Chapter 12: Electrochemistry

Drop a strip of zinc metal into a beaker of blue copper sulfate solution. Within minutes, the solution fades and a reddish film of copper coats the zinc. Zinc atoms are handing electrons directly to copper ions - a spontaneous redox reaction. The problem? All the energy is wasted as heat. None of it does useful work.

A galvanic cell solves this by physically separating the two half-reactions. Instead of electrons jumping directly from zinc to copper in the same beaker, the zinc sits in one container and the copper sits in another. The only path for electrons is through an external wire - and on the way, those electrons can power a lightbulb, a motor, or your phone.

The Daniell Cell - The Classic Example

The Daniell cell uses zinc and copper. Here is what happens:

At the anode (zinc side): Zinc atoms lose electrons and dissolve into solution as Zn2+ ions. The zinc electrode gradually shrinks.

Zn(s) -> Zn2+(aq) + 2e-

At the cathode (copper side): Cu2+ ions in solution gain electrons and plate out as solid copper on the electrode. The copper electrode gradually grows.

Cu2+(aq) + 2e- -> Cu(s)

Overall: Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)

The reaction is spontaneous because copper has a higher reduction potential than zinc. Copper ions “want” electrons more than zinc ions do, so electrons flow from zinc to copper through the wire.

Why Separation Matters

If you simply dropped zinc into a copper sulfate solution, electrons would transfer directly at the metal surface. You would see the reaction happen, but you could not capture any electrical energy. By separating the half-cells and connecting them with a wire, you force every electron to travel through the external circuit. That electron flow is electric current, and it can do work.

Key Features of Galvanic Cells

Every electrochemical cell has four essential parts: two electrodes, the electrolyte solutions they sit in, a salt bridge connecting the solutions internally, and a wire connecting the electrodes externally. Remove any one of these, and the cell stops working.

Electrodes - Where the Action Happens

The anode is the electrode where oxidation occurs. In a galvanic cell, the anode is labeled as the negative terminal because electrons are generated there and pushed out through the wire.

The cathode is the electrode where reduction occurs. In a galvanic cell, the cathode is the positive terminal because it attracts the electrons flowing through the wire.

Electrodes can be active or inert:

• Active electrodes participate in the reaction. In the Daniell cell, the zinc anode dissolves and the copper cathode grows. The electrode material is a reactant or product.
• Inert electrodes (platinum or graphite) do not react. They simply provide a surface for electron transfer and a conduction path. Inert electrodes are used when the reacting species are ions in solution or gases.

Electrolyte Solutions

Each half-cell contains an aqueous solution with dissolved ions. The anode compartment contains ions of the anode metal (e.g., ZnSO4 for a zinc anode). The cathode compartment contains ions of the cathode metal (e.g., CuSO4 for a copper cathode).

These ions are the chemical species that gain or lose electrons during the reaction.

The Salt Bridge - The Peacekeeper

As the reaction proceeds, a charge imbalance develops. The anode compartment gains positive ions (metal dissolves), and the cathode compartment loses positive ions (they plate out as metal). Without correction, this charge buildup would immediately stop the reaction - the anode side would become too positive to release more cations, and the cathode side would become too negative to accept more.

The salt bridge fixes this. It is typically a U-tube filled with a concentrated solution of an inert salt like KCl or KNO3 (or a gel saturated with these salts).

What flows through the salt bridge:

• Anions (like Cl- or NO3-) migrate toward the anode to balance the excess positive charge from dissolved metal ions
• Cations (like K+ or Na+) migrate toward the cathode to replace the positive charge lost as metal ions plate out

The salt bridge completes the internal circuit. Without it, the cell potential drops to zero almost instantly.

The External Circuit

A conductive wire connects the two electrodes externally. Electrons flow through this wire from anode to cathode. Any device placed in this circuit (lightbulb, motor, phone) is powered by this electron flow.

Putting It All Together

Drawing a full galvanic cell diagram every time would be exhausting. Cell notation is a shorthand that captures the same information in a single line of text. The MCAT uses this notation in passages, so you need to read it fluently.

The Rules

Cell notation always reads left to right, anode to cathode - the same direction electrons flow through the external circuit.

The Daniell Cell in Cell Notation

Zn(s) | Zn2+(1 M) || Cu2+(1 M) | Cu(s)

Reading from left to right:

1. Zn(s) - solid zinc anode (the electrode)
2. | - phase boundary between the solid electrode and the solution
3. Zn2+(1 M) - zinc ion solution in the anode compartment
4. || - salt bridge
5. Cu2+(1 M) - copper ion solution in the cathode compartment
6. | - phase boundary between solution and solid electrode
7. Cu(s) - solid copper cathode

Examples with Different Setups

A cell with an inert electrode and a gas:

Pt(s) | H2(g) | H+(aq) || Ag+(aq) | Ag(s)

Here, the anode is a platinum electrode where H2 gas is oxidized to H+ ions. The platinum does not react - it just provides a surface.

A cell where both species are in solution (requires inert electrode):

Pt(s) | Fe2+(aq), Fe3+(aq) || MnO4-(aq), Mn2+(aq), H+(aq) | Pt(s)

A comma separates species in the same phase. Fe2+ and Fe3+ are both in the anode solution. Multiple species in the cathode solution are also comma-separated.

Common Mistakes to Avoid

• Do not reverse the order. Anode is always on the left, cathode on the right.
• Do not confuse | and ||. A single line is a phase boundary within a half-cell. The double line is the salt bridge between half-cells.
• Do not forget phase labels. The MCAT expects you to know that electrodes are typically (s), solutions are (aq), and some half-reactions involve (g).

The standard reduction potential table is the single most important tool in electrochemistry. It ranks every half-reaction by how strongly that species “wants” electrons. Learn to read it, and you can predict which reactions are spontaneous, identify the anode and cathode, and calculate cell voltages - all from one table.

What the Table Shows

Each entry in the table is a half-reaction written as a reduction (electrons on the left side):

Cu2+(aq) + 2e- -> Cu(s)   E° = +0.34 V

Zn2+(aq) + 2e- -> Zn(s)   E° = -0.76 V

The E° value (standard reduction potential) tells you how strongly that species pulls electrons toward itself under standard conditions (25°C, 1 M solutions, 1 atm for gases).

The Reference Point - SHE

All reduction potentials are measured relative to the standard hydrogen electrode (SHE), which is assigned E° = 0.00 V by definition:

2H+(aq) + 2e- -> H2(g)   E° = 0.00 V

A species with a positive E° is reduced more easily than H+. A species with a negative E° is reduced less easily than H+.

Reading the Table

Key half-reactions to know for the MCAT:

Predicting Spontaneous Reactions

A spontaneous reaction occurs when the species with the higher (more positive) E° is reduced and the species with the lower (more negative) E° is oxidized.

Example: Will zinc reduce copper ions spontaneously?

• Cu2+ + 2e- -> Cu   E° = +0.34 V (higher, so Cu2+ is reduced)
• Zn2+ + 2e- -> Zn   E° = -0.76 V (lower, so Zn is oxidized)

Yes. Copper has the higher reduction potential, so it “wins” the electrons. Zinc gives them up. The reaction Zn + Cu2+ -> Zn2+ + Cu is spontaneous.

Will copper reduce zinc ions? No. That would require zinc to be reduced (+0.34 V species giving electrons to -0.76 V species), which is nonspontaneous.

Oxidizing and Reducing Agents

• Strong oxidizing agents are species that are easily reduced (high E°). They take electrons from others. F2, MnO4-, Cr2O72- are strong oxidizing agents.
• Strong reducing agents are species that are easily oxidized (low E°). They give electrons to others. Li, Na, Zn are strong reducing agents.

Flipping to Oxidation Potentials

The table lists reduction potentials. If you need the oxidation potential for a half-reaction, simply reverse the sign:

• Reduction: Cu2+ + 2e- -> Cu   E°red = +0.34 V
• Oxidation: Cu -> Cu2+ + 2e-   E°ox = -0.34 V

The MCAT almost always provides reduction potentials, but you should be comfortable flipping them mentally.

Calculating the standard cell potential is the most common quantitative task in MCAT electrochemistry. The formula is straightforward, and every calculation follows the same pattern.

The Master Equation

Important: Both E°cathode and E°anode are the standard reduction potentials from the table. You do not need to reverse any signs before plugging in. The subtraction does the work.

Step-by-Step Process

1. Identify the cathode and anode. The species with the higher (more positive) E° is reduced at the cathode. The species with the lower E° is oxidized at the anode.
2. Look up both reduction potentials from the table.
3. Subtract: E°cell = E°cathode - E°anode.
4. Interpret the sign. Positive = spontaneous (galvanic). Negative = nonspontaneous (requires external energy).

Worked Example 1: Daniell Cell

Half-reactions from the table:

• Cu2+ + 2e- -> Cu   E° = +0.34 V
• Zn2+ + 2e- -> Zn   E° = -0.76 V

Copper has the higher E°, so it is the cathode (reduction). Zinc has the lower E°, so it is the anode (oxidation).

E°cell = E°cathode - E°anode = (+0.34) - (-0.76) = +1.10 V

The positive value confirms this is a spontaneous reaction.

Worked Example 2: Iron-Silver Cell

Half-reactions:

• Ag+ + e- -> Ag   E° = +0.80 V
• Fe2+ + 2e- -> Fe   E° = -0.44 V

E°cell = (+0.80) - (-0.44) = +1.24 V

Silver is reduced, iron is oxidized. Spontaneous.

Worked Example 3: Testing a Nonspontaneous Direction

What if someone proposes: Cu + Zn2+ -> Cu2+ + Zn?

That would make zinc the cathode and copper the anode:

E°cell = E°cathode - E°anode = (-0.76) - (+0.34) = -1.10 V

Negative E°cell means this direction is nonspontaneous. You would need to supply at least 1.10 V of external energy to force this reaction (making it an electrolytic cell).

Critical Rule: Do NOT Multiply E° by Coefficients

When balancing a cell reaction, you may need to multiply one half-reaction to equalize electrons. Never multiply the E° value.

Example: Building a cell from Al and Cu:

• Cu2+ + 2e- -> Cu   E° = +0.34 V
• Al3+ + 3e- -> Al   E° = -1.66 V

To balance electrons, you multiply the copper half-reaction by 3 and the aluminum half-reaction by 2:

• 3Cu2+ + 6e- -> 3Cu   E° is still +0.34 V (not 3 x 0.34)
• 2Al3+ + 6e- -> 2Al   E° is still -1.66 V (not 2 x -1.66)

E°cell = (+0.34) - (-1.66) = +2.00 V

Three quantities tell you the same story about a reaction - whether it is spontaneous, at equilibrium, or nonspontaneous. They are connected by equations you must know cold for the MCAT.

The Three Master Equations

The Triangle of Relationships

These three equations form a triangle. If you know any one of E°, ΔG°, or K, you can calculate the other two.

The Sign Relationships

All three quantities agree on the direction of spontaneity, but with opposite sign conventions for E° versus ΔG°:

Why the Negative Sign in ΔG° = -nFE°?

The negative sign ensures consistency: E° and ΔG° carry opposite signs for the same reaction direction. A spontaneous galvanic cell has a positive E°cell (electron flow is favorable) but a negative ΔG° (free energy decreases). The negative sign in the equation makes this work out mathematically.

Worked Example

For the Daniell cell (Zn-Cu), E°cell = +1.10 V and n = 2 (two electrons transferred).

Calculate ΔG°:

ΔG° = -nFE° = -(2)(96,485)(1.10) = -212,267 J = -212.3 kJ

Negative ΔG° confirms the reaction is spontaneous.

Calculate K:

ΔG° = -RTlnK

-212,267 = -(8.314)(298)lnK

lnK = 212,267 / 2477.6 = 85.7

$K = e^{85.7} \approx$ $1.5 \times 10^{37}$

This enormous K tells you the reaction essentially goes to completion - products are overwhelmingly favored.

Quick Reference: Faraday’s Constant

Faraday’s constant (F) = 96,485 C/mol. This is the total charge carried by one mole of electrons.

For quick MCAT estimation: $F$ is approximately $10^5$ C/mol. So $\Delta G° = -nFE°$ is approximately $-n(10^5)(E°)$ joules.

Standard cell potentials (E°) assume everything is at 1 M concentration, 1 atm pressure, and 25°C. Real cells almost never operate under those conditions. The Nernst equation tells you the actual cell voltage when concentrations deviate from standard.

The Nernst Equation

What the Equation Tells You

The Nernst equation is a GPS for electrochemistry. E° tells you where you would be under standard conditions. The correction term -(RT/nF)lnQ adjusts for where you actually are.

Key Predictions from the Nernst Equation

The critical insight: As Q increases (products build up), the cell voltage decreases. This makes intuitive sense - as the reaction approaches equilibrium, there is less driving force to push electrons.

What Happens at Equilibrium

At equilibrium, $Q = K$ and $E = 0$:

$0 = E° - \dfrac{RT}{nF}\ln K$

Rearranging:

$\displaystyle E° = \dfrac{RT}{nF}\ln K$

This is the equation from the previous section that connects E° and K. The Nernst equation at equilibrium derives it naturally.

Worked Example

For the Daniell cell under non-standard conditions:

Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)

Given: E° = +1.10 V, n = 2, [Zn2+] = 2.0 M, [Cu2+] = 0.010 M, T = 25°C.

Q = [Zn2+]/[Cu2+] = $\frac{2.0}{0.010}$ = 200

Using the simplified Nernst equation:

E = 1.10 - ($\frac{0.0592}{2}$)log(200)

E = 1.10 - (0.0296)(2.30)

E = 1.10 - 0.068 = +1.03 V

The cell voltage is lower than E° because products (Zn2+) are concentrated and reactants (Cu2+) are dilute. The reaction quotient is large (Q > 1), reducing the driving force.

How Concentration Changes Affect E

Understanding how concentration changes affect cell potential connects directly to Le Chatelier’s principle:

• Increasing reactant concentration (e.g., more Cu2+) decreases Q, which makes the correction term less negative, increasing E
• Increasing product concentration (e.g., more Zn2+) increases Q, making the correction term more negative, decreasing E
• Diluting products decreases Q and increases E

What if you built a cell where both electrodes were made of the same metal, dipped into solutions of the same ion at different concentrations? E° would be zero - the same half-reaction on both sides cancels perfectly. Yet the cell still produces a voltage. How?

The Concept

A concentration cell uses identical electrodes and identical ions but at different concentrations in the two half-cells. The driving force comes entirely from the concentration difference, not from a difference in reduction potentials.

The system is trying to equalize concentrations - just like how gases diffuse from high to low concentration. Nature drives toward equilibrium, and the electron flow is the mechanism.

How It Works

Consider a copper concentration cell:

• Anode (dilute side): Cu(s) | Cu2+(0.010 M)
• Cathode (concentrated side): Cu2+(1.0 M) | Cu(s)

At the anode (dilute side): Copper metal oxidizes to add Cu2+ ions to the dilute solution, increasing its concentration.

Cu(s) -> Cu2+(aq) + 2e-

At the cathode (concentrated side): Cu2+ ions are reduced to solid copper, decreasing the concentration.

Cu2+(aq) + 2e- -> Cu(s)

Net effect: The dilute solution gets more concentrated, and the concentrated solution gets more dilute. The system drives toward equal concentrations on both sides.

Calculating the Voltage

Since E° = 0 (identical half-reactions), the Nernst equation gives:

E = 0 - (0.0592/n)log(Q)

For the copper concentration cell with [Cu2+]dilute = 0.010 M and [Cu2+]concentrated = 1.0 M:

Q = [Cu2+]anode / [Cu2+]cathode = 0.010 / 1.0 = 0.010

E = -($\frac{0.0592}{2}$)log(0.010) = -(0.0296)(-2) = +0.0592 V

The voltage is small but positive, confirming the reaction is spontaneous. As the concentrations equalize, Q approaches 1, log(Q) approaches 0, and E approaches 0.

Key Features of Concentration Cells

A galvanic cell converts chemical energy into electrical energy. An electrolytic cell does the reverse - it uses electrical energy to drive a nonspontaneous chemical reaction. If a galvanic cell is a ball rolling downhill, an electrolytic cell is pushing the ball back up.

The Basic Setup

An electrolytic cell requires an external power source (a battery or DC power supply) that forces electrons to flow in the “wrong” direction - against the thermodynamic preference of the reaction.

Key Differences from Galvanic Cells

The Charge Sign Flip

This is the trickiest part for students. In both cell types, oxidation still occurs at the anode and reduction at the cathode (An Ox, Red Cat - always). But the charge signs on the electrodes are reversed:

• In a galvanic cell, the anode is (-) because electrons are generated there and pushed away.
• In an electrolytic cell, the anode is (+) because the external battery pulls electrons away from it. The battery’s positive terminal connects to the anode, making it positive.

Common Electrolysis Reactions

Electrolysis of water:

2H2O(l) -> 2H2(g) + O2(g)

This is nonspontaneous (ΔG° = +474 kJ). An external voltage decomposes water into hydrogen gas at the cathode and oxygen gas at the anode.

Electrolysis of molten NaCl:

At the cathode: 2Na+(l) + 2e- -> 2Na(l) (sodium metal produced)

At the anode: 2Cl-(l) -> Cl2(g) + 2e- (chlorine gas produced)

Overall: 2NaCl(l) -> 2Na(l) + Cl2(g)

This is how sodium metal and chlorine gas are produced industrially.

Electroplating:

Electroplating deposits a thin layer of metal onto an object. The object to be plated is made the cathode (where reduction and metal deposition occur). The plating metal is dissolved in solution or used as a sacrificial anode.

For silver plating: Ag+(aq) + e- -> Ag(s) occurs at the cathode (the object being plated).

Minimum Voltage Required

To drive an electrolytic reaction, the external voltage must exceed the magnitude of the (negative) E°cell for the nonspontaneous reaction. For the electrolysis of water, you need at least 1.23 V (and typically more due to overpotential - extra voltage needed to overcome kinetic barriers at the electrode surface).

If you run an electrolytic cell for a known time at a known current, exactly how much metal will plate out? Faraday’s laws give you a direct calculation chain from amps and seconds to grams of product.

The Key Relationships

The Calculation Chain

Every Faraday’s law problem follows the same four-step chain:

Current and time -> Charge -> Moles of electrons -> Moles of substance -> Grams

1. q = It - Convert current (amps) and time (seconds) to charge (coulombs)
2. n(e-) = q/F - Convert charge to moles of electrons
3. Use stoichiometry - The half-reaction tells you the ratio of moles of electrons to moles of substance
4. m = n x M - Convert moles of substance to grams using molar mass

Worked Example

How many grams of copper are deposited when a 3.00 A current runs through a CuSO4 solution for 2.00 hours?

Half-reaction: Cu2+ + 2e- -> Cu (n = 2 electrons per copper atom)

Step 1: Convert time to seconds: 2.00 h x 3600 s/h = 7200 s

Step 2: Calculate charge: q = It = (3.00 A)(7200 s) = 21,600 C

Step 3: Moles of electrons: n(e-) = q/F = 21,$\frac{600}{96}$,485 = 0.2239 mol e-

Step 4: Moles of Cu: 0.2239 mol e- x (1 mol Cu / 2 mol e-) = 0.1120 mol Cu

Step 5: Grams of Cu: 0.1120 mol x 63.55 g/mol = 7.12 g Cu

Important Unit Relationships

Comparing Two Electrolytic Cells in Series

When two electrolytic cells are connected in series (same current flows through both), the same charge passes through each. The mass deposited at each cathode depends on the molar mass of the metal and the number of electrons in its half-reaction.

Example: Cells containing AgNO3 and CuSO4 are connected in series. The same current flows through both.

• Silver: Ag+ + 1e- -> Ag (1 electron per atom, M = 107.9 g/mol)
• Copper: Cu2+ + 2e- -> Cu (2 electrons per atom, M = 63.55 g/mol)

For the same charge: more silver is deposited because each silver atom requires only 1 electron, while each copper atom requires 2.

The MCAT loves asking you to compare galvanic and electrolytic cells. Some features stay the same between the two, and some flip. Knowing exactly which is which is worth easy points on test day.

What Stays the Same (Always True)

These facts are true in every electrochemical cell, no matter what type:

What Changes Between Cell Types

Why the Electrode Signs Flip

In a galvanic cell, the anode spontaneously produces electrons. Electrons accumulate there, making it negative. The cathode consumes electrons, making it positive.

In an electrolytic cell, the external battery forces the process. The battery’s positive terminal connects to the anode, pulling electrons away from it (making the anode positive). The battery’s negative terminal pushes electrons toward the cathode (making the cathode negative).

The Rechargeable Battery - Both Types in One Device

A rechargeable battery beautifully demonstrates both cell types:

• Discharging (using your phone): The battery operates as a galvanic cell. Spontaneous redox produces current. E > 0, ΔG < 0.
• Charging (plugging in your phone): The charger forces the battery to operate as an electrolytic cell. External energy drives the reverse reaction. E < 0, ΔG > 0.

The anode and cathode actually swap when switching between charging and discharging, because the direction of the reaction reverses.

Quick Decision Flowchart

1. Is the reaction spontaneous? -> Galvanic cell (E° > 0, ΔG < 0)
2. Is external energy required? -> Electrolytic cell (E° < 0, ΔG > 0)
3. Where does oxidation occur? -> Always the anode
4. Where does reduction occur? -> Always the cathode
5. What is the sign of the anode?
   • Galvanic: Negative (electrons generated)
   • Electrolytic: Positive (external battery pulls electrons away)

Everything you have learned in this chapter plays out in two everyday phenomena: batteries that power your devices and corrosion that slowly destroys metal structures. Both are electrochemistry in action.

Types of Batteries

Batteries are classified by whether they can be recharged:

Lead-Acid Battery (Car Battery)

The lead-acid battery is the classic example of a secondary (rechargeable) battery.

Anode (oxidation): Pb(s) + SO42-(aq) -> PbSO4(s) + 2e-

Cathode (reduction): PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- -> PbSO4(s) + 2H2O(l)

Overall: Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)

Key features:

• Both electrodes form PbSO4 during discharge
• The sulfuric acid electrolyte is consumed (density decreases as the battery drains)
• A single cell produces about 2 V; six cells in series give the standard 12 V car battery
• Fully reversible when charged

Nickel-Cadmium (NiCd) Battery

Nickel-cadmium batteries are secondary cells that use nickel oxide hydroxide (NiOOH) as the cathode and cadmium metal (Cd) as the anode, with a KOH electrolyte. They are rechargeable and can deliver high current, but are largely being phased out because cadmium is toxic. The MCAT may reference them as a classic example of a rechargeable battery alongside lead-acid cells.

The key concept: NiCd batteries suffer from a “memory effect” where partial discharge cycles reduce their effective capacity. This is not something you need to know in detail, but it illustrates that real-world batteries do not behave as perfectly reversible electrochemical cells.

Lithium-Ion Battery (Phone/Laptop Battery)

Lithium-ion batteries dominate modern portable electronics because of their high energy density and rechargeability.

During discharge, lithium ions move from the graphite anode to the metal oxide cathode through the electrolyte. During charging, an external voltage forces them back.

The key concept for the MCAT: lithium-ion batteries are secondary cells that alternate between galvanic (discharging) and electrolytic (charging) modes.

Fuel Cells

A fuel cell is a galvanic cell that runs continuously as long as fuel (usually hydrogen) is supplied. Unlike a battery, the reactants are not stored inside the cell - they are fed in from an external source.

Hydrogen fuel cell:

Anode: 2H2(g) -> 4H+(aq) + 4e-

Cathode: O2(g) + 4H+(aq) + 4e- -> 2H2O(l)

Overall: 2H2(g) + O2(g) -> 2H2O(l)

The only product is water. This is why fuel cells are considered “clean” energy.

Corrosion - Unwanted Electrochemistry

Corrosion is the electrochemical destruction of a metal by reaction with substances in its environment. Rusting of iron is the most familiar example.

The Rusting Process

Iron corrosion happens in stages:

1. Anodic region: Fe(s) -> Fe2+(aq) + 2e- (iron dissolves)
2. Cathodic region: O2(g) + 2H2O(l) + 4e- -> 4OH-(aq) (oxygen is reduced)
3. Rust formation: Fe2+ further oxidizes to Fe3+ and combines with oxygen and water to form hydrated iron(III) oxide (rust, Fe2O3 · nH2O)

Corrosion requires both water and oxygen. Iron immersed in pure water without dissolved oxygen does not rust. Iron in dry air does not rust. Both must be present.

Salt accelerates corrosion by increasing the conductivity of the electrolyte solution (dissolved ions carry charge more efficiently). This is why cars rust faster in coastal areas and where roads are salted in winter.

Preventing Corrosion

Cathodic Protection - Sacrificial Anodes

Cathodic protection exploits the activity series. A more active metal (lower reduction potential) is attached to the iron structure. Because the active metal is more easily oxidized, it corrodes instead of the iron.

Example: Zinc blocks attached to a steel ship hull. Zinc (E° = -0.76 V) is oxidized preferentially over iron (E° = -0.44 V). The iron becomes the cathode in this galvanic couple and is protected from corrosion. The zinc blocks are periodically replaced.

Galvanic Corrosion

When two dissimilar metals are in electrical contact in the presence of an electrolyte, the more active metal corrodes faster than it would alone. This is galvanic corrosion.

Example: An iron nail wrapped in copper wire in salt water. Iron (E° = -0.44 V) is more active than copper (E° = +0.34 V), so iron becomes the anode and corrodes rapidly. The copper acts as the cathode and is protected.

This is why plumbers avoid connecting copper pipes directly to steel pipes without a dielectric union.