Electrochemistry

Electrochemistry

Part 1 of 4 — Galvanic Cells & Cell Potential

Redox in One Breath

• Oxidation = LOSS of electrons; reduction = GAIN (OIL RIG).
• The species that gets oxidized is the reducing agent; the species that gets reduced is the oxidizing agent — agents are named for what they do to the OTHER species.

The Two Iron Laws of Electrodes

In EVERY electrochemical cell, galvanic or electrolytic:

• AN OX: oxidation happens at the ANODE.
• RED CAT: reduction happens at the CATHODE.

These never flip. What flips between cell types (Part 2) is the SIGN of each electrode.

Anatomy of a Galvanic (Voltaic) Cell

A galvanic cell harvests a SPONTANEOUS redox reaction as electrical energy — chemical energy out as current. Classic build (the Daniell cell): a zinc strip in $\text{Zn}^{2+}$ solution, a copper strip in $\text{Cu}^{2+}$ solution, wire between the metals, salt bridge between the solutions.

• Zinc is easier to oxidize: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ at the anode; the zinc electrode slowly DISSOLVES.
• Electrons travel through the WIRE from anode to cathode.
• At the cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$; copper PLATES onto the electrode and the blue $\text{Cu}^{2+}$ solution fades.
• In a galvanic cell the anode is labeled $(-)$ and the cathode $(+)$: electrons pour OUT of the anode terminal.

The Salt Bridge

Without it, charge would pile up in each beaker and the cell would die instantly. The bridge lets inert ions flow to keep both solutions neutral:

• Anions migrate toward the ANODE compartment (where positive ions are being produced).
• Cations migrate toward the CATHODE compartment (where positive ions are being consumed).

Ions complete the circuit in solution; electrons never swim.

Standard Cell Potential

Every half-reaction has a standard REDUCTION potential $E^\circ$; more positive means more eager to be reduced. The half-cell with the higher reduction potential runs as the cathode:

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$

(both taken from the reduction-potential table, no sign gymnastics). A spontaneous (galvanic) cell has $E^\circ_{cell} > 0$, which corresponds to $\Delta G < 0$ via $\Delta G^\circ = -nFE^\circ_{cell}$.

Never multiply $E^\circ$ by stoichiometric coefficients — potential is an intensive property, energy per unit charge.

Worked Example — Building the Daniell Cell from a Table

Given standard reduction potentials: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$, $E^\circ = +0.34\;\text{V}$; and $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$, $E^\circ = -0.76\;\text{V}$.

Step 1 — Assign the electrodes. Copper's reduction potential is higher, so copper is REDUCED: copper = cathode, zinc = anode.

Step 2 — Cell potential.

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

Positive, confirming spontaneity — this is a working battery.

Step 3 — Trace every mover. Electrons: zinc strip, through the wire, to the copper strip. Sulfate anions in the salt bridge: toward the zinc (anode) beaker. Mass: zinc electrode shrinks, copper electrode grows.

Worked Example — Predicting Whether a Reaction Goes

Will $\text{Ni}^{2+}$ oxidize solid silver? ($E^\circ$: $\text{Ag}^+/\text{Ag} = +0.80\;\text{V}$, $\text{Ni}^{2+}/\text{Ni} = -0.25\;\text{V}$)

Step 1 — Proposed roles. $\text{Ni}^{2+}$ would be reduced (cathode role); Ag would be oxidized (anode role).

Step 2 — Compute. $E^\circ_{cell} = -0.25 - 0.80 = -1.05\;\text{V}$.

Negative → nonspontaneous. Silver stays untouched in nickel solution. (The REVERSE reaction, $+1.05\;\text{V}$, is the one that runs: silver ion plates onto nickel.)

Galvanic Cells & Cell Potential 🎯

Key Takeaways — Part 1

• OIL RIG; oxidizing agent gets reduced, reducing agent gets oxidized
• AN OX and RED CAT hold in every cell, always
• Galvanic: spontaneous, chemical to electrical; anode $(-)$ dissolves, cathode $(+)$ plates; electrons wire-only, anode to cathode
• Salt bridge: anions to anode, cations to cathode — keeps beakers neutral
• $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ (reduction values as-is); positive means spontaneous ($\Delta G^\circ = -nFE^\circ < 0$); never scale $E^\circ$ by coefficients

Electrochemistry

Part 2 of 4 — Electrolytic Cells & Faraday's Laws

Electrolytic Cells — Redox Driven Backward

An electrolytic cell uses an EXTERNAL power source to force a NONSPONTANEOUS reaction ($E_{cell} < 0$, $\Delta G > 0$). Electrical energy in, chemical energy stored. Applications: electroplating jewelry, purifying copper, splitting water, producing aluminum and chlorine industrially.

The Sign Flip (and What Never Flips)

• AN OX / RED CAT: universal.
• Electrons ALWAYS enter the cell at the cathode — also universal.
• The SIGNS flip because in an electrolytic cell the power supply's $+$ terminal grabs the anode (yanking electrons out to force oxidation), and its $-$ terminal feeds the cathode.

Quick mental check for electrolytic cells: cations in solution drift to the NEGATIVE cathode and get reduced there — plating always happens at the cathode.

Electrolysis Logic

The power supply must provide at least $|E_{cell}|$ volts to force the reaction. In molten salt electrolysis (say $\text{NaCl}$), sodium metal forms at the cathode, chlorine gas at the anode. In AQUEOUS solutions, water competes: species are oxidized/reduced in order of ease, which is why electrolyzing salt water makes $\text{H}_2$ at the cathode instead of sodium metal.

Faraday's Laws — Counting Electrons Like Moles

Charge and chemical change are locked together by the Faraday constant:

$F = 96{,}485\;\text{C/mol}\;e^- \approx 96{,}500$

The working chain for every quantitative electrolysis problem:

$\text{moles of } e^- = \frac{It}{F}$

then use the half-reaction's stoichiometry: depositing one mole of $\text{M}^{n+}$ as metal takes $n$ moles of electrons. So

$\text{mass deposited} = \frac{It}{F} \times \frac{M_{molar}}{n}$

Same charge deposits HALF as many moles of a $2+$ ion as of a $1+$ ion — the charge on the ion is the exchange rate.

Worked Example — Copper Electroplating (Passage Style)

A technician plates copper from $\text{CuSO}_4$ solution ($\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$, molar mass $63.5\;\text{g/mol}$) using a $2.0\;\text{A}$ current for $965\;\text{s}$.

Step 1 — Total charge.

$Q = It = (2.0)(965) = 1930\;\text{C}$

Step 2 — Moles of electrons.

$n_{e^-} = \frac{1930}{96{,}500} = 0.020\;\text{mol}$

(MCAT arithmetic: $1930$ is $2\%$ of $96{,}500$.)

Step 3 — Moles and mass of copper. Two electrons per copper:

$n_{Cu} = \frac{0.020}{2} = 0.010\;\text{mol}, \quad m = (0.010)(63.5) = 0.635\;\text{g}$

The copper appears on the CATHODE — the electrode wired to the supply's negative terminal.

Doubling check: doubling current, time, or both scales the deposit linearly; switching to a $1+$ metal at the same charge would double the moles deposited.

Worked Example — Which Electrode Is Which?

A student electrolyzes molten $\text{NaCl}$. Where does each product form?

Step 1 — Identify the redox. $\text{Na}^+ + e^- \rightarrow \text{Na}$ is reduction; $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$ is oxidation.

Step 2 — Apply the iron laws. Reduction at the cathode → shiny sodium at the NEGATIVE electrode. Oxidation at the anode → green chlorine gas bubbling off the POSITIVE electrode.

Step 3 — Sanity. Both products are higher-energy than the salt — exactly what "pumping energy in" should produce. Left alone, they would happily react back.

Electrolytic Cells & Faraday's Laws 🎯

Key Takeaways — Part 2

• Electrolytic = driven, nonspontaneous ($\Delta G > 0$); electrical energy stored as chemical energy
• AN OX, RED CAT, and electrons-enter-at-cathode never change; electrode SIGNS flip (electrolytic anode $+$, cathode $-$)
• Plating and metal deposition happen at the cathode; gas from anion oxidation at the anode
• $F \approx 96{,}500\;\text{C/mol}$; moles $e^- = It/F$; divide by $n$ for moles of metal, then multiply by molar mass
• Aqueous electrolysis: water can out-compete the salt's ions at either electrode

Electrochemistry

Part 3 of 4 — Batteries, Concentration Cells & the Nernst Equation

Off-Standard Conditions: the Nernst Equation

Standard potentials assume $1\;\text{M}$ everything. Real cells drift as they run. At $25^\circ\text{C}$:

$E = E^\circ - \frac{0.0592}{n}\log Q$

where $Q$ is the reaction quotient (products over reactants) and $n$ the electrons transferred. Qualitative readings the MCAT actually tests:

• Products pile up ($Q$ rises) → $E$ FALLS. Every battery dies this way.
• Reactants enriched ($Q < 1$, $\log Q < 0$) → $E$ climbs ABOVE $E^\circ$.
• At equilibrium, $Q = K$ and $E = 0$: a dead battery is a cell at equilibrium.

Concentration Cells

Build a cell from two IDENTICAL half-cells — same metal, same ion — differing only in concentration. Then $E^\circ_{cell} = 0$: all the driving force is entropic, pushing the two concentrations toward equality.

• Reduction happens in the CONCENTRATED compartment (consuming its excess ions); oxidation in the dilute one (topping it up).
• The voltage is small — from the Nernst equation, $E = \frac{0.0592}{n}\log\frac{[\text{conc}]}{[\text{dilute}]}$ — about $59\;\text{mV}$ per tenfold ratio per electron.
• The cell runs until concentrations match, then dies at $E = 0$.

Biology is powered by this. The resting membrane potential of a neuron is concentration-cell physics: the $\text{K}^+$ gradient across the membrane sets a Nernst potential near $-90\;\text{mV}$; the $\text{Na}^+$ gradient, near $+60\;\text{mV}$. The MCAT loves this bridge between chapters.

Real Batteries

A battery is one or more galvanic cells packaged with practical trade-offs:

• Lead-acid (car battery): six ~$2\;\text{V}$ cells in series for $12\;\text{V}$; both electrodes are lead compounds in sulfuric acid; RECHARGEABLE; heavy but delivers huge surge current.
• Nickel-cadmium / NiMH / lithium-ion: rechargeable chemistries; lithium-ion wins on energy per kilogram, which is why it owns phones and pacemakers.
• Alkaline (standard disposables): about $1.5\;\text{V}$; not designed for recharging.

Recharging = running the cell as an electrolytic cell: the charger overpowers the cell's own EMF and forces current backward, regenerating reactants. Each electrode keeps its polarity sign, but its ROLE swaps (the discharge cathode becomes the charging anode).

Link back to circuits: a battery's internal resistance makes its terminal voltage sag under load ($V = \varepsilon - Ir$) — chemistry supplies the EMF, physics collects the tax.

Worked Example — A Copper Concentration Cell

Two copper electrodes sit in $\text{Cu}^{2+}$ solutions of $0.010\;\text{M}$ and $1.0\;\text{M}$, joined by wire and salt bridge.

Step 1 — Assign the electrodes. Reduction runs in the CONCENTRATED ($1.0\;\text{M}$) beaker — that side is the cathode. The dilute side oxidizes its copper electrode, releasing more $\text{Cu}^{2+}$.

Step 2 — Voltage. With $n = 2$ and a $100$-fold ratio:

$E = \frac{0.0592}{2}\log\frac{1.0}{0.010} = (0.0296)(2) = 0.0592\;\text{V}$

About $59\;\text{mV}$ — real but tiny; concentration cells are weak batteries.

Step 3 — Endgame. The concentrated side dilutes, the dilute side concentrates; at equal concentrations, $E = 0$ and the current stops.

Worked Example — Reading a Dying Battery with Nernst

A Daniell cell ($E^\circ = 1.10\;\text{V}$, $n = 2$, $Q = [\text{Zn}^{2+}]/[\text{Cu}^{2+}]$) has run until $[\text{Zn}^{2+}] = 1.0\;\text{M}$ and $[\text{Cu}^{2+}] = 0.01\;\text{M}$.

Step 1 — Reaction quotient. $Q = 1.0/0.01 = 100$.

Step 2 — Nernst correction.

$E = 1.10 - \frac{0.0592}{2}\log(100) = 1.10 - (0.0296)(2) = 1.10 - 0.06 = 1.04\;\text{V}$

Step 3 — Interpret. Product ion accumulating and reactant ion depleting shaved $0.06\;\text{V}$ off the standard voltage — and the slide continues until equilibrium ($E = 0$, battery dead).

Batteries, Concentration Cells & Nernst 🎯

Key Takeaways — Part 3

• Nernst at $25^\circ$C: $E = E^\circ - (0.0592/n)\log Q$; products up → voltage down; equilibrium = dead ($E = 0$)
• Concentration cell: $E^\circ = 0$; reduction on the concentrated side; about $59\;\text{mV}$ per decade per electron; runs until equal
• Neuron resting potential = membrane concentration cell ($\text{K}^+$ near $-90\;\text{mV}$)
• Lead-acid: six 2 V cells in series, rechargeable, big surge current; lithium-ion: best energy density
• Recharging runs the cell electrolytically — roles swap, polarity signs stay; terminal voltage still obeys $V = \varepsilon - Ir$

Electrochemistry

Part 4 of 4 — MCAT Integration: Sign Conventions, Cell Diagrams & Passage Strategy

Electrochemistry passages hand you a table of reduction potentials and a cell sketch, then probe whether your sign conventions survive contact. The playbook:

Strategy 1 — The Master Sign Table (Memorize Once)

Chemistry (AN OX, RED CAT) is invariant; only signs and spontaneity flip. When a question feels ambiguous, first decide WHICH cell type you are in.

Strategy 2 — Read Cell Diagram Notation Left to Right

$\text{Zn}(s)\;|\;\text{Zn}^{2+}(aq)\;||\;\text{Cu}^{2+}(aq)\;|\;\text{Cu}(s)$

• ANODE on the LEFT, cathode on the right — by convention, always.
• Single bar $|$ = phase boundary; double bar $||$ = salt bridge.
• Electrons flow left to right through the external wire.

Strategy 3 — Work the Potential Table Like a Ladder

Reduction potentials sorted from most positive to most negative:

• Best oxidizing agent = species on the LEFT of the topmost (most positive) half-reaction — it is the most eager electron thief.
• Best reducing agent = species on the RIGHT of the bottommost (most negative) half-reaction — most eager electron donor.
• Biggest possible battery = pair the top half-reaction (as cathode) with the bottom one (as anode): $E^\circ_{cell} = E^\circ_{top} - E^\circ_{bottom}$.

Strategy 4 — Potentials Are Intensive

Doubling a half-reaction's coefficients does NOT double its $E^\circ$ — volts are joules PER coulomb, already normalized. Only $\Delta G = -nFE$ scales with $n$. Passages that multiply a half-reaction by 2 and invite you to double the potential are running the classic trap.

Strategy 5 — Connect the Chapters

• Cell EMF drives circuit analysis: current through an external resistor is $I = \varepsilon/(R + r)$, power delivered $I^2R$ — circuits vocabulary, electrochemical source.
• Electrolysis stoichiometry is dimensional analysis: $\text{A}\cdot\text{s} \rightarrow \text{C} \rightarrow \text{mol}\;e^- \rightarrow \text{mol metal} \rightarrow \text{grams}$. Write the chain; never freelance.
• Membrane potentials, electrophoresis, and galvanic corrosion are the same ideas wearing biology, lab-methods, and materials costumes.

Worked Example — Interrogating a Potential Table

A passage provides standard reduction potentials: $\text{F}_2/\text{F}^-$ $+2.87\;\text{V}$; $\text{Ag}^+/\text{Ag}$ $+0.80\;\text{V}$; $\text{Zn}^{2+}/\text{Zn}$ $-0.76\;\text{V}$; $\text{Li}^+/\text{Li}$ $-3.04\;\text{V}$.

Best oxidizing agent: $\text{F}_2$ — it owns the most positive reduction potential, so it is reduced most readily.

Best reducing agent: solid $\text{Li}$ — the product side of the most NEGATIVE half-reaction; reversing that half-reaction (oxidation of Li) is the most favorable electron donation. Note it is Li metal, not $\text{Li}^+$: the ion has no electron to give.

Largest cell voltage available:

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 2.87 - (-3.04) = 5.91\;\text{V}$

pairing fluorine reduction with lithium oxidation — which is why lithium chemistry (tamed) powers batteries.

Worked Example — Galvanic or Electrolytic? Follow the Wire

A passage shows a cell wired to a DC power supply, silver bar on one side, spoon on the other, in $\text{Ag}^+$ solution. The spoon is connected to the supply's negative terminal. What happens?

Step 1 — Cell type. External supply drives it → electrolytic.

Step 2 — Electrode roles. Negative terminal feeds electrons to the spoon → the spoon is the cathode → $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ deposits silver onto the spoon. That is electroplating.

Step 3 — The other side. The silver bar (anode, $+$) oxidizes: $\text{Ag} \rightarrow \text{Ag}^+ + e^-$, replenishing the solution. The bar shrinks at the same rate the spoon gains — the solution concentration holds steady.

MCAT Integration: Electrochemistry 🎯

Key Takeaways — Part 4

• Master table: chemistry fixed (AN OX, RED CAT), signs flip (galvanic anode $-$, electrolytic anode $+$)
• Cell notation: anode left, cathode right; $|$ phase boundary, $||$ salt bridge
• Table ladder: most positive $E^\circ$ = best oxidizer (that species gets reduced); most negative couple's METAL = best reducer; max voltage = top minus bottom
• $E^\circ$ is intensive — never doubled with coefficients; only $\Delta G = -nFE$ scales with $n$
• Follow the wire to classify a cell: external supply = electrolytic; plating happens where electrons enter (cathode)
• Battery in a circuit: $I = \varepsilon/(R + r)$, terminal voltage sags by $Ir$