Chemistry Fundamentals: Significant Figures & Redox Reactions

Significant figures matter in all your chemistry calculations. Remember these key rules:

• All non-zero digits are significant
• Zeros between non-zero digits are significant
• Zeros to the right of a decimal point but to the left of a non-zero digit are NOT significant
• Zeros to the right of a non-zero digit AND to the right of a decimal point ARE significant

Understanding oxidation-reduction reactions is crucial for chemistry success. Remember these simple memory aids:

• OIL RIG: Oxidation Is Loss of electrons, Reduction Is Gain of electrons
• The reducing agent gains electrons (gets oxidized)
• The oxidizing agent loses electrons (gets reduced)

When balancing redox reactions, follow these steps:

1. Identify and write out the separate oxidation and reduction half-reactions
2. Balance each half-reaction for atoms and charge
3. Multiply each half-reaction by factors to make electrons equal
4. Combine the half-reactions and simplify

Pro Tip: Pay attention to whether the solution is acidic $H+ present$ or basic $OH- present$ when balancing redox reactions—this affects how you balance oxygen and hydrogen atoms!

Electrochemistry connects chemical reactions with electricity through concepts like current, charge, and potential difference. Key units to know include:

• Electric current (I): measured in amperes (A)
• Electric charge (Q): measured in coulombs (C) where Q = I × t
• Potential difference (V): measured in volts (V) where 1 V = 1 J/C

Electrochemistry Essentials

Ohm's Law is a fundamental relationship in electrochemistry: V = I × R

• Where V is voltage (volts), I is current (amperes), and R is resistance (ohms)
• You can rearrange this to find current: I = V/R
• Power can be calculated as: P = I × R or P = I × V

When working with electrochemical cells, you need to understand their notation:

• Cells are written as: anode || cathode
• Example: Zn | Zn²⁺ || Cu²⁺ | Cu
• Oxidation always occurs at the anode: Zn → Zn²⁺ + 2e⁻
• Reduction always occurs at the cathode: Cu²⁺ + 2e⁻ → Cu

Faraday's Laws of Electrolysis describe how the amount of substance produced during electrolysis relates to the quantity of electricity used:

• The mass of substance produced is proportional to the charge: m ∝ Q or m ∝ It
• Faraday's Constant is 9.65 × 10⁴ C/mol

Remember This! One electron equals -1.602 × 10⁻¹⁹ C, and one mole of electrons equals 96,500 C (Faraday's constant). These values are crucial for solving electrochemistry problems!

When analyzing significant figures, always check trailing zeros carefully. For example:

• In 1.200, all four digits are significant $zeros after decimal point and after non-zero digit$
• In 1200, without a decimal point, the zeros may or may not be significant (would need more context)

Standard Cell Potentials and Nuclear Chemistry

To calculate the standard cell potential, use one of these formulas:

• E°ₑₗₗ = E°ᵣₑₐ(cathode) + E°ᵣₑₐ(anode) (change the sign)
• OR E°ₑₗₗ = E°ₑₐₜₕₒₐₑ - E°ₐₙₒₐₑ (use values as they are)

For example, with Ni/Ni²⁺ || Cl₂/Cl⁻:

• Ni → Ni²⁺ + 2e⁻ (anode)
• Cl₂ + 2e⁻ → 2Cl⁻ (cathode)
• E°ₑₗₗ = 0.236 V + 1.360 V = 1.596 V

Nuclear chemistry focuses on the nucleus and its particles. Key concepts include:

• Mass number (A) = protons + neutrons
• Atomic number (Z) = protons only (determines element identity)
• For example, ¹²C has 6 protons and 6 neutrons (12-6=6)

When determining particles in an atom or ion:

• Protons = atomic number
• Neutrons = mass number - atomic number
• Electrons = protons - charge (for positive ions) or protons + charge (for negative ions)

Quick Tip: The most stable nuclei have an even number of both protons and neutrons. Nuclei with odd numbers of both are the least stable (only 4 exist)!

Radioactivity occurs when unstable nuclei emit particles to become more stable. The main types are:

• Alpha (α) emission: ²₃₈U → ²₃₄Th + α (loses 2 protons, 2 neutrons)
• Beta (β) emission: ¹³¹I → ¹³¹Xe + ⁰₍₋₁₎e (neutron converts to proton)
• Positron (β⁺) emission: ⁴⁰K → ⁴⁰Ar + ⁰₍₊₁₎e (proton converts to neutron)
• Gamma (γ) emission: high-energy electromagnetic radiation
• Electron capture: ¹⁹⁷Hg + e⁻ → ¹⁹⁷Au (electron combines with proton)

Nuclear Binding Energy and Chemical Stoichiometry

Nuclear binding energy explains the energy that holds the nucleus together:

• Calculated using Einstein's equation: E = Δm × c²
• Where Δm is the mass defect (difference between actual mass and sum of parts)
• The speed of light (c) is 3 × 10⁸ m/s

To calculate binding energy, follow these steps:

1. Find the mass difference between the nucleus and its components
2. Convert this mass to kilograms $1 amu = 1.66054 × 10⁻²⁷ kg$
3. Apply Einstein's equation and convert to your desired energy units

Radioactive decay follows predictable patterns:

• Half-life equation: kt₁/₂ = 0.693 (where k is the rate constant)
• Activity equation: ln$A/A₀$ = -kt (where A is final activity, A₀ is initial activity)
• Activity is measured in Becquerels $1 Bq = 1 atom/second$ or Curies

Important: When analyzing radioactive decay, remember that the number of radioactive atoms decreases by exactly half after each half-life. This is true no matter how much material you start with!

Stoichiometry connects different parts of chemical equations. For empirical and molecular formulas:

1. Convert mass percentages to moles
2. Divide by the smallest mole value to get the simplest ratio
3. If needed, multiply by a whole number to get integers
4. For molecular formula, divide the actual molecular weight by the empirical formula weight

For combustion reactions, remember:

• Complete combustion produces CO₂, H₂O, and SO₂
• For hydrocarbons: CₘHₙ + $m + n/4$O₂ → mCO₂ + $n/2$H₂O
• Incomplete combustion produces CO instead of CO₂

Combustion Calculations

For combustion problems, you need to calculate the theoretical oxygen required:

• Method I: Total O₂ required - O₂ in the fuel
• Method II: $O₂ required by C + O₂ required by H + O₂ required by S$ - O₂ in fuel

Let's see how this works with pure carbon combustion:

200 kg C + O₂ → CO₂

Theoretical O₂ needed:

200 kg C × (1 kmol C/12 kg C) × (1 kmol O₂/1 kmol C) = 16.667 kmol O₂

For more complex fuels, analyze each component separately. For example, with blast furnace gas (25% CO, 10% CO₂, 5% H₂, 10% CH₄, 45% N₂, 5% O₂):

1. Calculate O₂ required for each component:

   • CO requires 0.5 mol O₂ per mol CO
   • H₂ requires 0.5 mol O₂ per mol H₂
   • CH₄ requires 2 mol O₂ per mol CH₄

2. Calculate O₂ already in the fuel (from CO₂, CO, and free O₂)

3. Subtract to find theoretical O₂ needed

Pro Tip: When solving combustion problems, always start by writing balanced equations for each component that burns. This makes it much easier to track oxygen requirements!

When excess air is used (common in real combustion systems), calculate:

• O₂ free = Excess O₂ + Unused O₂
• N₂ = $Theoretical O₂ + Excess O₂$ × (79/21)

For example, with 25% excess air and pure carbon:

• Theoretical O₂ = 8.33 kmol
• Excess O₂ = 8.33 × 0.25 = 2.08 kmol
• N₂ = (8.33 + 2.08) × (79/21) = 39.17 kmol

The final composition would be 16.8% CO₂, 79% N₂, and 4.2% O₂.

Hydrocarbon Combustion and Incomplete Combustion

When burning pure hydrocarbons with excess air, follow these steps to find the flue gas composition:

For complete combustion of C₂H₄ with 40% excess air:

1. Write the balanced equation: C₂H₄ + 3.5O₂ → 2CO₂ + 2H₂O
2. Calculate products:
   • 100 kmol C₂H₄ produces 200 kmol CO₂
   • 100 kmol C₂H₄ produces 300 kmol H₂O
3. Calculate theoretical O₂: 100 kmol C₂H₄ × 3.5 = 350 kmol O₂
4. Calculate excess O₂: 350 kmol × 0.4 = 140 kmol O₂ (free)
5. Calculate N₂: (350 + 140) × (79/21) = 1843.33 kmol N₂
6. Find composition percentages: CO₂ (8.07%), H₂O (12.08%), N₂ (74.23%), O₂ (5.64%)

For incomplete combustion, you need to account for CO formation:

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O  (complete)
C₂H₄ + 2O₂ → 2CO + 2H₂O   (incomplete)

If 90% of carbon burns to CO₂ and 10% to CO:

1. Calculate CO₂: 100 kmol C₂H₄ × 2 × 0.9 = 180 kmol CO₂
2. Calculate CO: 100 kmol C₂H₄ × 2 × 0.1 = 20 kmol CO
3. Calculate H₂O: 100 kmol C₂H₄ × 2 = 200 kmol H₂O
4. Calculate theoretical O₂ needed
5. Calculate N₂ and free O₂
6. Find final composition

Remember: In real combustion processes, incomplete combustion is common. CO formation represents wasted fuel and can be dangerous, so understanding these calculations helps with both efficiency and safety!

The key difference between complete and incomplete combustion is oxygen availability. With limited oxygen, carbon forms CO (carbon monoxide) instead of CO₂ (carbon dioxide). This changes both the products formed and the amount of oxygen required for the reaction.