Redox Reactions
Define Oxidation 3 ways
Oxidization (LEO)
- Loss of electrons
- Oxidization number increase
- Oxygen is gained / Hydrogen is lost
Define Oxidizing Agent AND State What Happens to It
Oxidizing Agent:
Define ΔG AND state what sign means spontaneous
ΔG = change in Gibbs free energy
Define Ksp AND explain why solid is excluded
This value represents how much of a solid dissolves into IONS in solution at equilibrium.
For AB(s), write dissociation AND Ksp
AB(s) ⇌ A⁺ + B⁻; Ksp = [A⁺][B⁻]
Define Reduction 3 Ways
Reduction (GER)
- Electron is gained
- Oxidization number decreases
- Oxygen is lost / Hydrogen is gained
Define Reducing Agent AND State What Happens to It
Reducing Agent:
Define ΔH AND explain sign for exothermic
Enthalpy- Heat absorbed or released by a system at constant pressure
- ΔH < 0 because heat released
Write dissociation + Ksp for CaF₂
CaF₂ ⇌ Ca²⁺ + 2F⁻; Ksp = [Ca²⁺][F⁻]²
For CaF₂, write [ions] in terms of s AND Ksp expression
[Ca²⁺]=s
[F⁻]=2s
Ksp = s(2s)². or 4s3
In Na + Cl₂ → NaCl, explain electron transfer AND identify oxidation/reduction
Na loses e⁻ (oxidized); Cl₂ gains e⁻ (reduced)
In H₂ + Cl₂ → HCl, identify agents AND explain WHY
H₂ reducing agent (loses e⁻); Cl₂ oxidizing agent (gains e⁻)
Given ΔH > 0 (+), ΔS > 0 (+), determine spontaneity AND explain using ΔG
Spontaneous at high T; TΔS overcomes ΔH
Given Q < Ksp, explain what happens AND why (equilibrium shift)
No precipitate; shifts right to form more ions
For Al(OH)₃, write dissociation, ion expressions, AND Ksp in terms of s
Al(OH)₃ ⇌ Al³⁺ + 3OH⁻; [Al³⁺]=s, [OH⁻]=3s; Ksp=27s⁴
Explain difference between oxidation vs oxidizing agent
Oxidation = losing electrons; oxidizing agent = gains electrons and causes oxidation
In Fe + Cl₂ → Fe³⁺ + Cl⁻, identify ALL roles AND explain electron flow
Fe oxidized & reducing agent; Cl₂ reduced & oxidizing agent
Given ΔH < 0, ΔS < 0, explain temperature dependence step-by-step
Low T spontaneous; high T non spontaneous
1. Q>Ksp → ?
2. Q<Ksp → ?
1. Q>Ksp → precipitate
2. Q<Ksp → none
The Ksp of CaF₂ is 3.9×10⁻¹¹. Calculate the molar solubility (s). Then calculate [F⁻].
Molar solubility (s) = 2.14 × 10⁻⁴ M
[F⁻] = 4.28 × 10⁻⁴ M
Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s)
(a) Identify what is oxidized & reduced
(b) Identify what is reducing & oxidizing agent
(a) Oxidized: Fe(s) & Reduced: Ag⁺(aq)
(b) Reducing agent: Fe(s) & Oxidizing agent: Ag⁺(aq)
Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)
(a) Identify which species is oxidized and which is reduced
(b) Identify the oxidizing agent and reducing agent
(c) Explain why the oxidizing agent is defined the way it is
(a) Oxidation vs Reduction
(b) Agents
(c) Explanation
ΔH = –85.0 kJ/mol ΔS = –150.0 J·mol⁻¹·K⁻¹ Temperature = 298 K
(a) Calculate ΔG at 298 K
(b) Determine whether the reaction is spontaneous at this temperature
(a) ΔG = –40.3 kJ/mol
(b) The reaction is spontaneous (because ΔG < 0)
A solution has [Pb²⁺] = 2.0×10⁻⁴ M and [Cl⁻] = 3.0×10⁻³ M. Given Ksp(PbCl2)=1.7×10−5K_{sp}(\text{PbCl}_2)=1.7×10^{-5}Ksp(PbCl2)=1.7×10−5, calculate Q and determine if a precipitate forms.
Q=(2.0×10−4)(3.0×10−3)2 =1.8×10−9
Since Q < Ksp → no precipitate
Define the common ion effect
The common ion effect is the decrease in the solubility of a sparingly soluble solid when a solution already contains one of the ions from that solid, causing the equilibrium to shift left.