Cell Potential Under Nonstandard Conditions - AP Chemistry

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

9. What does $Q$ represent in the Nernst equation for an electrochemical cell?

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

10. What is the cell potential at equilibrium, and what is the corresponding value of $Q$?

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

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

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

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

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

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

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

14. What does $n$ represent in the Nernst equation for a galvanic cell?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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