Cell Potential Under Nonstandard Conditions

Emily Wilson

8 min read

Next Topic - Electrolysis and Faraday's Law

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Study Guide Overview

This study guide covers non-standard conditions and electrochemical cells, focusing on the relationship between E°_cell (standard cell potential), E_cell (non-standard cell potential), and the Nernst Equation. It explains how concentration affects cell potential and how to use the Nernst Equation to predict changes in E_cell under non-standard conditions, including at equilibrium. The guide also connects E_cell, Gibbs Free Energy (ΔG), and the equilibrium constant (K). Finally, it provides practice questions and exam tips.

#Non-Standard Conditions & Electrochemical Cells 🔋

Hey there, future AP Chem master! Let's dive into the world of electrochemical cells outside of standard conditions. It's all about how concentration, cell potential, and equilibrium play together. Get ready to make some awesome predictions!

#Comparing E_cell and E°_cell

Remember, standard conditions are 298.15 K, 1 atm, and 1 M concentrations (Q = 1). But what happens when we're not at these perfect conditions? That's where things get interesting!

E°_cell is the cell potential under standard conditions.
E_cell is the cell potential under non-standard conditions.

Key Concept

A running galvanic cell is not at equilibrium. It's moving towards equilibrium. The further from equilibrium, the greater the cell potential. When the cell reaches equilibrium, E_cell = 0 (dead battery!).

Memory Aid

Think of a battery like a spring. When it's fully wound (far from equilibrium), it has lots of potential (high voltage). As it unwinds (moves towards equilibrium), the potential decreases until it's completely unwound (dead battery, E_cell = 0).

#The Nernst Equation

The Nernst Equation is our go-to tool for understanding how non-standard conditions affect cell potential. It relates E_cell to E°_cell and the reaction quotient, Q.

$E_{cell} = E^\circ_{cell} - \frac{RT}{nF}lnQ$

$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n}logQ$

R = 8.314 J/mol·K (gas constant)
T = Temperature in Kelvin
n = moles of electrons transferred in the balanced redox reaction
F = Faraday's constant (96485 C/mol e-)

Exam Tip

On the AP exam, you won't need to calculate exact values using the Nernst Equation. Instead, focus on making predictions about how changes in Q, E°_cell, and E_cell will affect each other.

#Concentration and Cell Potential

Concentration is a major player in determining cell potential under non-standard conditions. Remember that:

$Q = \frac{[Products]^n}{[Reactants]^m}$

where n and m are the stoichiometric coefficients from the balanced reaction.

If Q > 1 (too many products), E_cell < E°_cell.
If Q < 1 (too many reactants), E_cell > E°_cell.

Quick Fact

At standard conditions, Q = 1 because all concentrations are 1 M.

#Example

Consider the reaction:

$2Al(s) + 3Mn^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Mn(s)$

$Q = \frac{[Al^{3+}]^2}{[Mn^{2+}]^3}$

If [Al³⁺] = 1.5 M and [Mn²⁺] = 1.0 M, then Q = (1.5)²/(1.0)³ = 2.25 > 1. Therefore, E_cell < E°_cell.
If [Al³⁺] = 1.0 M and [Mn²⁺] = 1.5 M, then Q = (1.0)²/(1.5)³ = 0.296 < 1. Therefore, E_cell > E°_cell.
If [Al³⁺] = 1.5 M and [Mn²⁺] = 1.5 M, then Q = (1.5)²/(1.5)³ = 0.667 < 1. Therefore, E_cell > E°_cell.

#Multiple Choice Example

Multiple Choice Example

Common Mistake

Changing the size of an electrode does not change the concentration of the ions in solution. Therefore, it does not affect Q. However, removing an electrode does change the system.

For the reaction: $Cd(s) + 2Ag^+(aq) \rightarrow Cd^{2+}(aq) + 2Ag(s)$

$Q = \frac{[Cd^{2+}]}{[Ag^+]^2}$

The correct answer is D: Voltage does not change. Because the size of the silver electrode does not change the concentration of the ions, Q remains constant.

#Using the Nernst Equation

Let's take a closer look at the Nernst Equation and its applications:

Nernst Equation

#Equilibrium and the Nernst Equation

At equilibrium, E_cell = 0, and Q = K (equilibrium constant). We can modify the Nernst Equation to find E°_cell at equilibrium:

Equilibrium and Nernst Equation

Understanding the relationship between E, G, and K is crucial. Remember the "Big Triangle of Chemistry":

Big Triangle of Chemistry

ΔG = Gibbs Free Energy (energy available to do work)
ΔG = -nFE_cell
ΔG = -RTlnK

#Applications of the Nernst Equation

The Nernst Equation is incredibly versatile. It helps us:

Predict cell potential at non-standard temperatures.
Determine the equilibrium constant (K) at a given temperature.
Calculate concentrations at equilibrium.
Understand the effect of temperature changes on cell potential.
Predict cell potential in non-standard concentration solutions.

Memory Aid

Remember, the Nernst Equation is all about predicting how changes in conditions affect the cell's drive to reach equilibrium. It's a dynamic relationship!

#Final Exam Focus

Okay, it's crunch time! Here are the key takeaways for your exam:

Nernst Equation: Focus on predicting changes in E_cell, not calculations.
Q and E_cell: Understand how Q relates to E_cell (Q > 1, E_cell < E°_cell and vice versa).
Equilibrium: Know that at equilibrium, E_cell = 0 and Q = K.
Big Triangle: Be able to connect E, G, and K.
Concentration: Changes in concentration affect Q and, therefore, E_cell.

#Last-Minute Tips

Time Management: Don't get bogged down in calculations. Focus on understanding the concepts.
Common Pitfalls: Be careful with the stoichiometry when calculating Q. Remember that changing electrode size doesn't change concentration.
FRQs: Clearly explain your reasoning. Use terms like "shift towards products/reactants" and "increase/decrease cell potential".

#Practice Questions

Practice Question

Multiple Choice Questions:

A voltaic cell is constructed using the following reaction:

$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

What is the effect on the cell voltage if the concentration of $Cu^{2+}$ is increased?

(A) The cell voltage increases.
(B) The cell voltage decreases.
(C) The cell voltage remains the same.
(D) The cell voltage becomes zero.

2. Consider the following electrochemical cell:

$Pb(s) | Pb^{2+}(1.0 M) || Ag^+(1.0 M) | Ag(s)$

Which of the following changes will increase the cell potential?

(A) Increasing the size of the Pb electrode.
(B) Increasing the concentration of <math-inline>Pb^{2+}</math-inline>.
(C) Increasing the concentration of <math-inline>Ag^+</math-inline>.
(D) Adding water to both half-cells.

Free Response Question:

A voltaic cell is constructed using a standard hydrogen electrode (SHE) and a copper electrode in a 1.0 M solution of $Cu^{2+}$ . The standard reduction potential for $Cu^{2+}$ is +0.34 V.

(a) Write the balanced overall reaction for the cell. (b) Calculate the standard cell potential, E°_cell. (c) If the concentration of $Cu^{2+}$ is decreased to 0.10 M, what is the effect on the cell potential? Explain your reasoning. (d) Calculate the cell potential, E_cell, at 298 K when the concentration of $Cu^{2+}$ is 0.10 M.

Answer Key and Scoring Guidelines

Multiple Choice

(A) Increasing the concentration of a reactant ( $Cu^{2+}$ ) will shift the equilibrium towards the products, increasing the cell voltage.
(C) Increasing the concentration of $Ag^+$ will increase the reaction quotient, leading to a higher cell potential.

Free Response Question

(a) $H_2(g) + Cu^{2+}(aq) \rightarrow 2H^+(aq) + Cu(s)$

*   1 point for correct reactants and products
*   1 point for correct balancing

(b) E°_cell = E°_cathode - E°_anode = 0.34 V - 0.00 V = 0.34 V

*   1 point for using the correct formula
*   1 point for correct calculation

(c) Decreasing the concentration of $Cu^{2+}$ will decrease the cell potential. According to the Nernst equation, Q will increase, making E_cell < E°_cell.

*   1 point for stating that the cell potential decreases
*   1 point for correct reasoning using Q or Nernst.

(d) $E_{cell} = E^\circ_{cell} - \frac{0.0592}{n}logQ$

<math-block>Q = \frac{[H^+]^2}{[Cu^{2+}]}</math-block>

<math-block>Q = \frac{1^2}{0.10} = 10</math-block>

<math-block>E\_{cell} = 0.34 - \frac{0.0592}{2}log(10)</math-block>

<math-block>E\_{cell} = 0.34 - 0.0296 = 0.31 V</math-block>

*   1 point for correct Q expression
*   1 point for calculating Q correctly
*   1 point for correct substitution and calculation

You've got this! Remember to stay calm, think through each problem carefully, and use all the tools you've learned. Good luck on your AP Chemistry exam! 🚀