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Wyatt's Notes — IB

Redox Reactions (Advanced)

1. Electrochemical Cells

Galvanic (Voltaic) Cells

A galvanic cell converts chemical energy from spontaneous redox reactions into electrical Energy.

Cell Components

ComponentFunction
AnodeSite of oxidation; negative electrode
CathodeSite of reduction; positive electrode
Salt bridgeMaintains electrical neutrality by allowing ion flow
External circuitAllows electron flow from anode to cathode
ElectrolyteProvides ions for internal conduction

Half-Reactions and Cell Notation

The cell is written with the anode on the left and cathode on the right:

Zn(s)∣Zn2+(aq)∥Cu2+(aq)∣Cu(s)\mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(aq) \parallel \mathrm{Cu}^{2+}(aq) \mid \mathrm{Cu}(s)
  • Single vertical line (∣\mid): phase boundary
  • Double vertical line (∥\parallel): salt bridge
  • Anode (oxidation) is on the left; cathode (reduction) is on the right

Standard Electrode Potentials (E∘E^\circ)

Each half-reaction has a standard electrode potential measured relative to the standard hydrogen Electrode (SHE), assigned E∘=0.00VE^\circ = 0.00\mathrm{ V}.

2H+(aq,1M)+2e−⇌H2(g,100kPa)E∘=0.00V2\mathrm{H}^+(aq, 1\mathrm{ M}) + 2e^- \rightleftharpoons \mathrm{H}_2(g, 100\mathrm{ kPa}) \qquad E^\circ = 0.00\mathrm{ V}

Standard conditions: 298K298\mathrm{ K}, 1M1\mathrm{ M} solutions, 100kPa100\mathrm{ kPa} gas pressure.

Standard Reduction Potentials (Selected)

Half-reactionE∘E^\circ (V)
F2+2e−→2F−\mathrm{F}_2 + 2e^- \to 2\mathrm{F}^-+2.87+2.87
MnO4−+8H++5e−→Mn2++4H2O\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}+1.51+1.51
Cl2+2e−→2Cl−\mathrm{Cl}_2 + 2e^- \to 2\mathrm{Cl}^-+1.36+1.36
Ag++e−→Ag\mathrm{Ag}^+ + e^- \to \mathrm{Ag}+0.80+0.80
Cu2++2e−→Cu\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}+0.34+0.34
2H++2e−→H22\mathrm{H}^+ + 2e^- \to \mathrm{H}_20.000.00
Fe2++2e−→Fe\mathrm{Fe}^{2+} + 2e^- \to \mathrm{Fe}−0.44-0.44
Zn2++2e−→Zn\mathrm{Zn}^{2+} + 2e^- \to \mathrm{Zn}−0.76-0.76
Na++e−→Na\mathrm{Na}^+ + e^- \to \mathrm{Na}−2.71-2.71

Calculating Standard Cell Potential

Ecell∘=Ecathode∘−Eanode∘E_{\mathrm{cell}}^\circ = E_{\mathrm{cathode}}^\circ - E_{\mathrm{anode}}^\circ

Example — Daniell Cell

Zn∣Zn2+∥Cu2+∣Cu\mathrm{Zn} \mid \mathrm{Zn}^{2+} \parallel \mathrm{Cu}^{2+} \mid \mathrm{Cu}

Anode (oxidation): Zn→Zn2++2e−E∘=−0.76V\mathrm{Zn} \to \mathrm{Zn}^{2+} + 2e^- \qquad E^\circ = -0.76\mathrm{ V}

Cathode (reduction): Cu2++2e−→CuE∘=+0.34V\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu} \qquad E^\circ = +0.34\mathrm{ V}

Ecell∘=0.34−(−0.76)=+1.10VE_{\mathrm{cell}}^\circ = 0.34 - (-0.76) = +1.10\mathrm{ V}

Predicting Spontaneity

ConditionResult
Ecell∘>0E_{\mathrm{cell}}^\circ \gt 0Spontaneous (galvanic cell)
Ecell∘=0E_{\mathrm{cell}}^\circ = 0At equilibrium
Ecell∘<0E_{\mathrm{cell}}^\circ \lt 0Non-spontaneous (electrolytic cell required)

The more positive E∘E^\circ value, the stronger the oxidizing agent. The more negative E∘E^\circ Value, the stronger the reducing agent.

Common Pitfalls

  • Always subtract Eanode∘E_{\mathrm{anode}}^\circ from Ecathode∘E_{\mathrm{cathode}}^\circ — never reverse the sign of a reduction potential.
  • E∘E^\circ values are intensive properties: they do not depend on the stoichiometric coefficients.
  • The SHE is the reference, not necessarily the actual anode or cathode.

2. The Nernst Equation

For non-standard conditions, the cell potential is given by the Nernst equation:

E_{\mathrm{cell}} = E_{\mathrm{cell}}^\circ - \frac`\{RT}``\{nF}`\ln Q

At 298K298\mathrm{ K}This simplifies to:

Ecell=Ecell∘−0.0592nlog⁡QE_{\mathrm{cell}} = E_{\mathrm{cell}}^\circ - \frac{0.0592}{n}\log Q

Where:

  • nn = number of moles of electrons transferred
  • F=96485C/molF = 96485\mathrm{ C/mol} (Faraday constant)
  • QQ = reaction quotient (activities of products / activities of reactants)

Relationship to Gibbs Free Energy

ΔG∘=−nFEcell∘\Delta G^\circ = -nFE_{\mathrm{cell}}^\circΔG=−nFEcell\Delta G = -nFE_{\mathrm{cell}}

At equilibrium, Ecell=0E_{\mathrm{cell}} = 0 and ΔG=0\Delta G = 0So:

E_{\mathrm{cell}}^\circ = \frac`\{RT}``\{nF}`\ln K = \frac{0.0592}{n}\log K

Example

For the Daniell cell: Ecell∘=1.10VE_{\mathrm{cell}}^\circ = 1.10\mathrm{ V}, n=2n = 2.

log⁡K=nE∘0.0592=2×1.100.0592=37.2\log K = \frac{nE^\circ}{0.0592} = \frac{2 \times 1.10}{0.0592} = 37.2K=1037.2=1.6×1037K = 10^{37.2} = 1.6 \times 10^{37}

The extremely large KK confirms that the reaction proceeds essentially to completion.

3. Electrolysis

Principles

Electrolysis uses electrical energy to drive a non-spontaneous redox reaction. The electrolytic Cell has:

  • Anode: positive electrode (oxidation)
  • Cathode: negative electrode (reduction)
  • External power source drives electrons from anode to cathode

Faraday”s Laws

First law: The amount of substance produced at an electrode is proportional to the charge Passed.

M=Q⋅Mn⋅F=It⋅Mn⋅FM = \frac{Q \cdot M}{n \cdot F} = \frac{It \cdot M}{n \cdot F}

Where:

  • Q=ItQ = It (charge = current ×\times time)
  • MM = molar mass (g/mol\mathrm{g/mol})
  • nn = number of moles of electrons per mole of product
  • F=96485C/molF = 96485\mathrm{ C/mol}

Second law: The masses of different substances produced by the same charge are proportional to Their equivalent masses (M/nM/n).

Example

Calculate the mass of copper deposited when a current of 2.00A2.00\mathrm{ A} is passed through CuSO4\mathrm{CuSO}_4 solution for 30.030.0 minutes.

Q=It=2.00×30.0×60=3600CQ = It = 2.00 \times 30.0 \times 60 = 3600\mathrm{ C}M=3600×63.552×96485=228780192970=1.19gM = \frac{3600 \times 63.55}{2 \times 96485} = \frac{228780}{192970} = 1.19\mathrm{ g}

Electrolysis of Aqueous Solutions

When electrolysing aqueous solutions, water can also be oxidized or reduced:

Atcathode(reduction):2H2O+2e−→H2+2OH−E∘=−0.83V\mathrm{At cathode (reduction): } 2\mathrm{H}_2\mathrm{O} + 2e^- \to \mathrm{H}_2 + 2\mathrm{OH}^- \qquad E^\circ = -0.83\mathrm{ V}Atanode(oxidation):2H2O→O2+4H++4e−E∘=−1.23V\mathrm{At anode (oxidation): } 2\mathrm{H}_2\mathrm{O} \to \mathrm{O}_2 + 4\mathrm{H}^+ + 4e^- \qquad E^\circ = -1.23\mathrm{ V}

Discharge Rules (in Aqueous Solution)

At the cathode: the species with the least negative (most positive) E∘E^\circ is reduced.

Cathode productsCondition
MetalIf the metal is less reactive than hydrogen (E∘>−0.83VE^\circ \gt -0.83\mathrm{ V})
HydrogenIf the metal is more reactive than hydrogen

At the anode: the species with the least positive (most negative) E∘E^\circ is oxidized.

Anode productsCondition
Halogen (Cl2_2Br2_2I2_2)If halide ions present (except F−^-)
OxygenOtherwise (from water oxidation)

Example — Electrolysis of CuSO4\mathrm{CuSO}_4(aq) with inert electrodes

Cathode: Cu2++2e−→Cu\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu} (E∘=+0.34VE^\circ = +0.34\mathrm{ V}More positive than Water’s −0.83V-0.83\mathrm{ V})

Anode: 2H2O→O2+4H++4e−2\mathrm{H}_2\mathrm{O} \to \mathrm{O}_2 + 4\mathrm{H}^+ + 4e^- (sulfate is not oxidised; Water is oxidised instead)

The solution becomes acidic as H+\mathrm{H}^+ accumulates at the anode.

Electrolysis of Molten Salts

No water present, so only the ions from the salt are discharged.

Example — Molten NaCl\mathrm{NaCl}

Cathode: Na++e−→Na(l)\mathrm{Na}^+ + e^- \to \mathrm{Na}(l)

Anode: 2Cl−→Cl2(g)+2e−2\mathrm{Cl}^- \to \mathrm{Cl}_2(g) + 2e^-

This is the Downs process for industrial sodium production.

Common Pitfalls

  • In aqueous solutions, water may be preferentially discharged instead of the expected ion.
  • Faraday’s constant is 96485C/mol96485\mathrm{ C/mol}Not 9650096500 (though 9650096500 is often used for estimation).
  • The anode in an electrolytic cell is positive (opposite to a galvanic cell).

4. Applications of Electrochemistry

Fuel Cells

A fuel cell converts chemical energy from a fuel ( H2\mathrm{H}_2) directly into electricity. The overall reaction:

2H2+O2→2H2O2\mathrm{H}_2 + \mathrm{O}_2 \to 2\mathrm{H}_2\mathrm{O}

Advantages over combustion engines:

  • Higher efficiency (no Carnot limit on electrochemical conversion).
  • No CO2\mathrm{CO}_2 emissions (only H2O\mathrm{H}_2\mathrm{O} as product).
  • Quiet operation.

Limitations:

  • Hydrogen storage and transport.
  • Cost of platinum catalysts.
  • Infrastructure for hydrogen refuelling.

Corrosion

Rusting of iron is an electrochemical process:

Anode (oxidation): Fe→Fe2++2e−\mathrm{Fe} \to \mathrm{Fe}^{2+} + 2e^-

Cathode (reduction): O2+2H2O+4e−→4OH−\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} + 4e^- \to 4\mathrm{OH}^-

Overall: 4Fe+3O2+6H2O→4Fe(OH)34\mathrm{Fe} + 3\mathrm{O}_2 + 6\mathrm{H}_2\mathrm{O} \to 4\mathrm{Fe(OH)}_3 (rust)

Prevention methods:

MethodPrinciple
Sacrificial anode (Zn)Zn is more reactive, corrodes preferentially
Galvanization (Zn coating)Same principle; Zn acts as sacrificial anode
Painting/oilingBarrier between Fe and O2_2/H2_2O
Alloying (stainless steel)Cr forms protective oxide layer

Practice Problems

Problem 1

A cell is constructed from Fe∣Fe2+\mathrm{Fe} \mid \mathrm{Fe}^{2+} and Ag∣Ag+\mathrm{Ag} \mid \mathrm{Ag}^+. Calculate Ecell∘E_{\mathrm{cell}}^\circ and write the overall equation.

Solution:

Fe2++2e−→Fe\mathrm{Fe}^{2+} + 2e^- \to \mathrm{Fe}: E∘=−0.44VE^\circ = -0.44\mathrm{ V} (anode, oxidation)

Ag++e−→Ag\mathrm{Ag}^+ + e^- \to \mathrm{Ag}: E∘=+0.80VE^\circ = +0.80\mathrm{ V} (cathode, reduction)

Ecell∘=0.80−(−0.44)=+1.24VE_{\mathrm{cell}}^\circ = 0.80 - (-0.44) = +1.24\mathrm{ V}

Overall: Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)\mathrm{Fe}(s) + 2\mathrm{Ag}^+(aq) \to \mathrm{Fe}^{2+}(aq) + 2\mathrm{Ag}(s)

Problem 2

What current is required to produce 5.00g5.00\mathrm{ g} of aluminium from molten Al2O3\mathrm{Al}_2\mathrm{O}_3 in 2.002.00 hours?

Solution:

Al3++3e−→Al,n=3,M=27.0g/mol\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}, \quad n = 3, \quad M = 27.0\mathrm{ g/mol}N(Al)=5.0027.0=0.185molN(\mathrm{Al}) = \frac{5.00}{27.0} = 0.185\mathrm{ mol}Q=n(Al)×3×F=0.185×3×96485=53600CQ = n(\mathrm{Al}) \times 3 \times F = 0.185 \times 3 \times 96485 = 53600\mathrm{ C}I=Qt=536002.00×3600=536007200=7.44AI = \frac{Q}{t} = \frac{53600}{2.00 \times 3600} = \frac{53600}{7200} = 7.44\mathrm{ A}
Problem 3

Use the Nernst equation to calculate the cell potential for Zn∣Zn2+(0.010M)∥Cu2+(1.0M)∣Cu\mathrm{Zn} \mid \mathrm{Zn}^{2+}(0.010\mathrm{ M}) \parallel \mathrm{Cu}^{2+}(1.0\mathrm{ M}) \mid \mathrm{Cu} At 298K298\mathrm{ K}.

Solution:

Ecell∘=0.34−(−0.76)=1.10VE_{\mathrm{cell}}^\circ = 0.34 - (-0.76) = 1.10\mathrm{ V}Q=[Zn2+][Cu2+]=0.0101.0=0.010Q = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]} = \frac{0.010}{1.0} = 0.010Ecell=1.10−0.05922log⁡(0.010)=1.10−0.05922×(−2)=1.10+0.059=1.16VE_{\mathrm{cell}} = 1.10 - \frac{0.0592}{2}\log(0.010) = 1.10 - \frac{0.0592}{2} \times (-2) = 1.10 + 0.059 = 1.16\mathrm{ V}

The lower concentration of Zn2+\mathrm{Zn}^{2+} drives the reaction further (Le Chatelier), increasing The cell potential.

Problem 4

Explain why a block of zinc attached to an iron ship hull prevents the iron from rusting.

Solution:

Zinc has a more negative standard reduction potential (E∘=−0.76VE^\circ = -0.76\mathrm{ V}) than iron (E∘=−0.44VE^\circ = -0.44\mathrm{ V}), making zinc a stronger reducing agent. When connected electrically, Zinc acts as the anode and is preferentially oxidised:

Zn→Zn2++2e−\mathrm{Zn} \to \mathrm{Zn}^{2+} + 2e^-

The electrons flow to the iron, which becomes the cathode where oxygen reduction occurs. Since the Iron is not oxidised, it does not corrode. The zinc is the “sacrificial anode” and is periodically Replaced.

Worked Examples

Worked Example: Determining the Anode and Cathode from Potentials

Given the following half-reactions and their standard reduction potentials, construct the spontaneous galvanic cell. Write the cell notation, overall equation, and calculate Ecell∘E_{\mathrm{cell}}^\circ.

Sn2++2e−→SnE∘=−0.14V\mathrm{Sn}^{2+} + 2e^- \to \mathrm{Sn} \qquad E^\circ = -0.14\mathrm{ V}Pb2++2e−→PbE∘=−0.13V\mathrm{Pb}^{2+} + 2e^- \to \mathrm{Pb} \qquad E^\circ = -0.13\mathrm{ V}
Solution

The more positive E∘E^\circ value corresponds to the stronger oxidizing agent (cathode). The more negative E∘E^\circ corresponds to the stronger reducing agent (anode).

  • Anode (oxidation): Sn→Sn2++2e−\mathrm{Sn} \to \mathrm{Sn}^{2+} + 2e^- (E∘=−0.14VE^\circ = -0.14\mathrm{ V})
  • Cathode (reduction): Pb2++2e−→Pb\mathrm{Pb}^{2+} + 2e^- \to \mathrm{Pb} (E∘=−0.13VE^\circ = -0.13\mathrm{ V})

Cell notation:

Sn(s)∣Sn2+(aq)∥Pb2+(aq)∣Pb(s)\mathrm{Sn}(s) \mid \mathrm{Sn}^{2+}(aq) \parallel \mathrm{Pb}^{2+}(aq) \mid \mathrm{Pb}(s)

Overall equation: Sn(s)+Pb2+(aq)→Sn2+(aq)+Pb(s)\mathrm{Sn}(s) + \mathrm{Pb}^{2+}(aq) \to \mathrm{Sn}^{2+}(aq) + \mathrm{Pb}(s)

Ecell∘=Ecathode∘−Eanode∘=(−0.13)−(−0.14)=+0.01VE_{\mathrm{cell}}^\circ = E_{\mathrm{cathode}}^\circ - E_{\mathrm{anode}}^\circ = (-0.13) - (-0.14) = +0.01\mathrm{ V}

The small positive value means the reaction is spontaneous but only marginally so. The equilibrium lies close to the centre.

Worked Example: Electrolysis Product Prediction

Predict the products at each electrode when aqueous NaBr\mathrm{NaBr} is electrolysed using inert platinum electrodes. Write half-equations and the overall equation.

Solution

Cathode (reduction): The cations present are Na+\mathrm{Na}^+ and H+\mathrm{H}^+ (from water). Since sodium is more reactive than hydrogen (E∘(Na+/Na)=−2.71VE^\circ(\mathrm{Na}^+/\mathrm{Na}) = -2.71\mathrm{ V}, E∘(H+/H2)=0.00VE^\circ(\mathrm{H}^+/\mathrm{H}_2) = 0.00\mathrm{ V}), hydrogen is discharged preferentially:

2H2O(l)+2e−→H2(g)+2OH−(aq)2\mathrm{H}_2\mathrm{O}(l) + 2e^- \to \mathrm{H}_2(g) + 2\mathrm{OH}^-(aq)

Anode (oxidation): The anions present are Br−\mathrm{Br}^- and OH−\mathrm{OH}^- (from water). Since bromide is a halide (other than fluoride), it is preferentially discharged over hydroxide:

2Br−(aq)→Br2(aq)+2e−2\mathrm{Br}^-(aq) \to \mathrm{Br}_2(aq) + 2e^-

Overall equation (combining and balancing):

2H2O(l)+2NaBr(aq)→H2(g)+2NaOH(aq)+Br2(aq)2\mathrm{H}_2\mathrm{O}(l) + 2\mathrm{NaBr}(aq) \to \mathrm{H}_2(g) + 2\mathrm{NaOH}(aq) + \mathrm{Br}_2(aq)

Observation: colourless gas at the cathode (H2\mathrm{H}_2) and an orange-brown solution at the anode (Br2\mathrm{Br}_2). The solution near the cathode becomes alkaline due to OH−\mathrm{OH}^- production.

Worked Example: Faraday’s Law — Time to Deposit a Metal

How long must a current of 3.50A3.50\mathrm{ A} be passed through AgNO3\mathrm{AgNO}_3 solution to deposit 0.500g0.500\mathrm{ g} of silver on the cathode?

Solution

Reduction half-reaction: Ag++e−→Ag\mathrm{Ag}^+ + e^- \to \mathrm{Ag}So n=1n = 1.

Molar mass of silver: M=107.87g/molM = 107.87\mathrm{ g/mol}.

N(Ag)=mM=0.500107.87=0.00464molN(\mathrm{Ag}) = \frac{m}{M} = \frac{0.500}{107.87} = 0.00464\mathrm{ mol}Q=n×ne×F=0.00464×1×96485=448CQ = n \times n_e \times F = 0.00464 \times 1 \times 96485 = 448\mathrm{ C}T=QI=4483.50=128s≈2.13minutesT = \frac{Q}{I} = \frac{448}{3.50} = 128\mathrm{ s} \approx 2.13\mathrm{ minutes}

Worked Example: Equilibrium Constant from Cell Potential

Calculate the equilibrium constant at 298K298\mathrm{ K} for the reaction:

MnO4−(aq)+8H+(aq)+5Fe2+(aq)→Mn2+(aq)+4H2O(l)+5Fe3+(aq)\mathrm{MnO}_4^-(aq) + 8\mathrm{H}^+(aq) + 5\mathrm{Fe}^{2+}(aq) \to \mathrm{Mn}^{2+}(aq) + 4\mathrm{H}_2\mathrm{O}(l) + 5\mathrm{Fe}^{3+}(aq)

Given: E∘(MnO4−/Mn2+)=+1.51VE^\circ(\mathrm{MnO}_4^-/\mathrm{Mn}^{2+}) = +1.51\mathrm{ V}, E∘(Fe3+/Fe2+)=+0.77VE^\circ(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = +0.77\mathrm{ V}.

Solution

Cathode (reduction): MnO4−+8H++5e−→Mn2++4H2O\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}, E∘=+1.51VE^\circ = +1.51\mathrm{ V}

Anode (oxidation): Fe2+→Fe3++e−\mathrm{Fe}^{2+} \to \mathrm{Fe}^{3+} + e^-, E∘=+0.77VE^\circ = +0.77\mathrm{ V}

Ecell∘=1.51−0.77=+0.74VE_{\mathrm{cell}}^\circ = 1.51 - 0.77 = +0.74\mathrm{ V}

Number of electrons transferred: n=5n = 5.

log⁡K=nE∘0.0592=5×0.740.0592=3.700.0592=62.5\log K = \frac{nE^\circ}{0.0592} = \frac{5 \times 0.74}{0.0592} = \frac{3.70}{0.0592} = 62.5K=1062.5K = 10^{62.5}

This astronomically large KK means the reaction proceeds essentially to completion under standard conditions.

Worked Example: Non-Standard Conditions with the Nernst Equation

A voltaic cell consists of a Zn\mathrm{Zn} electrode in 0.50M0.50\mathrm{ M} ZnSO4\mathrm{ZnSO}_4 and a Cu\mathrm{Cu} electrode in 0.10M0.10\mathrm{ M} CuSO4\mathrm{CuSO}_4 at 298K298\mathrm{ K}. Calculate the cell potential.

SolutionEcell∘=ECu2+/Cu∘−EZn2+/Zn∘=0.34−(−0.76)=+1.10VE_{\mathrm{cell}}^\circ = E_{\mathrm{Cu}^{2+}/\mathrm{Cu}}^\circ - E_{\mathrm{Zn}^{2+}/\mathrm{Zn}}^\circ = 0.34 - (-0.76) = +1.10\mathrm{ V}

Overall reaction: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)\mathrm{Zn}(s) + \mathrm{Cu}^{2+}(aq) \to \mathrm{Zn}^{2+}(aq) + \mathrm{Cu}(s)

Q=[Zn2+][Cu2+]=0.500.10=5.0Q = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]} = \frac{0.50}{0.10} = 5.0Ecell=Ecell∘−0.0592nlog⁡Q=1.10−0.05922log⁡(5.0)E_{\mathrm{cell}} = E_{\mathrm{cell}}^\circ - \frac{0.0592}{n}\log Q = 1.10 - \frac{0.0592}{2}\log(5.0)log⁡(5.0)=0.699\log(5.0) = 0.699Ecell=1.10−0.05922×0.699=1.10−0.0207=1.08VE_{\mathrm{cell}} = 1.10 - \frac{0.0592}{2} \times 0.699 = 1.10 - 0.0207 = 1.08\mathrm{ V}

The cell potential is slightly lower than standard because the higher concentration of product (Zn2+\mathrm{Zn}^{2+}) relative to reactant (Cu2+\mathrm{Cu}^{2+}) pushes the reaction slightly back toward equilibrium.

Common Pitfalls

  • Reversing E∘E^\circ values incorrectly: When identifying the anode, the half-reaction is written as oxidation, but the E∘E^\circ value used in the Ecell∘E_{\mathrm{cell}}^\circ calculation is always the reduction potential. Never flip the sign of a reduction potential — just subtract Eanode∘E_{\mathrm{anode}}^\circ from Ecathode∘E_{\mathrm{cathode}}^\circ.

  • Multiplying E∘E^\circ by stoichiometric coefficients: E∘E^\circ is an intensive property. Doubling the half-reaction Ag++e−→Ag\mathrm{Ag}^+ + e^- \to \mathrm{Ag} does not double E∘E^\circ to +1.60V+1.60\mathrm{ V}; it remains +0.80V+0.80\mathrm{ V}. Only ΔG∘\Delta G^\circ scales with stoichiometry.

  • Confusing galvanic and electrolytic cell polarity: In a galvanic cell the anode is negative; in an electrolytic cell the anode is positive. The mnemonic “anox red cat” (anode = oxidation, cathode = reduction) applies to both, but the polarity flips.

  • Using Faraday constant with wrong units: F=96485C/molF = 96485\mathrm{ C/mol}. If current is in milliamperes, convert to amperes first. If time is in minutes, convert to seconds. If mass is in milligrams, convert to grams.

  • Forgetting water as a competitor in aqueous electrolysis: When electrolysing NaCl(aq)\mathrm{NaCl}(aq), H2\mathrm{H}_2 is produced at the cathode (not Na\mathrm{Na}), because Na+\mathrm{Na}^+ is too reactive. Similarly, when electrolysing Na2SO4(aq)\mathrm{Na}_2\mathrm{SO}_4(aq), O2\mathrm{O}_2 is produced at the anode (not from sulfate oxidation).

  • Nernst equation sign errors: The Nernst equation is E=E∘−(0.0592/n)log⁡QE = E^\circ - (0.0592/n)\log Q. If you use ln⁡\ln instead of log⁡\logThe factor is RT/nF=0.0257/nRT/nF = 0.0257/nNot 0.0592/n0.0592/n. Mixing these up yields a wrong answer.

Exam-Style Problems

  1. [Medium] Write the cell notation for a galvanic cell in which Ni(s)\mathrm{Ni}(s) is oxidised to Ni2+(aq)\mathrm{Ni}^{2+}(aq) and Ag+(aq)\mathrm{Ag}^+(aq) is reduced to Ag(s)\mathrm{Ag}(s). Calculate Ecell∘E_{\mathrm{cell}}^\circ. (E∘(Ni2+/Ni)=−0.25VE^\circ(\mathrm{Ni}^{2+}/\mathrm{Ni}) = -0.25\mathrm{ V}, E∘(Ag+/Ag)=+0.80VE^\circ(\mathrm{Ag}^+/\mathrm{Ag}) = +0.80\mathrm{ V})

  2. [Hard] A current of 4.00A4.00\mathrm{ A} is passed through molten Al2O3\mathrm{Al}_2\mathrm{O}_3 for 45.045.0 minutes. Calculate: (a) the mass of aluminium produced, (b) the volume of oxygen gas produced at STP (22.7L/mol22.7\mathrm{ L/mol}). (Mr(Al)=27.0M_r(\mathrm{Al}) = 27.0)

  3. [Medium] In an H2/O2\mathrm{H}_2/\mathrm{O}_2 fuel cell, the standard cell potential is +1.23V+1.23\mathrm{ V}. Calculate ΔG∘\Delta G^\circ for the overall reaction 2H2+O2→2H2O(l)2\mathrm{H}_2 + \mathrm{O}_2 \to 2\mathrm{H}_2\mathrm{O}(l) and comment on the efficiency relative to combusting hydrogen (ΔH=−572kJ/mol\Delta H = -572\mathrm{ kJ/mol} for H2\mathrm{H}_2).

  4. [Hard] A cell is constructed with standard hydrogen electrode and a half-cell containing Fe3+\mathrm{Fe}^{3+} and Fe2+\mathrm{Fe}^{2+} at equal concentrations. The measured cell potential is +0.77V+0.77\mathrm{ V}. Write the spontaneous cell reaction and calculate the equilibrium constant.

  5. [Medium] Explain why, during the electrolysis of concentrated NaCl(aq)\mathrm{NaCl}(aq)The product at the anode is Cl2\mathrm{Cl}_2 rather than O2\mathrm{O}_2Even though the standard reduction potential for O2/H2O\mathrm{O}_2/\mathrm{H}_2\mathrm{O} (+1.23V+1.23\mathrm{ V}) is less positive than that of Cl2/Cl−\mathrm{Cl}_2/\mathrm{Cl}^- (+1.36V+1.36\mathrm{ V}).

  6. [Hard] Using the Nernst equation, calculate the potential of a half-cell consisting of a silver electrode in a solution where [Ag+]=0.0010M[\mathrm{Ag}^+] = 0.0010\mathrm{ M} at 298K298\mathrm{ K}. (E∘(Ag+/Ag)=+0.80VE^\circ(\mathrm{Ag}^+/\mathrm{Ag}) = +0.80\mathrm{ V})

  7. [Medium] A student sets up an electrolytic cell to copper-plate a spoon. The spoon is placed at the cathode. Describe the process, including the half-equations, and calculate the time required to deposit a 0.020mm0.020\mathrm{ mm} thick layer of copper over a surface area of 50.0cm250.0\mathrm{ cm}^2 using a current of 0.500A0.500\mathrm{ A}. (Density of copper =8.96g/cm3= 8.96\mathrm{ g/cm}^3, Mr(Cu)=63.55M_r(\mathrm{Cu}) = 63.55)

  8. [Hard] For the cell Zn∣Zn2+(1.0M)∥Fe2+(0.010M)∣Fe\mathrm{Zn} \mid \mathrm{Zn}^{2+}(1.0\mathrm{ M}) \parallel \mathrm{Fe}^{2+}(0.010\mathrm{ M}) \mid \mathrm{Fe}Determine: (a) Ecell∘E_{\mathrm{cell}}^\circ(b) EcellE_{\mathrm{cell}} at the given concentrations, (c) ΔG\Delta G for the reaction under these non-standard conditions. (E∘(Zn2+/Zn)=−0.76VE^\circ(\mathrm{Zn}^{2+}/\mathrm{Zn}) = -0.76\mathrm{ V}, E∘(Fe2+/Fe)=−0.44VE^\circ(\mathrm{Fe}^{2+}/\mathrm{Fe}) = -0.44\mathrm{ V})

Worked Examples (Expanded)

Worked Example: Concentration Cell

A concentration cell is constructed from two Ag\mathrm{Ag} electrodes. One half-cell contains AgNO3\mathrm{AgNO}_3 at 0.100  M0.100\;\mathrm{M} and the other contains AgNO3\mathrm{AgNO}_3 at 0.0010  M0.0010\;\mathrm{M}. Calculate the cell potential at 298  K298\;\mathrm{K}.

Solution

A concentration cell has identical half-reactions but different ion concentrations. The half-cell with the lower concentration acts as the anode (oxidation) because the equilibrium Ag++e−⇌Ag\mathrm{Ag}^+ + e^- \rightleftharpoons \mathrm{Ag} shifts left more readily when [Ag+][\mathrm{Ag}^+] is low.

Anode (oxidation): Ag(s)→Ag+(aq,0.0010  M)+e−\mathrm{Ag}(s) \to \mathrm{Ag}^+(aq, 0.0010\;\mathrm{M}) + e^-

Cathode (reduction): Ag+(aq,0.100  M)+e−→Ag(s)\mathrm{Ag}^+(aq, 0.100\;\mathrm{M}) + e^- \to \mathrm{Ag}(s)

Since Ecell∘=0E_{\mathrm{cell}}^\circ = 0 (same half-reaction), the Nernst equation gives:

Ecell=0−0.05921log⁡ ⁣([Ag+]anode[Ag+]cathode)E_{\mathrm{cell}} = 0 - \frac{0.0592}{1}\log\!\left(\frac{[\mathrm{Ag}^+]_{\mathrm{anode}}}{[\mathrm{Ag}^+]_{\mathrm{cathode}}}\right)

Ecell=−0.0592log⁡ ⁣(0.00100.100)=−0.0592log⁡(0.010)=−0.0592×(−2)=+0.118  VE_{\mathrm{cell}} = -0.0592\log\!\left(\frac{0.0010}{0.100}\right) = -0.0592\log(0.010) = -0.0592 \times (-2) = +0.118\;\mathrm{V}

The cell potential is +0.118  V+0.118\;\mathrm{V}. The reaction spontaneously proceeds to equalise the concentrations, and the cell potential decreases to zero as the concentrations converge.

Worked Example: Electrolytic Cell — Current Efficiency and Side Reactions

A current of 5.00  A5.00\;\mathrm{A} is passed through NaCl(aq)\mathrm{NaCl}(aq) for 2.002.00 hours. The expected product at the anode is Cl2\mathrm{Cl}_2 (from 2Cl−→Cl2+2e−2\mathrm{Cl}^- \to \mathrm{Cl}_2 + 2e^-), but some O2\mathrm{O}_2 is also produced (from 4OH−→O2+2H2O+4e−4\mathrm{OH}^- \to \mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} + 4e^-). If 8.90  g8.90\;\mathrm{g} of Cl2\mathrm{Cl}_2 is collected (instead of the theoretical 13.1  g13.1\;\mathrm{g}), calculate the current efficiency for Cl2\mathrm{Cl}_2 production and the mass of O2\mathrm{O}_2 produced as a side product.

Solution

Theoretical mass of Cl2\mathrm{Cl}_2:

Q=It=5.00×7200=36000  CQ = It = 5.00 \times 7200 = 36000\;\mathrm{C}

n(e−)=QF=3600096485=0.373  mol  e−n(e^-) = \frac{Q}{F} = \frac{36000}{96485} = 0.373\;\mathrm{mol\;e^-}

For Cl2\mathrm{Cl}_2: 2  mol  e−2\;\mathrm{mol\;e^-} per mol Cl2\mathrm{Cl}_2:

n(Cl2)theoretical=0.3732=0.187  moln(\mathrm{Cl}_2)_{\mathrm{theoretical}} = \frac{0.373}{2} = 0.187\;\mathrm{mol}

m(Cl2)theoretical=0.187×71.0=13.2  gm(\mathrm{Cl}_2)_{\mathrm{theoretical}} = 0.187 \times 71.0 = 13.2\;\mathrm{g}

Current efficiency:

Efficiency=8.9013.2×100=67.4%\mathrm{Efficiency} = \frac{8.90}{13.2} \times 100 = 67.4\%

Electrons used for Cl2\mathrm{Cl}_2:

n(e−)Cl2=2×8.9071.0=0.251  mol  e−n(e^-)_{\mathrm{Cl}_2} = 2 \times \frac{8.90}{71.0} = 0.251\;\mathrm{mol\;e^-}

Remaining electrons used for O2\mathrm{O}_2:

n(e−)O2=0.373−0.251=0.122  mol  e−n(e^-)_{\mathrm{O}_2} = 0.373 - 0.251 = 0.122\;\mathrm{mol\;e^-}

For O2\mathrm{O}_2: 4  mol  e−4\;\mathrm{mol\;e^-} per mol O2\mathrm{O}_2:

n(O2)=0.1224=0.0305  moln(\mathrm{O}_2) = \frac{0.122}{4} = 0.0305\;\mathrm{mol}

m(O2)=0.0305×32.0=0.976  gm(\mathrm{O}_2) = 0.0305 \times 32.0 = 0.976\;\mathrm{g}

Worked Example: Gibbs Free Energy, Cell Potential, and Equilibrium

For the reaction Cu(s)+2Ag+(aq)⇌Cu2+(aq)+2Ag(s)\mathrm{Cu}(s) + 2\mathrm{Ag}^+(aq) \rightleftharpoons \mathrm{Cu}^{2+}(aq) + 2\mathrm{Ag}(s): (a) Calculate Ecell∘E_{\mathrm{cell}}^\circ, ΔG∘\Delta G^\circAnd KK at 298  K298\;\mathrm{K}. (b) Calculate ΔG\Delta G when [Ag+]=0.010  M[\mathrm{Ag}^+] = 0.010\;\mathrm{M} and [Cu2+]=0.10  M[\mathrm{Cu}^{2+}] = 0.10\;\mathrm{M}. (E∘(Ag+/Ag)=+0.80  VE^\circ(\mathrm{Ag}^+/\mathrm{Ag}) = +0.80\;\mathrm{V}, E∘(Cu2+/Cu)=+0.34  VE^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\;\mathrm{V})

Solution

(a) Standard conditions:

Ecell∘=Ecathode∘−Eanode∘=0.80−0.34=+0.46  VE_{\mathrm{cell}}^\circ = E_{\mathrm{cathode}}^\circ - E_{\mathrm{anode}}^\circ = 0.80 - 0.34 = +0.46\;\mathrm{V}

ΔG∘=−nFEcell∘=−2×96485×0.46=−88 766  J/mol=−88.8  kJ/mol\Delta G^\circ = -nFE_{\mathrm{cell}}^\circ = -2 \times 96485 \times 0.46 = -88\,766\;\mathrm{J/mol} = -88.8\;\mathrm{kJ/mol}

log⁡K=nE∘0.0592=2×0.460.0592=15.5\log K = \frac{nE^\circ}{0.0592} = \frac{2 \times 0.46}{0.0592} = 15.5

K=1015.5=3.2×1015K = 10^{15.5} = 3.2 \times 10^{15}

(b) Non-standard conditions:

Q=[Cu2+][Ag+]2=0.10(0.010)2=0.100.00010=1000Q = \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^+]^2} = \frac{0.10}{(0.010)^2} = \frac{0.10}{0.00010} = 1000

Ecell=Ecell∘−0.0592nlog⁡Q=0.46−0.05922log⁡(1000)=0.46−0.05922×3=0.46−0.0888=0.371  VE_{\mathrm{cell}} = E_{\mathrm{cell}}^\circ - \frac{0.0592}{n}\log Q = 0.46 - \frac{0.0592}{2}\log(1000) = 0.46 - \frac{0.0592}{2} \times 3 = 0.46 - 0.0888 = 0.371\;\mathrm{V}

ΔG=−nFEcell=−2×96485×0.371=−71 600  J/mol=−71.6  kJ/mol\Delta G = -nFE_{\mathrm{cell}} = -2 \times 96485 \times 0.371 = -71\,600\;\mathrm{J/mol} = -71.6\;\mathrm{kJ/mol}

The reaction is still spontaneous (ΔG<0\Delta G < 0, Ecell>0E_{\mathrm{cell}} > 0) but less strongly so because the high product-to-reactant ratio pushes the reaction back toward equilibrium.

Exam-Style Problems (Expanded)

Problem 9: Quantitative -- Lead-Acid Battery Discharge

A 12  V12\;\mathrm{V} car battery contains six lead-acid cells in series. The overall cell reaction is: Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)\mathrm{Pb}(s) + \mathrm{PbO}_2(s) + 2\mathrm{H}_2\mathrm{SO}_4(aq) \to 2\mathrm{PbSO}_4(s) + 2\mathrm{H}_2\mathrm{O}(l)

(a) Calculate Ecell∘E_{\mathrm{cell}}^\circ per cell given that K=1.0×1081K = 1.0 \times 10^{81} at 298  K298\;\mathrm{K}. (b) If the battery can deliver 4.0  A4.0\;\mathrm{A} for 60  minutes60\;\mathrm{minutes} before needing recharge, calculate the total charge transferred and the mass of Pb\mathrm{Pb} consumed at the anode. (Mr(Pb)=207.2M_r(\mathrm{Pb}) = 207.2)

Problem 10: Extended Response -- Overpotential and Industrial Electrolysis

In the industrial chlor-alkali process (membrane cell), the theoretical voltage required to electrolyse concentrated NaCl(aq)\mathrm{NaCl}(aq) is approximately 2.2  V2.2\;\mathrm{V}But in practice the cell operates at 3.53.5—4.0  V4.0\;\mathrm{V}. (a) Define overpotential and explain its origin at the electrode surfaces. (b) Identify two additional sources of voltage loss in an industrial cell (besides overpotential). (c) Calculate the energy cost per kg of Cl2\mathrm{Cl}_2 produced if the cell operates at 3.8  V3.8\;\mathrm{V} and 85%85\% current efficiency. (1  kWh=3.6  MJ1\;\mathrm{kWh} = 3.6\;\mathrm{MJ})

Problem 11: Quantitative -- $E^\circ$ from $\Delta G_f^\circ$ Values

Given the following standard Gibbs free energies of formation: ΔGf∘(Zn2+,aq)=−147  kJ/mol\Delta G_f^\circ(\mathrm{Zn}^{2+}, aq) = -147\;\mathrm{kJ/mol} ΔGf∘(Cu2+,aq)=+65  kJ/mol\Delta G_f^\circ(\mathrm{Cu}^{2+}, aq) = +65\;\mathrm{kJ/mol} ΔGf∘(Zn,s)=0  kJ/mol\Delta G_f^\circ(\mathrm{Zn}, s) = 0\;\mathrm{kJ/mol} ΔGf∘(Cu,s)=0  kJ/mol\Delta G_f^\circ(\mathrm{Cu}, s) = 0\;\mathrm{kJ/mol}

(a) Calculate ΔG∘\Delta G^\circ for the reaction Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)\mathrm{Zn}(s) + \mathrm{Cu}^{2+}(aq) \to \mathrm{Zn}^{2+}(aq) + \mathrm{Cu}(s). (b) Calculate Ecell∘E_{\mathrm{cell}}^\circ for the Daniell cell. (c) Compare your result with the standard reduction potential values (E∘(Zn2+/Zn)=−0.76  VE^\circ(\mathrm{Zn}^{2+}/\mathrm{Zn}) = -0.76\;\mathrm{V}, E∘(Cu2+/Cu)=+0.34  VE^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\;\mathrm{V}).

Problem 12: Extended Response -- Corrosion Mechanism and Prevention Economics

A steel pipeline (Fe\mathrm{Fe}) buried in soil is protected using a sacrificial anode of magnesium. (a) Write the half-equations for the corrosion of iron and the protection reaction of magnesium. (b) Calculate the minimum mass of magnesium required to protect 1.00  tonne1.00\;\mathrm{tonne} (1000  kg1000\;\mathrm{kg}) of iron from complete corrosion. (Mr(Mg)=24.3M_r(\mathrm{Mg}) = 24.3, Mr(Fe)=55.8M_r(\mathrm{Fe}) = 55.8) (c) Explain why the pipeline must be electrically connected to the magnesium block. (d) Discuss why impressed-current cathodic protection (using an external DC power supply and an inert anode) may be preferred for large structures.

Problem 13: Quantitative -- pH Change During Electrolysis

During the electrolysis of CuSO4(aq)\mathrm{CuSO}_4(aq) with inert electrodes, O2\mathrm{O}_2 is produced at the anode and Cu\mathrm{Cu} is deposited at the cathode. The overall reaction is: 2Cu2+(aq)+2H2O(l)→2Cu(s)+4H+(aq)+O2(g)2\mathrm{Cu}^{2+}(aq) + 2\mathrm{H}_2\mathrm{O}(l) \to 2\mathrm{Cu}(s) + 4\mathrm{H}^+(aq) + \mathrm{O}_2(g)

A current of 2.50  A2.50\;\mathrm{A} is passed through 500  mL500\;\mathrm{mL} of 0.200  M0.200\;\mathrm{M} CuSO4\mathrm{CuSO}_4 solution for 30.0  minutes30.0\;\mathrm{minutes}. (a) Calculate the concentration of H+\mathrm{H}^+ produced. (b) Calculate the pH of the solution after electrolysis. (c) Calculate the volume of O2\mathrm{O}_2 gas produced at STP (22.7  L/mol22.7\;\mathrm{L/mol}).

Problem 14: Extended Response -- Fuel Cell Efficiency

An H2/O2\mathrm{H}_2/\mathrm{O}_2 fuel cell operates at 80°C80\degree\mathrm{C} with Ecell=1.15  VE_{\mathrm{cell}} = 1.15\;\mathrm{V} and a current density of 0.5  A/cm20.5\;\mathrm{A/cm}^2. (a) Calculate ΔG\Delta G for the cell reaction at 80°C80\degree\mathrm{C} (353  K353\;\mathrm{K}). (b) The enthalpy change for H2\mathrm{H}_2 combustion is ΔH=−286  kJ/mol\Delta H = -286\;\mathrm{kJ/mol}. Calculate the thermodynamic efficiency of the fuel cell (ΔG/ΔH\Delta G / \Delta H). (c) Explain why the actual operating voltage (1.15  V1.15\;\mathrm{V}) is less than the theoretical Ecell∘=1.23  VE_{\mathrm{cell}}^\circ = 1.23\;\mathrm{V}And calculate the voltage efficiency.

Common Pitfalls (Expanded)

  • Forgetting to convert between volts and joules in ΔG\Delta G calculations: ΔG=−nFE\Delta G = -nFE requires FF in C/mol\mathrm{C/mol} and EE in V\mathrm{V}. Since 1  V=1  J/C1\;\mathrm{V} = 1\;\mathrm{J/C}The product nFEnFE is automatically in joules. However, if you need the answer in kJ/mol\mathrm{kJ/mol}You must divide by 10001000.

  • Confusing QQ (reaction quotient) with KK (equilibrium constant): QQ uses the current (non-equilibrium) concentrations, while KK uses equilibrium concentrations. In the Nernst equation, QQ is used. At equilibrium, Q=KQ = K and Ecell=0E_{\mathrm{cell}} = 0.

  • Applying Faraday’s law without accounting for current efficiency: In real electrolysis, not all current goes to the desired product. Side reactions (e.g., O2\mathrm{O}_2 instead of Cl2\mathrm{Cl}_2 evolution) consume some charge. Always check whether the problem specifies a current efficiency before assuming 100%100\%.

  • Using the standard hydrogen electrode incorrectly in cell diagrams: The SHE is always written on the side (anode or cathode) that makes Ecell∘E_{\mathrm{cell}}^\circ positive. If the unknown half-cell has E∘>0E^\circ > 0The SHE is the anode. If E∘<0E^\circ < 0The SHE is the cathode.

  • Neglecting the stoichiometry of electrons when using ΔG=−nFE\Delta G = -nFE: The nn must be the total number of electrons transferred in the balanced equation, not per mole of a specific reactant. For Cu2++2e−→Cu\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}, n=2n = 2. For 2Fe3++2e−→2Fe2+2\mathrm{Fe}^{3+} + 2e^- \to 2\mathrm{Fe}^{2+}, nn is still 22 (not 2×2=42 \times 2 = 4).

  • Assuming standard conditions apply to real batteries: A car battery at 12  V12\;\mathrm{V} (open circuit) delivers less than 12  V12\;\mathrm{V} under load due to internal resistance. The terminal voltage is V=Ecell−IrinternalV = E_{\mathrm{cell}} - Ir_{\mathrm{internal}}Where II is the current drawn.

If You Get These Wrong, Revise:

Summary

This topic covers the essential chemistry of redox reactions (advanced), including key reactions, underlying theories, and practical applications.

Key concepts include:

  • standard electrode potentials
  • electrochemical cells
  • electrolysis and Faraday’s laws
  • corrosion and prevention
  • fuel cells

Mastery of these concepts requires both theoretical understanding and the ability to apply knowledge to unfamiliar contexts, particularly in calculation and practical questions.