Module 9: Electrochemistry

General Chemistry – Module 9: Electrochemistry (Main Lesson)

🚀 Scope of This Module

This module covers the principles that govern how chemical reactions produce electricity (and vice versa). You’ll explore:

• Oxidation–reduction (redox) reactions
• Galvanic (voltaic) cells and electrolytic cells
• Standard reduction potentials
• Cell notation and Nernst equation
• Electroplating and Faraday’s Law

Electrochemistry is highly testable on the MCAT, especially in passage-based questions that require interpreting cell diagrams, half-reactions, and thermodynamic/electrical relationships.

Oxidation–Reduction (Redox) Reactions

Electrochemistry is powered by electron transfer, and this transfer occurs through oxidation–reduction (redox) reactions. Redox reactions involve two key processes: one substance loses electrons (oxidation), and another gains electrons (reduction). These reactions are the basis of how batteries function, how electroplating works, and how the body handles essential biochemical electron transport.

What Is Oxidation?

Oxidation refers to the loss of electrons by an atom, ion, or molecule. As a result, the species that is oxidized becomes more positively charged.

• Occurs at the anode in electrochemical cells.
• Electrons are liberated and can do electrical work (e.g., light a bulb).
• Often (but not always) involves gaining oxygen or losing hydrogen in biological settings.

Example:

Zn(s) → Zn²⁺ + 2e⁻
(Zinc metal is oxidized to zinc ion.)

What Is Reduction?

Reduction refers to the gain of electrons. The species that is reduced becomes less positively charged (or more negative).

• Occurs at the cathode in electrochemical cells.
• Accepts electrons produced by the oxidized species.
• Often associated with the gain of hydrogen or loss of oxygen in biochem.

Example:

Cu²⁺ + 2e⁻ → Cu(s)
(Copper ion is reduced to solid copper.)

Mnemonic to Remember:

• OIL RIG = Oxidation Is Loss, Reduction Is Gain
• LEO says GER = Lose Electrons: Oxidation, Gain Electrons: Reduction

Paired Processes

A redox reaction always involves both oxidation and reduction happening simultaneously — one species gives up electrons, and another takes them.

Species Role	Definition
Reducing agent	The species that loses electrons and is oxidized
Oxidizing agent	The species that gains electrons and is reduced

Example Full Redox Reaction:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
• Zn is oxidized (reducing agent)
• Cu²⁺ is reduced (oxidizing agent)

Common MCAT Traps

• Don’t confuse the oxidizing/reducing agents with what happens to them. The oxidizing agent gets reduced, and the reducing agent gets oxidized.
• Electron bookkeeping is essential. The number of electrons lost = number of electrons gained.
• You may be asked to balance redox reactions in acidic or basic solution (especially for half-reactions in electrochemical cells).

Oxidation vs. Reduction

Term	Definition	Mnemonic
Oxidation	Loss of electrons	OIL (Oxidation Is Loss)
Reduction	Gain of electrons	RIG (Reduction Is Gain)

Example:

Oxidation (loss of electrons):
Zn(s) → Zn²⁺ + 2e⁻

Reduction (gain of electrons):
Cu²⁺ + 2e⁻ → Cu(s)

Quick Diagnostic Checklist

Use this table to rapidly identify redox elements in a reaction:

Clue	Interpretation
Element goes from 0 → + charge	It’s oxidized (lost e⁻)
Ion goes from + charge → 0 or lower	It’s reduced (gained e⁻)
Species gives off e⁻	Reducing agent
Species accepts e⁻	Oxidizing agent

Redox Reaction = Electron Transfer

Electrons always move from the substance being oxidized to the substance being reduced.

• The species that gets reduced is the oxidizing agent.
• The species that gets oxidized is the reducing agent.

MCAT Tip: Be able to identify oxidized/reduced species, agents, and the number of electrons transferred.

Summary: Identifying Redox Roles

Species	Role
Loses e⁻	Oxidized
Gains e⁻	Reduced
Causes oxidation	Reducing agent
Causes reduction	Oxidizing agent

Balancing Redox Reactions (Half-Reaction Method)

Balancing redox reactions ensures that mass and charge are conserved. On the MCAT, you’ll often see these reactions in acidic or basic solution, and the half-reaction method is the most systematic way to handle them.

Step-by-Step: Half-Reaction Method (in Acidic Solution)

Let’s walk through how to balance the redox reaction:

Example:
MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (in acidic solution)

Step 1: Split into Half-Reactions

Split the overall reaction into two half-reactions:

• One for oxidation (loss of electrons)
• One for reduction (gain of electrons)

Oxidation (Fe²⁺ → Fe³⁺):
Fe²⁺ → Fe³⁺

Reduction (MnO₄⁻ → Mn²⁺):
MnO₄⁻ → Mn²⁺

Step 2: Balance Atoms Other Than O and H

Fe is already balanced.
Mn is already balanced.

Step 3: Balance Oxygen by Adding H₂O

MnO₄⁻ → Mn²⁺ has 4 oxygen atoms on the left.

Add 4 H₂O to the right:
MnO₄⁻ → Mn²⁺ + 4H₂O

Step 4: Balance Hydrogen by Adding H⁺

You now have 4 H₂O → 8 H atoms on the right.
Add 8 H⁺ to the left:

8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O

Step 5: Balance Charge by Adding Electrons

Oxidation half-reaction:
Fe²⁺ → Fe³⁺ + 1e⁻

Reduction half-reaction:
Left side: 8H⁺ (8+) + MnO₄⁻ (−1) → total +7
Right side: Mn²⁺ (+2)
So, to go from +7 to +2, add 5 electrons to the left:

5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O

Step 6: Equalize Electron Transfer

Oxidation gives off 1 electron, reduction requires 5 electrons.
So multiply the entire oxidation half-reaction by 5:

5Fe²⁺ → 5Fe³⁺ + 5e⁻

Step 7: Add Half-Reactions Together

Now combine:

Oxidation:
5Fe²⁺ → 5Fe³⁺ + 5e⁻

Reduction:
5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O

—

Overall Balanced Reaction:
5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Mass and charge are both conserved.

Acidic vs. Basic Solutions

For basic solutions, follow the same steps as above, but:

• After adding H⁺, neutralize each H⁺ by adding OH⁻ to both sides.
• H⁺ + OH⁻ → H₂O, so you’ll end up replacing H⁺ with water molecules.
• Finally, simplify any waters that appear on both sides.

Redox in Basic Solution — Step-by-Step Example

Example Reaction (Unbalanced):
ClO⁻ + Cr(OH)₃ → Cl⁻ + CrO₄²⁻ (in basic solution)

We’ll balance this reaction in a basic aqueous environment.

Step 1: Separate into Half-Reactions

Oxidation: ClO⁻ → Cl⁻
Reduction: Cr(OH)₃ → CrO₄²⁻

Step 2: Balance Atoms Other Than O and H

Both Cl and Cr are already balanced.

Step 3: Balance O by Adding H₂O

Oxidation Half-Reaction:
ClO⁻ → Cl⁻
(1 oxygen on left → add 1 H₂O to the right)
ClO⁻ → Cl⁻ + H₂O

Reduction Half-Reaction:
Cr(OH)₃ → CrO₄²⁻
Cr(OH)₃ has 3 oxygen atoms; CrO₄²⁻ has 4.
You need 1 more O on the right → add 1 H₂O to the left:

H₂O + Cr(OH)₃ → CrO₄²⁻

Now total O atoms:
Left: 1 (H₂O) + 3 (from Cr(OH)₃) = 4
Right: 4 (in CrO₄²⁻)

Step 4: Balance H by Adding OH⁻ (not H⁺)

Oxidation:
Right side has 2 H from H₂O. Add 2 OH⁻ to left:
ClO⁻ + 2OH⁻ → Cl⁻ + H₂O

Reduction:
Left has 2 H from H₂O and 3 H from Cr(OH)₃ = 5 total
Right side has none. Add 5 OH⁻ to right:
H₂O + Cr(OH)₃ → CrO₄²⁻ + 5OH⁻

Step 5: Balance Charge by Adding Electrons

Oxidation Half-Reaction:
Left: ClO⁻ (−1) + 2 OH⁻ (−2) = −3
Right: Cl⁻ (−1) + H₂O (neutral) = −1
To go from −3 to −1, add 2 e⁻ to the right:

ClO⁻ + 2OH⁻ → Cl⁻ + H₂O + 2e⁻

Reduction Half-Reaction:
Left: H₂O + Cr(OH)₃ = neutral
Right: CrO₄²⁻ (−2) + 5OH⁻ (−5) = −7
To go from 0 to −7, add 7 e⁻ to the left:

7e⁻ + H₂O + Cr(OH)₃ → CrO₄²⁻ + 5OH⁻

Step 6: Equalize Electrons

To eliminate electrons, find the least common multiple of 2 and 7:

• Multiply oxidation half-reaction ×7
• Multiply reduction half-reaction ×2

Oxidation ×7:
7ClO⁻ + 14OH⁻ → 7Cl⁻ + 7H₂O + 14e⁻

Reduction ×2:
14e⁻ + 2H₂O + 2Cr(OH)₃ → 2CrO₄²⁻ + 10OH⁻

Step 7: Add Half-Reactions

7ClO⁻ + 14OH⁻ + 2H₂O + 2Cr(OH)₃ →
7Cl⁻ + 7H₂O + 2CrO₄²⁻ + 10OH⁻

Simplify H₂O and OH⁻ if needed:
Left: 14OH⁻ + 2H₂O
Right: 10OH⁻ + 7H₂O

Subtract:

• OH⁻: 14 − 10 = 4 OH⁻ on left
• H₂O: 7 − 2 = 5 H₂O on right

Final Balanced Reaction:

7ClO⁻ + 2Cr(OH)₃ + 4OH⁻ → 7Cl⁻ + 2CrO₄²⁻ + 5H₂O

---

Key MCAT Takeaways

• Always balance atoms, then charge.
• Redox is often tested in electrochemistry cells, so recognize the half-reactions.
• Know that electrons flow from anode to cathode in galvanic cells, and vice versa in electrolytic cells.
• Use OH⁻ and H₂O in basic solutions, not H⁺.
• Practice spotting half-reactions in cell diagrams or redox passage descriptions.
• Understand charge flow (electrons) and mass balance intuitively.
• Knowing how to balance redox helps with electrolytic cell and galvanic cell analysis later.

Galvanic vs. Electrolytic Cells

Electrochemical cells are devices that either produce or consume electrical energy through redox reactions. The MCAT focuses primarily on two types:

• Galvanic (voltaic) cells – spontaneous redox reactions that generate electricity
• Electrolytic cells – nonspontaneous redox reactions that require external energy input

Though they operate differently, both types share common elements: a redox reaction, an anode, a cathode, and some form of electron transfer path (wire) and ion flow path (salt bridge or membrane).

Galvanic (Voltaic) Cells

These cells harness the free energy from spontaneous redox reactions (ΔG < 0) to produce an electric current. This is how batteries work!

Key Features:

• Spontaneous (E°cell > 0, ΔG < 0)
• Converts chemical energy → electrical energy
• Electrons flow from anode to cathode (just like in all cells)
• Anode = oxidation (loss of e⁻)
• Cathode = reduction (gain of e⁻)
• Requires salt bridge to maintain charge neutrality

Example:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
Half-reactions:

• Zn(s) → Zn²⁺ + 2e⁻ (oxidation at anode)
• Cu²⁺ + 2e⁻ → Cu(s) (reduction at cathode)

Electrons flow through the wire from Zn to Cu.

Electrolytic Cells

Electrolytic cells use electrical energy from an external source to drive a nonspontaneous redox reaction (ΔG > 0).

These are used in:

• Electroplating (coating metals)
• Electrolysis (e.g., splitting water into H₂ and O₂)
• Rechargeable batteries (when charging)

Key Features:

• Nonspontaneous (E°cell < 0, ΔG > 0)
• Requires external voltage source
• Electrons still flow from anode to cathode
• But now the anode is positive, and cathode is negative (reversed polarity)
• No salt bridge — often occurs in one container

Example:

Electrolysis of molten NaCl:

• Na⁺ is reduced at the cathode: Na⁺ + e⁻ → Na(s)
• Cl⁻ is oxidized at the anode: 2Cl⁻ → Cl₂(g) + 2e⁻

Comparison Table: Galvanic vs. Electrolytic Cells

Feature	Galvanic Cell	Electrolytic Cell
Spontaneity	Spontaneous reaction (ΔG∘ < 0)	Nonspontaneous reaction (ΔG∘ > 0)
E°cell	Positive (E∘cell > 0)	Negative (E∘cell < 0)
Energy Conversion	Chemical → Electrical (produces electricity)	Electrical → Chemical (requires external power)
Electron Flow	Anode → Cathode	Anode → Cathode (same direction)
Site of Oxidation	Anode	Anode (same)
Site of Reduction	Cathode	Cathode (same)
Anode Charge	Negative (source of e⁻)	Positive (pulls e⁻ from species)
Cathode Charge	Positive (accepts e⁻)	Negative (supplied with e⁻ by power source)
External Power Source?	No	Yes
Example	Zn–Cu battery	Electrolysis of molten NaCl

MCAT Tip: Despite differences, electrons always flow from anode to cathode, and oxidation always happens at the anode.

Common Pitfalls

• Don’t confuse sign conventions:
  • In galvanic cells, the anode is negative and the cathode is positive.
  • In electrolytic cells, the external power source forces positive to anode and negative to cathode — reversed!
• The species with the more positive reduction potential is reduced (acts as cathode).

Electrochemical Cell Notation (Line Notation)

Cell notation (also called line notation) is a shorthand way of describing the components and half-reactions in a galvanic or electrolytic cell.

🧪 General Format:

Anode | Anode solution (concentration) || Cathode solution (concentration) | CathodeStructure Breakdown:

• Electrons always flow left → right (anode → cathode)
• Single vertical line ( | ) separates different phases (solid, liquid, aqueous, gas)
• Double vertical line ( || ) indicates the salt bridge or phase boundary between two half-cells
• Species are written from left (anode) to right (cathode)

Example: Zn–Cu Galvanic Cell

Reaction:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Cell Notation:

Zn(s) | Zn²⁺(1.0 M) || Cu²⁺(1.0 M) | Cu(s)

• Zn is the anode: it is oxidized to Zn²⁺.
• Cu²⁺ is the cathode: it is reduced to Cu(s).
• The salt bridge separates the two half-cells.

Example: Inert Electrode (for Ions Only)

If there’s no solid metal to act as an electrode (e.g., in redox of Fe³⁺/Fe²⁺), you use an inert electrode like platinum (Pt).

Reaction:
Fe³⁺ + e⁻ → Fe²⁺

Cell Notation (with Pt):

Pt(s) | Fe²⁺(1.0 M), Fe³⁺(1.0 M) || …

• The comma separates species in the same phase (both Fe ions are aqueous).
• Pt(s) is required as a conducting surface for electron flow.

MCAT Strategy Tips:

• Read left to right: Identify what is being oxidized at the anode and what is being reduced at the cathode.
• Double-check the direction of electron flow: From the substance being oxidized (losing e⁻) to the one being reduced (gaining e⁻).
• Don’t forget concentration values if provided, as they can affect Nernst equation calculations.

Standard Electrode Potentials (E°) and Cell Voltage

The standard electrode potential (E°) is a measure of the tendency of a half-cell to be reduced. It’s like an “electric pressure” that pushes or pulls electrons in redox reactions. These values are measured under standard conditions:

• 25°C (298 K)
• 1 atm pressure (for gases)
• 1.0 M concentrations (for aqueous solutions)

The MCAT provides standard reduction potentials in volts (V) on a table — typically referenced against the standard hydrogen electrode (SHE), which is arbitrarily assigned an E° of 0.00 V.

What Does E° Mean?

• A positive E° = strong tendency to gain electrons (be reduced)
• A negative E° = less likely to gain electrons; may be oxidized instead
• The more positive the E°, the stronger the oxidizing agent
• The more negative the E°, the stronger the reducing agent

Calculating E°cell for a Redox Reaction

To calculate the overall voltage (standard cell potential) of an electrochemical cell:

Key Equation:

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}$$

Or think of it as:

Reduction potential of what gets reduced – Reduction potential of what gets oxidized

Example:

Reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

From the standard reduction potential table:

• Cu²⁺ + 2e⁻ → Cu  E° = +0.34 V (reduction at cathode)
• Zn²⁺ + 2e⁻ → Zn  E° = –0.76 V
    → But Zn is oxidized, so this is the anode

Apply the formula:

$$E^\circ_{\text{cell}} = 0.34\,\text{V} - (-0.76\,\text{V}) = 1.10\,\text{V}$$

Spontaneous reaction → galvanic cell

Alternate: Sum of Half-Reactions

Some students prefer to reverse the oxidation half-reaction and add E° values instead:

• Cu²⁺ + 2e⁻ → Cu  E° = +0.34 V
• Zn → Zn²⁺ + 2e⁻  E° = +0.76 V (reverse of Zn²⁺ + 2e⁻ → Zn)

E°cell = 0.34 + 0.76 = 1.10 V

Same result.

MCAT Warnings:

• Do not multiply E° values by coefficients, even if the half-reactions require multiple electrons.
  • Voltage is an intensive property.
• Always ensure electrons cancel when combining half-reactions.
• If E°cell is positive, the reaction is spontaneous (galvanic).
• If E°cell is negative, the reaction is nonspontaneous (electrolytic).

Summary Table

Concept	Symbol	Meaning
Standard electrode potential	E°	Tendency to be reduced
Cell potential	E°cell	Voltage generated by the redox reaction
E°cell > 0	—	Reaction is spontaneous (galvanic)
E°cell < 0	—	Reaction is nonspontaneous (electrolytic)

Gibbs Free Energy and Electrochemistry

When a redox reaction occurs in an electrochemical cell, it either releases or requires energy. That energy change is captured by ΔG (Gibbs free energy) — and is directly related to the cell voltage (E°cell).

Key Equation:

$$
\Delta G^\circ = -n F E^\circ_{\text{cell}}
$$

Where:

$$
\Delta G^\circ = \text{standard Gibbs free energy change (J or kJ)}
$$

$$
n = \text{number of moles of electrons transferred}
$$

$$
F = \text{Faraday’s constant} = 96,!485 \ \text{C/mol e}^-
$$

$$
E^\circ_{\text{cell}} = \text{standard cell potential (Volts)}
$$

Unit Check:
1 V = 1 J/C → ΔG (J) = (mol e⁻) × (C/mol) × (J/C)

MCAT Conceptual Implications:

$$
\text{If } E^\circ_{\text{cell}} > 0, \text{ then } \Delta G^\circ < 0 \Rightarrow \text{spontaneous}
$$

$$
\text{If } E^\circ_{\text{cell}} < 0, \text{ then } \Delta G^\circ > 0 \Rightarrow \text{nonspontaneous}
$$

$$
\text{When } \Delta G^\circ = 0, \text{ the system is at equilibrium}
$$

Example Problem:

Given: A galvanic cell has E^∘_cell=1.10 V and transfers 2 electrons.
Find ΔG^∘ in kJ/mol.

$$
n = 2 \
F = 96,!485 \ \text{C/mol e}^- \
E^\circ_{\text{cell}} = 1.10 \ \text{V}
$$

Plug into the equation:

$$
\Delta G^\circ = -n F E^\circ_{\text{cell}} = -(2)(96,!485)(1.10) = -212,!267 \ \text{J}
$$

Convert to kJ:

ΔG^∘ =−212.3 kJ/mol

This reaction is spontaneous because ΔG^∘ < 0.

MCAT Tips:

• Be sure to identify n correctly — it’s the total number of electrons transferred in the balanced redox reaction.
• Units matter! If the question asks for kJ, don’t forget to convert.
• You can expect questions linking E°, ΔG, and reaction spontaneity.
• Some passage-based questions may require you to calculate ΔG from a table of reduction potentials using:

$$
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}} \Rightarrow \Delta G^\circ = -n F E^\circ_{\text{cell}}
$$

The Nernst Equation – Non-Standard Conditions

So far, we’ve worked with standard cell potential E^∘_cell​, which assumes:

• All solutes at 1 M concentration
• All gases at 1 atm pressure
• Temperature = 25°C (298 K)

But real biological and lab systems rarely meet these exact conditions. That’s where the Nernst equation comes in — it adjusts the cell potential based on concentration (and, if needed, temperature).

The Nernst Equation

$$
E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0592}{n} \log Q
$$

Where:

$$
E_{\text{cell}} = \text{actual cell potential under current conditions}
$$

$$
E^\circ_{\text{cell}} = \text{standard cell potential}
$$

$$
n = \text{moles of electrons transferred}
$$

$$
Q = \text{reaction quotient (same form as } K\text{, but for current concentrations)}
$$

$$
0.0592 = \text{a constant derived from } \frac{RT}{nF} \text{ at 25}^\circ \text{C}
$$

The logarithm is base 10 on the MCAT version (log, not ln)

What is Q?

The reaction quotient (Q) is calculated like an equilibrium expression — but using current, possibly non-equilibrium, concentrations.

For a reaction like:

$$
aA + bB \rightarrow cC + dD
$$

Then:

$$
Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}
$$

Solids and liquids are excluded from Q (just like in equilibrium expressions).

Interpreting Nernst Equation Results

• If

  • 

    Diamond Membership

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  • 

    Gold Membership

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  ↑, then Q ↑, so E_cell ↓:
  → Cell potential decreases as products accumulate.
• If [reactants] ↑, then Q ↓, so E_cell ​↑:
  → Cell potential increases when more reactants are present.
• At equilibrium, Q = K, and E_cell ​= 0 (no driving force).

MCAT Tips

• Focus on qualitative reasoning:
  Can you predict how changing ion concentrations shifts EcellE_{\text{cell}}Ecell​?
• Use 0.0592/n when working at 25°C
• Know that Ecell = 0 at equilibrium and link it to ΔG=0.

Electrolytic Cells and Faraday’s Law

While galvanic cells (also called voltaic cells) generate electricity from a spontaneous redox reaction, electrolytic cells use electricity to drive a nonspontaneous reaction. This is the fundamental distinction tested on the MCAT.

What Is an Electrolytic Cell?

An electrolytic cell requires an external power source (like a battery) to force a redox reaction to occur in the non-spontaneous direction.

Key Features:

• ΔG^∘ > 0 (nonspontaneous)
• E^∘_cell < 0
• Requires input of energy
• Electrons still flow from anode → cathode
• Anode is positive, cathode is negative (opposite of galvanic cells)

Example: Electrolysis of Molten NaCl

When molten NaCl is electrolyzed:

• At the cathode (reduction):
  • Na⁺ + e⁻ → Na(s)
• At the anode (oxidation):
  • 2Cl⁻ → Cl₂(g) + 2e⁻

This process requires electrical energy input because sodium doesn’t spontaneously separate from chloride.

Faraday’s Law of Electrolysis

This law links electrical charge (current) with the amount of substance produced at each electrode.

🔧 Faraday’s Law Equation:

$$
\text{mol product} = \frac{It}{nF}
$$

Where:

• I = current in amperes (A)
• t = time in seconds (s)
• n = moles of electrons per mole of product
• F = Faraday’s constant = 96,485 C/mol
• mol product = moles of metal (or gas) deposited

1 amp = 1 coulomb per second

MCAT Strategy Notes:

• You’ll often be asked to calculate the amount of metal plated or gas produced after passing a certain current for a specific time.
• Key conversions:
  • From current × time → total charge (in coulombs)
  • From charge → moles of e⁻ → moles of product

Worked Example – Faraday’s Law of Electrolysis

Problem:
A current of 2.00 A is passed through a solution of CuSO₄ for 1.50 hours. How many grams of copper metal (Cu) will be deposited at the cathode?

Step 1: Identify what’s happening

Cathode reaction (reduction):

Cu²⁺ + 2e⁻ → Cu(s)

This means:

• n = 2 moles of electrons are required to reduce 1 mole of Cu²⁺

Step 2: Use Faraday’s Law

Faraday’s Law:

$$
\text{mol of metal} = \frac{It}{nF}
$$

Given:

• I=2.00 A
• t=1.50 hr=1.50×3600=5400 s
• n=2 mol e−/ mol Cu
• F=96, ⁣485 C/mol

Plug in:

$$
\text{mol Cu} = \frac{(2.00)(5400)}{2 \times 96,!485} = \frac{10,!800}{192,!970} \approx 0.0560 \, \text{mol}
$$

Step 2: Mass of copper deposited

$$
\text{grams Cu} = 0.0560 \times 63.55 = 3.56 \, \text{g}
$$

Final Answer:

3.56 grams of copper

---

Electrochemistry Module Recap

Core Concepts:

• Redox reactions involve the transfer of electrons.
  • Oxidation = loss of electrons
  • Reduction = gain of electrons
  • Use LEO the lion says GER as a mnemonic.
• Galvanic cells convert spontaneous chemical energy into electricity.
• Electrolytic cells require electricity to drive nonspontaneous reactions.
• Faraday’s Law connects charge (Q) to the moles of material produced or consumed.
• Cell potential (E°cell) determines spontaneity.
• Nernst Equation adjusts Ecell for nonstandard conditions.

📊 Galvanic vs. Electrolytic Summary:

Feature	Galvanic Cell	Electrolytic Cell
Spontaneity	Spontaneous (ΔG° < 0)	Nonspontaneous (ΔG° > 0)
E°cell	Positive	Negative
Energy Conversion	Chemical → Electrical	Electrical → Chemical
Anode / Cathode Roles	Anode = Ox, Cathode = Red	Same
Charge of Electrodes	Anode = –, Cathode = +	Anode = +, Cathode = –

Strategy Tips:

• MCAT = concepts over calculations: Understand what each variable means.
• Signs matter: Make sure you’re clear on electrode charges (especially in electrolytic cells).
• Always label redox reactions: Know which species is oxidized vs. reduced.
• Watch for Q and Le Châtelier: Changing ion concentrations will affect cell potential.
• Time + current = charge: Use this to calculate mass deposited (Faraday’s Law).
• Be fluent in cell notation: (anode | anode soln || cathode soln | cathode).