State the formula for cell potential under nonstandard conditions (Nernst equation).

Answer: E=E∘−RTnFln⁡QE=E^\circ-\frac{RT}{nF}\ln QE=E∘−nFRT​lnQ. Modified by reaction quotient QQQ and temperature; nnn is electrons transferred.

Flashcard 28: State the Nernst equation at 25∘C25^\circ\text{C}25∘C using base-10 logarithms.

Answer: E=E∘−0.0592nlog⁡QE=E^\circ-\frac{0.0592}{n}\log QE=E∘−n0.0592​logQ. Simplified form at 298 K using 2.303RTF=0.0592\frac{2.303RT}{F}=0.0592F2.303RT​=0.0592 V.

Flashcard 29: What does QQQ represent in the Nernst equation for an electrochemical cell?

Answer: Reaction quotient, products over reactants using activities. Measures reaction progress; equals KKK at equilibrium.

Flashcard 30: What does nnn represent in the Nernst equation for a galvanic cell?

Answer: Moles of electrons transferred in the balanced redox reaction. Count electrons in half-reaction balancing.