#Electrochemical Cells: A Comprehensive Guide ⚡

#Introduction to Cell Potential

In electrochemistry, cell potential is the driving force that pushes electrons through a circuit. It's the "oomph" behind redox reactions, measured in volts (V), and is also known as electromotive force (EMF). Think of it like the pressure in a water pipe – the higher the pressure, the faster the water flows. Similarly, the higher the cell potential, the more spontaneous the reaction.

Key Concept

Cell potential is a measure of the spontaneity of a redox reaction. A higher cell potential means a more spontaneous reaction.

#Key Concepts:

Anode: The electrode where oxidation occurs (loss of electrons). Remember: An Ox.
Cathode: The electrode where reduction occurs (gain of electrons). Remember: Red Cat.
Electrons flow from the anode to the cathode through an external circuit.
Cell potential (E°cell) is measured under standard conditions (1 M, 298 K, 1 atm).

#Calculating Standard Cell Potential

The standard cell potential (E°cell) is calculated using the following equation:

$E°_{cell} = E°_{cathode} - E°_{anode}$

Where:

E°cathode is the standard reduction potential of the cathode.
E°anode is the standard reduction potential of the anode.

Exam Tip

Always use reduction potentials from the table, even when calculating for oxidation. If a reaction is an oxidation, you will need to negate the reduction potential from the table.

#Example Calculation

Let's revisit the example from the notes:

Reaction: 2AgBr + 2Hg → 2Ag + Hg2Br2

Given Half-Reactions:

Hg2Br2 + 2e- → 2Hg + 2Br- (E° = +0.140 V)
2AgBr + 2e- → 2Ag + 2Br- (E° = +0.071 V)

Identify Cathode and Anode:
- AgBr is reduced, so it's at the cathode.
- Hg is oxidized, so it's at the anode.
Apply the Formula:

E°cell = E°cathode - E°anode

E°cell = 0.071 V - 0.140 V = -0.069 V

Common Mistake

Remember to subtract the anode potential from the cathode potential. It's easy to mix them up! Also, be careful with signs – a negative cell potential indicates a non-spontaneous reaction.

#Alternative Method

Another way to calculate cell potential is to:

Reverse the oxidation half-reaction (change the sign of its potential).
Add the potentials of the reduction and oxidation half-reactions.

In our example:

Reduction (cathode): 2AgBr + 2e- → 2Ag + 2Br- (E° = +0.071 V)
Oxidation (anode): 2Hg + 2Br- → Hg2Br2 + 2e- (E° = -0.140 V) (Note the sign change!)

E°cell = +0.071 V + (-0.140 V) = -0.069 V

#Standard Reduction Potentials

#Using Standard Reduction Potential Tables

Standard reduction potential tables list reduction half-reactions and their corresponding potentials. The AP exam will provide the necessary data, but understanding how to use these tables is crucial.

Here's a sample table:

Standard Reduction Potentials

Image From Grade12UChem

Quick Fact

The reduction of H+ to H2 is defined as 0 V. All other reduction potentials are relative to this value.

#Understanding Reduction Potentials

Positive reduction potential: The species is easily reduced (good oxidizing agent).
Negative reduction potential: The species is difficult to reduce (good reducing agent).

Memory Aid

Think of it this way: substances with large positive reduction potentials are "greedy" for electrons, while those with large negative potentials are "eager" to give them away.

#Cell Potential and Spontaneity

#Predicting Spontaneity

The sign of the cell potential (E°cell) tells us if a reaction is spontaneous:

E°cell > 0: The reaction is spontaneous (thermodynamically favorable), and ΔG° is negative.
E°cell < 0: The reaction is non-spontaneous (thermodynamically unfavorable), and ΔG° is positive.

Understanding the relationship between cell potential and spontaneity is crucial. Expect questions that ask you to predict spontaneity based on E°cell.

#Calculating ΔG° from E°cell

We can calculate the standard Gibbs Free Energy change (ΔG°) using the following equation:

$ΔG° = -nFE°_{cell}$

Where:

ΔG° is the standard Gibbs Free Energy change.
n is the number of moles of electrons transferred.
F is Faraday's constant (96,485 C/mol e-).
E°cell is the standard cell potential.

#Example Calculation

Given: E°cell = 1.02 V, n = 1 mol of electrons transferred

Calculate ΔG°:

ΔG° = - (1 mol e-) * (96,485 C/mol e-) * (1.02 V) = -98414.7 J = -98.414 kJ
Calculate K (Equilibrium Constant):

$K = e^{\frac{-ΔG°}{RT}}$

$K = e^{\frac{98414.7}{(8.314)(298)}} = 1.78 * 10^{17}$

Memory Aid

Remember the equation ΔG° = -nFE°cell using the mnemonic "Good News For Everyone", where G is Gibbs free energy, n is moles of electrons, F is Faraday's constant, and E is cell potential. Also, remember that a positive E°cell goes with a negative ΔG° (spontaneous) and vice-versa.

#Final Exam Focus

#High-Priority Topics

Calculating cell potential using standard reduction potentials.
Predicting spontaneity based on the sign of E°cell.
Calculating ΔG° from E°cell and vice versa.
Understanding the role of the anode and cathode in electrochemical cells.
Relating Gibbs Free Energy to cell potential and equilibrium constant.

#Common Question Types

Multiple-choice questions testing your understanding of cell potential, spontaneity, and the relationship between ΔG° and E°cell.
Free-response questions requiring you to calculate cell potentials, predict reaction spontaneity, and perform calculations involving ΔG° and K.
Questions that combine electrochemistry with thermodynamics and equilibrium concepts.

#Last-Minute Tips

Time Management: Quickly identify the cathode and anode, apply the formula, and double-check your calculations.
Common Pitfalls: Be careful with signs, especially when reversing reactions. Ensure you are using reduction potentials from the table correctly.
Strategies: Practice with various examples, focus on understanding the concepts, and use your time wisely during the exam.

#Practice Questions

Practice Question

#Multiple Choice Questions

A voltaic cell is constructed using the following half-reactions:

$Zn^{2+} (aq) + 2e^- \rightarrow Zn(s) \quad E° = -0.76 V$ $Cu^{2+} (aq) + 2e^- \rightarrow Cu(s) \quad E° = +0.34 V$

What is the standard cell potential for this voltaic cell?

(A) -1.10 V (B) -0.42 V (C) +0.42 V (D) +1.10 V
Which of the following statements is true for a spontaneous redox reaction?

(A) ΔG° > 0 and E°cell > 0 (B) ΔG° > 0 and E°cell < 0 (C) ΔG° < 0 and E°cell > 0 (D) ΔG° < 0 and E°cell < 0
Given the following standard reduction potentials:

$Ag^+ (aq) + e^- \rightarrow Ag(s) \quad E° = +0.80 V$ $Ni^{2+} (aq) + 2e^- \rightarrow Ni(s) \quad E° = -0.25 V$

What is the standard cell potential for the reaction: 2Ag+(aq) + Ni(s) → 2Ag(s) + Ni2+(aq)?

(A) -1.05 V (B) -0.55 V (C) +0.55 V (D) +1.05 V

#Free Response Question

Consider the following electrochemical cell:

$Zn(s) | Zn^{2+}(aq) || Ag^+(aq) | Ag(s)$

The standard reduction potentials are:

$Zn^{2+} (aq) + 2e^- \rightarrow Zn(s) \quad E° = -0.76 V$ $Ag^+ (aq) + e^- \rightarrow Ag(s) \quad E° = +0.80 V$

(a) Write the balanced overall cell reaction.

(b) Calculate the standard cell potential, E°cell.

(d) Calculate the equilibrium constant, K, for the reaction at 298 K.

#Scoring Breakdown

(a) (2 points)

1 point for correct oxidation half-reaction: Zn(s) → Zn2+(aq) + 2e-
1 point for correct reduction half-reaction: 2Ag+(aq) + 2e- → 2Ag(s)
Overall balanced reaction: Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)

(b) (1 point)

E°cell = E°cathode - E°anode = +0.80 V - (-0.76 V) = +1.56 V

1 point for correct substitution into the equation ΔG° = -nFE°cell
1 point for correct answer: ΔG° = -2 * 96485 * 1.56 = -301000 J = -301 kJ

(d) (2 points)

1 point for correct substitution into the equation K = e^(-ΔG°/RT)
1 point for correct answer: K = e^(301000/(8.314*298)) = 1.6 x 10^52

Cell Potential and Free Energy

Ethan Taylor

8 min read

Next Topic - Cell Potential Under Nonstandard Conditions

Listen to this study note

Study Guide Overview

This study guide covers electrochemical cells, focusing on cell potential (EMF) and its relationship to spontaneity. It explains how to calculate standard cell potential using reduction potentials, determine anode and cathode, and predict spontaneity using E°cell. Finally, it connects cell potential to Gibbs Free Energy (ΔG°) and the equilibrium constant (K), including example calculations and practice questions.

#Electrochemical Cells: A Comprehensive Guide ⚡

#Introduction to Cell Potential

Key Concept

Cell potential is a measure of the spontaneity of a redox reaction. A higher cell potential means a more spontaneous reaction.

#Key Concepts:

Anode: The electrode where oxidation occurs (loss of electrons). Remember: An Ox.
Cathode: The electrode where reduction occurs (gain of electrons). Remember: Red Cat.
Electrons flow from the anode to the cathode through an external circuit.
Cell potential (E°cell) is measured under standard conditions (1 M, 298 K, 1 atm).

#Calculating Standard Cell Potential

The standard cell potential (E°cell) is calculated using the following equation:

$E°_{cell} = E°_{cathode} - E°_{anode}$

Where:

E°cathode is the standard reduction potential of the cathode.
E°anode is the standard reduction potential of the anode.

Exam Tip

Always use reduction potentials from the table, even when calculating for oxidation. If a reaction is an oxidation, you will need to negate the reduction potential from the table.

#Example Calculation

Let's revisit the example from the notes:

Reaction: 2AgBr + 2Hg → 2Ag + Hg2Br2

Given Half-Reactions:

Hg2Br2 + 2e- → 2Hg + 2Br- (E° = +0.140 V)
2AgBr + 2e- → 2Ag + 2Br- (E° = +0.071 V)

Identify Cathode and Anode:
- AgBr is reduced, so it's at the cathode.
- Hg is oxidized, so it's at the anode.
Apply the Formula:

E°cell = E°cathode - E°anode

E°cell = 0.071 V - 0.140 V = -0.069 V

Common Mistake

Remember to subtract the anode potential from the cathode potential. It's easy to mix them up! Also, be careful with signs – a negative cell potential indicates a non-spontaneous reaction.

#Alternative Method

Another way to calculate cell potential is to:

Reverse the oxidation half-reaction (change the sign of its potential).
Add the potentials of the reduction and oxidation half-reactions.

In our example:

Reduction (cathode): 2AgBr + 2e- → 2Ag + 2Br- (E° = +0.071 V)
Oxidation (anode): 2Hg + 2Br- → Hg2Br2 + 2e- (E° = -0.140 V) (Note the sign change!)

E°cell = +0.071 V + (-0.140 V) = -0.069 V

#Standard Reduction Potentials

#Using Standard Reduction Potential Tables

Here's a sample table:

Standard Reduction Potentials

Image From Grade12UChem

Quick Fact

The reduction of H+ to H2 is defined as 0 V. All other reduction potentials are relative to this value.

#Understanding Reduction Potentials

Positive reduction potential: The species is easily reduced (good oxidizing agent).
Negative reduction potential: The species is difficult to reduce (good reducing agent).

Memory Aid

Think of it this way: substances with large positive reduction potentials are "greedy" for electrons, while those with large negative potentials are "eager" to give them away.

#Cell Potential and Spontaneity

#Predicting Spontaneity

The sign of the cell potential (E°cell) tells us if a reaction is spontaneous:

E°cell > 0: The reaction is spontaneous (thermodynamically favorable), and ΔG° is negative.
E°cell < 0: The reaction is non-spontaneous (thermodynamically unfavorable), and ΔG° is positive.

Understanding the relationship between cell potential and spontaneity is crucial. Expect questions that ask you to predict spontaneity based on E°cell.

#Calculating ΔG° from E°cell

We can calculate the standard Gibbs Free Energy change (ΔG°) using the following equation:

$ΔG° = -nFE°_{cell}$

Where:

ΔG° is the standard Gibbs Free Energy change.
n is the number of moles of electrons transferred.
F is Faraday's constant (96,485 C/mol e-).
E°cell is the standard cell potential.

#Example Calculation

Given: E°cell = 1.02 V, n = 1 mol of electrons transferred

Calculate ΔG°:

ΔG° = - (1 mol e-) * (96,485 C/mol e-) * (1.02 V) = -98414.7 J = -98.414 kJ
Calculate K (Equilibrium Constant):

$K = e^{\frac{-ΔG°}{RT}}$

$K = e^{\frac{98414.7}{(8.314)(298)}} = 1.78 * 10^{17}$

Memory Aid

#Final Exam Focus

#High-Priority Topics

Calculating cell potential using standard reduction potentials.
Predicting spontaneity based on the sign of E°cell.
Calculating ΔG° from E°cell and vice versa.
Understanding the role of the anode and cathode in electrochemical cells.
Relating Gibbs Free Energy to cell potential and equilibrium constant.

#Common Question Types

Multiple-choice questions testing your understanding of cell potential, spontaneity, and the relationship between ΔG° and E°cell.
Free-response questions requiring you to calculate cell potentials, predict reaction spontaneity, and perform calculations involving ΔG° and K.
Questions that combine electrochemistry with thermodynamics and equilibrium concepts.

#Last-Minute Tips

Time Management: Quickly identify the cathode and anode, apply the formula, and double-check your calculations.
Common Pitfalls: Be careful with signs, especially when reversing reactions. Ensure you are using reduction potentials from the table correctly.
Strategies: Practice with various examples, focus on understanding the concepts, and use your time wisely during the exam.

#Practice Questions

Practice Question

#Multiple Choice Questions

A voltaic cell is constructed using the following half-reactions:

$Zn^{2+} (aq) + 2e^- \rightarrow Zn(s) \quad E° = -0.76 V$ $Cu^{2+} (aq) + 2e^- \rightarrow Cu(s) \quad E° = +0.34 V$

What is the standard cell potential for this voltaic cell?

(A) -1.10 V (B) -0.42 V (C) +0.42 V (D) +1.10 V
Which of the following statements is true for a spontaneous redox reaction?

(A) ΔG° > 0 and E°cell > 0 (B) ΔG° > 0 and E°cell < 0 (C) ΔG° < 0 and E°cell > 0 (D) ΔG° < 0 and E°cell < 0
Given the following standard reduction potentials:

$Ag^+ (aq) + e^- \rightarrow Ag(s) \quad E° = +0.80 V$ $Ni^{2+} (aq) + 2e^- \rightarrow Ni(s) \quad E° = -0.25 V$

What is the standard cell potential for the reaction: 2Ag+(aq) + Ni(s) → 2Ag(s) + Ni2+(aq)?

(A) -1.05 V (B) -0.55 V (C) +0.55 V (D) +1.05 V

#Free Response Question

Consider the following electrochemical cell:

$Zn(s) | Zn^{2+}(aq) || Ag^+(aq) | Ag(s)$

The standard reduction potentials are:

$Zn^{2+} (aq) + 2e^- \rightarrow Zn(s) \quad E° = -0.76 V$ $Ag^+ (aq) + e^- \rightarrow Ag(s) \quad E° = +0.80 V$

(a) Write the balanced overall cell reaction.

(b) Calculate the standard cell potential, E°cell.

(d) Calculate the equilibrium constant, K, for the reaction at 298 K.

#Scoring Breakdown

(a) (2 points)

1 point for correct oxidation half-reaction: Zn(s) → Zn2+(aq) + 2e-
1 point for correct reduction half-reaction: 2Ag+(aq) + 2e- → 2Ag(s)
Overall balanced reaction: Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)

(b) (1 point)

E°cell = E°cathode - E°anode = +0.80 V - (-0.76 V) = +1.56 V

1 point for correct substitution into the equation ΔG° = -nFE°cell
1 point for correct answer: ΔG° = -2 * 96485 * 1.56 = -301000 J = -301 kJ

(d) (2 points)

1 point for correct substitution into the equation K = e^(-ΔG°/RT)
1 point for correct answer: K = e^(301000/(8.314*298)) = 1.6 x 10^52

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Previous Topic - Galvanic (Voltaic) and Electrolytic Cells Next Topic - Cell Potential Under Nonstandard Conditions

Question 1 of 10

What is cell potential also known as and what unit is it measured in? 🤔

Oxidation force, measured in Amperes (A)

Reduction potential, measured in Coulombs (C)

Electromotive force (EMF), measured in Volts (V)

Gibbs Free Energy, measured in Joules (J)