Topic 9.10 Notes – Cell Potential Under Nonstandard Conditions

1. What Cell Potential Under Nonstandard Conditions Means

$E^\circ_{\text{cell}}$ is the voltage under standard conditions:

• 1 M solutions
• 1 atm gases
• 298 K
• $Q = 1$

$E_{\text{cell}}$ is the actual voltage at whatever concentrations you have right now.

Here’s the big idea you need to internalize:

• A working galvanic cell is not at equilibrium.
• The cell potential is a driving force toward equilibrium.
• As the reaction proceeds, the system moves closer to equilibrium and the magnitude of $E_{\text{cell}}$ decreases.
• At equilibrium:
  • $Q = K$
  • $E_{\text{cell}} = 0$
  • The battery is “dead.”

That “dead battery” idea shows up in conceptual questions a lot. If voltage is zero, the reaction has reached equilibrium.

Now we need the tool that connects concentration to voltage. That tool is $Q$.

2. The Reaction Quotient Q in Electrochemical Cells

For a redox reaction,

$$Q = \frac{\text{products}}{\text{reactants}}$$

Each raised to their stoichiometric coefficients.
Solids and pure liquids are omitted.

At standard conditions:

• All aqueous concentrations are 1 M
• So $Q = 1$
• Therefore $E_{\text{cell}} = E^\circ_{\text{cell}}$

Now compare $Q$ to 1:

When $Q < 1$

• Fewer products relative to reactants
• Reaction is far from equilibrium
• The system has a strong “push” forward
• $E_{\text{cell}} > E^\circ_{\text{cell}}$

When $Q > 1$

• More products than at standard conditions
• Closer to equilibrium
• Smaller driving force
• $E_{\text{cell}} < E^\circ_{\text{cell}}$

As $Q \to K$

• The system approaches equilibrium
• $E_{\text{cell}} \to 0$

This pattern is one of the most tested conceptual relationships in Unit 9.

Important warning

Do not use Le Châtelier’s principle language here. A running electrochemical cell is not at equilibrium. On FRQs, using Le Châtelier reasoning instead of $Q$ reasoning can cost you points.

3. The Nernst Equation and What It Tells You

The Nernst equation makes the relationship precise:

$$E = E^\circ - \frac{RT}{nF}\ln Q$$

You are not expected to grind through heavy calculations on the AP exam. You are expected to reason from it.

Look at the structure:

• If $Q$ increases → $\ln Q$ increases → more is subtracted → $E$ decreases
• If $Q$ decreases → $\ln Q$ decreases → less is subtracted → $E$ increases

At equilibrium:

$$0 = E^\circ - \frac{RT}{nF}\ln K$$

This links everything together:

• $E^\circ > 0 \Rightarrow K > 1$ (products favored)
• $E^\circ < 0 \Rightarrow K < 1$

And don’t forget the energy relationship:

$$\Delta G = -nFE$$

• Positive $E$ → negative $\Delta G$ → spontaneous
  Voltage is just another measure of thermodynamic driving force.

4. How Concentration Changes Affect Cell Potential

When a question changes concentration, do this mentally:

1. Write the expression for $Q$.
2. Decide if the change makes $Q$ bigger or smaller.
3. Compare $Q$ to 1 or to $K$.
4. Predict how $E$ changes.

Example reasoning

If the reaction is:

$$\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}$$

$$Q = \frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]}$$

• Increase $[\ce{Zn^{2+}}]$ → $Q$ increases → $E$ decreases.
• Increase $[\ce{Cu^{2+}}]$ → $Q$ decreases → $E$ increases.
• Change the mass of Zn(s) → no effect on $Q$ → no change in $E$.

Students often miss that changing the size of a solid electrode does not change voltage. Only concentrations in $Q$ matter.

5. Concentration Cells and Direction of Electron Flow

A concentration cell uses the same half-reaction in both half-cells but at different ion concentrations.

Since the substances are the same:

• $E^\circ = 0$
• Voltage comes entirely from concentration differences

Here is a typical metal/metal-ion concentration cell using zinc:

Zinc concentration cell

Electrons flow in the direction that makes concentrations equal.

• From the lower ion concentration half-cell
• Toward the higher ion concentration half-cell

The reaction proceeds until:

• Concentrations equalize
• $Q = 1$
• $E = 0$

The system always runs toward equilibrium.