Topic 16

Topic 16 on a page

Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, follow the link back to the full page.

Definitions to learn word for word

Oxidation: loss of electrons, or an increase in oxidation number. Reduction: gain of electrons, or a decrease in oxidation number. (page 1)
Oxidising agent: an electron acceptor, itself reduced. Reducing agent: an electron donor, itself oxidised.
Standard electrode potential, $E^\ominus$ : the emf of a cell made from that half-cell and a standard hydrogen electrode, measured under standard conditions. (page 3)
Standard conditions: 298 K, 100 kPa pressure of gases, 1.00 mol dm⁻³ concentration of ions. (page 3)
Thermodynamically feasible: $E^\ominus_\text{cell}$ is positive. (page 9)
Disproportionation: the same element, in one species, is simultaneously oxidised and reduced. (page 12)
Fuel cell: uses the energy released on the reaction of a fuel with oxygen to generate a voltage; the fuel and oxygen are supplied continuously. (page 18)

The standard hydrogen electrode

Hydrogen gas at 100 kPa over a platinum electrode coated with platinum black, in 1.00 mol dm⁻³ H⁺(aq), at 298 K. Its potential is defined as 0.00 V. It is needed because the potential of a single electrode cannot be measured. (page 4)

Couple has	Half-cell
a solid metal	the metal in 1.00 mol dm⁻³ of its ions
a gas	the gas at 100 kPa over platinum, in 1.00 mol dm⁻³ of its ions
two ions of one element	platinum in a solution of both ions, each 1.00 mol dm⁻³

Electrode potentials used in these notes

All are written as reductions; the most negative is at the top. (page 5)

Half-equation	$E^\ominus$ / V
Mg²⁺ + 2e⁻ ⇌ Mg	−2.37
Zn²⁺ + 2e⁻ ⇌ Zn	−0.76
Fe²⁺ + 2e⁻ ⇌ Fe	−0.44
2H⁺ + 2e⁻ ⇌ H₂	0.00
S₄O₆²⁻ + 2e⁻ ⇌ 2S₂O₃²⁻	+0.09
Cu²⁺ + e⁻ ⇌ Cu⁺	+0.15
Cu²⁺ + 2e⁻ ⇌ Cu	+0.34
Cu⁺ + e⁻ ⇌ Cu	+0.52
I₂ + 2e⁻ ⇌ 2I⁻	+0.54
Fe³⁺ + e⁻ ⇌ Fe²⁺	+0.77
Ag⁺ + e⁻ ⇌ Ag	+0.80
Br₂ + 2e⁻ ⇌ 2Br⁻	+1.09
O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O	+1.23
MnO₂ + 4H⁺ + 2e⁻ ⇌ Mn²⁺ + 2H₂O	+1.23
Cl₂ + 2e⁻ ⇌ 2Cl⁻	+1.36
MnO₄⁻ + 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O	+1.51

Calculations and rules

To find	Use	Page
cell emf	$E^\ominus_\text{cell}$ = $E^\ominus_\text{right}$ − $E^\ominus_\text{left}$ = positive − negative	Cell emf
feasibility	positive $E^\ominus_\text{cell}$ means feasible	Feasibility
extent	$E^\ominus_\text{cell}$ is proportional to $\Delta S_\text{total}$ and ln K	Entropy and K
effect of conditions	write the electrode equilibrium and use Le Chatelier	Conditions
disproportionation	$E^\ominus$ (reduced in) − $E^\ominus$ (oxidised in) > 0	Disproportionation
percentage uncertainty	uncertainty ÷ measured value × 100	Uncertainty

Cell diagrams

Negative electrode on the left, positive on the right. | is a phase boundary, ‖ is the salt bridge, electrodes at the outer ends, oxidised forms next to the salt bridge, Pt(s) if there is no metal, state symbols included. (page 6)

\ce{Zn(s)} \mid \ce{Zn^2+(aq)} \parallel \ce{Cu^2+(aq)} \mid \ce{Cu(s)}

E^\ominus_\text{cell} = +1.10\ \text{V}

Limits of predictions

Kinetic stability: feasible but too slow, because the activation energy is high (H₂ and O₂, +1.23 V). (page 11)
Non-standard conditions: a negative $E^\ominus_\text{cell}$ can still react, for example MnO₂ with concentrated HCl (−0.13 V). (page 11)

Redox titrations

Reaction	Ratio	End point
MnO₄⁻ with Fe²⁺ (acid: dilute H₂SO₄)	MnO₄⁻ : Fe²⁺ = 1 : 5	first permanent pale pink
MnO₄⁻ with H₂O₂	MnO₄⁻ : H₂O₂ = 2 : 5	first permanent pale pink
I₂ with S₂O₃²⁻	I₂ : S₂O₃²⁻ = 1 : 2	blue-black to colourless (starch added near the end)
Cu²⁺ with excess I⁻, then thiosulfate	Cu²⁺ : S₂O₃²⁻ = 1 : 1	blue-black to colourless
IO₃⁻ with excess I⁻, then thiosulfate	IO₃⁻ : S₂O₃²⁻ = 1 : 6	blue-black to colourless
IO₄⁻ with excess I⁻, then thiosulfate	IO₄⁻ : S₂O₃²⁻ = 1 : 8	blue-black to colourless

Route: moles of titrant → mole ratio → moles in the portion → scale up → mass, concentration or percentage. (page 13)

Fuel cells

	Acidic electrolyte	Alkaline electrolyte
negative electrode	H₂ → 2H⁺ + 2e⁻	H₂ + 2OH⁻ → 2H₂O + 2e⁻
positive electrode	O₂ + 4H⁺ + 4e⁻ → 2H₂O	O₂ + 2H₂O + 4e⁻ → 4OH⁻
overall	2H₂ + O₂ → 2H₂O	2H₂ + O₂ → 2H₂O

$E^\ominus_\text{cell}$ = +1.23 V in both. Methanol cell overall: 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O. (page 19)

The ten traps

Calling zinc the oxidising agent in Zn + Cu²⁺: the species oxidised is the reducing agent.
Writing oxidation numbers as 2+ and ionic charges as +2.
Forgetting the 2 in 2Cr³⁺ when balancing the dichromate half-equation.
Missing 100 kPa or platinum black in a hydrogen electrode answer.
Putting an iron rod in a Fe³⁺/Fe²⁺ half-cell instead of platinum.
Doubling $E^\ominus$ because the equation has 2Ag⁺, or changing its sign when the half-equation is reversed.
Subtracting the wrong way round: $E^\ominus_\text{cell}$ is positive electrode minus negative electrode.
Using hydrochloric acid with manganate(VII): MnO₄⁻ oxidises Cl⁻ (+0.15 V), so the titre is too large.
Using the ratio 1 : 2 for iodate and thiosulfate (it is 1 : 6), or forgetting to scale 25.0 cm³ up to the whole solution.
Writing "clear" for the thiosulfate end point (it is blue-black to colourless), or reducing uncertainty by "repeating the titration".

Mixed test

Try these with no notes, then check the answers.

Calculate $E^\ominus_\text{cell}$ for the cell made from Fe²⁺/Fe (−0.44 V) and Cu²⁺/Cu (+0.34 V), and write the cell diagram.
Use $E^\ominus$ values to decide whether Br₂ oxidises Fe²⁺ to Fe³⁺, and whether I₂ does.
Show that Cu⁺ disproportionates and write the equation.
Give two reasons why a reaction with a positive $E^\ominus_\text{cell}$ may not be seen.
25.0 cm³ of Fe²⁺(aq) needs 23.60 cm³ of 0.0200 mol dm⁻³ KMnO₄. Calculate [Fe²⁺].
25.0 cm³ of 0.0200 mol dm⁻³ KIO₃ is treated with excess KI and acid. The iodine needs 30.00 cm³ of sodium thiosulfate. Calculate [S₂O₃²⁻], and the percentage uncertainty of the burette reading if each reading is ± 0.05 cm³.
List the features of the standard hydrogen electrode.
Write the electrode equations for a hydrogen-oxygen fuel cell with an alkaline electrolyte.

Answers cover these until you have tried the questions

$E^\ominus_\text{cell}$ = +0.34 − (−0.44) = +0.78 V. Iron is on the left:
$\ce{Fe(s)} \mid \ce{Fe^2+(aq)} \parallel \ce{Cu^2+(aq)} \mid \ce{Cu(s)}$
Br₂ is reduced: +1.09 − (+0.77) = +0.32 V, feasible. I₂ is reduced: +0.54 − (+0.77) = −0.23 V, not feasible. Br₂ does, I₂ does not.
Cu⁺ is reduced to Cu (+0.52 V) and oxidised to Cu²⁺ (+0.15 V). $E^\ominus_\text{cell}$ = +0.52 − (+0.15) = +0.37 V, positive. $\ce{2Cu+(aq) -> Cu(s) + Cu^2+(aq)}$
Kinetic stability (the activation energy is high, so the rate is too slow) and non-standard conditions (concentration, pressure or temperature changes E, which can alter the sign for a small value).
n(MnO₄⁻) = 0.0200 × 23.60 ÷ 1000 = 4.72 × 10⁻⁴ mol. n(Fe²⁺) = 5 × 4.72 × 10⁻⁴ = 2.36 × 10⁻³ mol. [Fe²⁺] = 2.36 × 10⁻³ ÷ 0.0250 = 0.0944 mol dm⁻³.
n(IO₃⁻) = 0.0200 × 25.0 ÷ 1000 = 5.00 × 10⁻⁴ mol. n(S₂O₃²⁻) = 6 × 5.00 × 10⁻⁴ = 3.00 × 10⁻³ mol. [S₂O₃²⁻] = 3.00 × 10⁻³ ÷ 0.03000 = 0.100 mol dm⁻³. Burette uncertainty ± 0.10 cm³: 0.10 ÷ 30.00 × 100 = 0.33%.
Hydrogen gas at 100 kPa; platinum electrode coated with platinum black; 1.00 mol dm⁻³ H⁺(aq); 298 K; potential defined as 0.00 V.
Negative: H₂ + 2OH⁻ → 2H₂O + 2e⁻. Positive: O₂ + 2H₂O + 4e⁻ → 4OH⁻. (Overall 2H₂ + O₂ → 2H₂O, $E^\ominus_\text{cell}$ = +1.23 V.)