AP Chemistry Flashcards: Cell Potential Under Nonstandard Conditions

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.

Flashcard 1: Identify the condition on $Q$ when $E=E^\circ$ at a fixed temperature.

Answer: $Q=1$. When all species at unit activity, $\ln(1)=0$.

Flashcard 2: State the relationship between $E^\circ$ and the equilibrium constant $K$ at $25^\circ\text{C}$.

Answer: $E^\circ=\frac{0.0592}{n}\log K$. Derived from $\Delta G^\circ=-RT\ln K=-nFE^\circ$.

Flashcard 3: State the formula for the cell potential under nonstandard conditions (Nernst equation).

Answer: $E=E^\circ-\frac{RT}{nF}\ln Q$. Modifies standard potential by concentration effects via $\ln Q$ term.

Flashcard 4: Identify how $E$ changes when product concentrations increase (others constant).

Answer: $Q$ increases, so $E$ decreases. More products favor reverse reaction, lowering potential.

Flashcard 5: State the Nernst equation at $25^\circ\text{C}$ using base-10 logarithms.

Answer: $E=E^\circ-\frac{0.0592}{n}\log Q$. Simplified form at 298 K using $\log$ instead of $\ln$.

Flashcard 6: Calculate $E$ at $25^\circ\text{C}$: $E^\circ=1.10\ \text{V}$, $n=2$, $Q=10^{-4}$.

Answer: $1.22\ \text{V}$. $E=1.10-\frac{0.0592}{2}\log(10^{-4})=1.10+0.118=1.22$ V.

Flashcard 7: What is the relationship between $E$ and $E^\circ$ when $Q=1$?

Answer: $E=E^\circ$. Standard conditions: all activities equal 1.

Flashcard 8: Identify the condition on $Q$ when $E=0$ for a cell at equilibrium.

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

Flashcard 9: What does $Q$ 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 10: What is the cell potential at equilibrium, and what is the corresponding value of $Q$?

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

Flashcard 11: Identify how $E$ changes when $Q$ increases (with $E^\circ$, $T$, and $n$ constant).

Answer: $E$ decreases. Larger $Q$ means more products, driving reverse reaction.

Flashcard 12: Calculate $E$ at $25^\circ\text{C}$: $E^\circ=0.80\ \text{V}$, $n=1$, $Q=10^{3}$.

Answer: $0.62\ \text{V}$. $E=0.80-\frac{0.0592}{1}\log(10^3)=0.80-0.178=0.62$ V.

Flashcard 13: Find $Q$ at $25^\circ\text{C}$ if $E^\circ=0.50\ \text{V}$, $E=0.44\ \text{V}$, and $n=2$.

Answer: $Q=4.0\times 10^2$. $0.44=0.50-\frac{0.0592}{2}\log Q$; solve for $Q$.

Flashcard 14: What does $n$ 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.

Flashcard 15: Calculate $E$ at $25^\circ\text{C}$ if $E^\circ=1.10\ \text{V}$, $n=2$, and $Q=10^{-4}$.

Answer: $E=1.22\ \text{V}$. $E=1.10-\frac{0.0592}{2}\log(10^{-4})=1.10+0.118$.

Flashcard 16: State the formula for cell potential under nonstandard conditions (Nernst equation).

Answer: $E=E^\circ-\frac{RT}{nF}\ln Q$. Modified by reaction quotient $Q$ and temperature; $n$ is electrons transferred.

Flashcard 17: What is the activity (effective concentration) of a pure solid or pure liquid in $Q$?

Answer: $1$. Pure solids/liquids have constant activity by definition.

Flashcard 18: Calculate $E^\circ$ at $25^\circ\text{C}$ if $K=1.0\times 10^6$ and $n=2$.

Answer: $E^\circ=0.178\ \text{V}$. $E^\circ=\frac{0.0592}{2}\log(10^6)=0.0296\times 6$.

Flashcard 19: Calculate $\log K$ at $25^\circ\text{C}$ if $E^\circ=0.30\ \text{V}$ and $n=3$.

Answer: $\log K\approx15.2$. $0.30=\frac{0.0592}{3}\log K$; $\log K=\frac{0.90}{0.0592}\approx15.2$.

Flashcard 20: Identify how $E$ changes when reactant concentrations increase (others constant).

Answer: $Q$ decreases, so $E$ increases. More reactants drive forward reaction, raising potential.

Flashcard 21: What is the activity used for a gas in $Q$ for AP Chemistry calculations?

Answer: Use partial pressure, typically $P$ in atm, in place of activity. Gas activity approximated by pressure in atmospheres.

Flashcard 22: Calculate $E$ at $25^\circ\text{C}$ if $E^\circ=0.80\ \text{V}$, $n=1$, and $Q=10^2$.

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

Flashcard 23: Identify which species are omitted from $Q$: pure solids, pure liquids, aqueous ions, or gases.

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

Flashcard 24: State the relationship between $\Delta G^\circ$ and the equilibrium constant $K$.

Answer: $\Delta G^\circ=-RT\ln K$. Fundamental thermodynamic relationship at equilibrium.

Flashcard 25: Calculate $K$ at $25^\circ\text{C}$ if $E^\circ=0.30\ \text{V}$ and $n=2$.

Answer: $K=1.4\times 10^{10}$. $0.30=\frac{0.0592}{2}\log K$; $\log K=10.14$.

Flashcard 26: Calculate $\Delta G$ if $n=2$ and $E=0.25\ \text{V}$ (use $F=96485\ \text{C mol}^{-1}$).

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

Flashcard 27: Find $Q$ at $25^\circ\text{C}$ if $E^\circ=0.50\ \text{V}$, $E=0.44\ \text{V}$, and $n=2$.

Answer: $Q\approx10^{2}$. $0.44=0.50-\frac{0.0592}{2}\log Q$; $\log Q=2.03$, so $Q\approx10^2$.

Flashcard 28: Identify the direction of spontaneity when $E<0$ for the cell reaction as written.

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

Flashcard 29: Identify the correct sign of $E$ for a spontaneous galvanic cell under given conditions.

Answer: $E>0$. Positive potential indicates spontaneous electron flow.

Flashcard 30: What is the sign of the Nernst correction when $Q>1$ (relative to $E^\circ$)?

Answer: $E<E^\circ$. Products favored, so potential decreases from standard.

Flashcard 31: State the general relationship between $\Delta G$ and cell potential $E$.

Answer: $\Delta G=-nFE$. Relates electrical work to thermodynamic spontaneity.

Flashcard 32: Identify the direction of spontaneity when $E>0$ for the cell reaction as written.

Answer: Spontaneous as written. Positive $E$ means $\Delta G<0$.

Flashcard 33: What does $Q$ represent in the Nernst equation for an electrochemical cell?

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

Flashcard 34: What is $n$ in the Nernst equation, and how is it determined from the reaction?

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

Flashcard 35: What is the sign of the Nernst correction when $Q<1$ (relative to $E^\circ$)?

Answer: $E>E^\circ$. Reactants favored, so potential increases from standard.

Flashcard 36: State the Nernst equation at $25^\circ\text{C}$ using base-10 logarithms.

Answer: $E=E^\circ-\frac{0.0592}{n}\log Q$. Simplified form at 298 K using $\frac{2.303RT}{F}=0.0592$ V.

Flashcard 37: State the relationship between standard cell potential and standard Gibbs free energy.

Answer: $\Delta G^\circ=-nFE^\circ$. Links thermodynamics to electrochemistry at standard state.

Flashcard 38: State the relationship between $E^\circ$ and the equilibrium constant $K$ at $25^\circ\text{C}$.

Answer: $E^\circ=\frac{0.0592}{n}\log K$. Derived from $E=0$ when $Q=K$ in Nernst equation.

Flashcard 39: State the relationship between cell potential and Gibbs free energy under any conditions.

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