Study Guide: Oxidation-Reduction Concepts

Core Definitions

Oxidation Number (Oxidation State): The apparent charge (positive or negative) that an atom would have within a compound.
Oxidation:
- The loss of electrons by a substance.
- An increase in oxidation number.
Reduction:
- The gain of electrons by a substance.
- A decrease in oxidation number.
Redox Reactions: Short for oxidation-reduction reactions; these occur when elements undergo a change in oxidation number simultaneously.

Identifying Redox Reactions

To determine if oxidation or reduction has occurred, follow these steps:

Assign Oxidation Numbers: Write the oxidation state above every atom in the chemical equation.
- Note: Elements in their elemental form (e.g., $Na$ or $Cl_2$) always have an oxidation number of zero.
Compare Changes:
- If the number goes up (e.g., $0$ to $+1$), the element is oxidized.
- If the number goes down (e.g., $0$ to $-1$), the element is reduced.

Examples of Redox Processes

1. Zinc and Copper Reaction

Oxidation: Zinc metal ($Zn$) loses two electrons to become $Zn^{2+}$.
Reduction: Copper ions ($Cu^{2+}$) gain two electrons to become copper metal ($Cu$).
Observation: When zinc is dipped in blue $CuSO_4$ solution, the blue color fades, and a brown copper layer forms on the zinc surface.

2. Combustion of Natural Gas ($CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$)

Carbon ($C$): Changes from $-4$ in $CH_4$ to $+4$ in $CO_2$ (Increase = Oxidized).
Oxygen ($O$): Changes from $0$ in $O_2$ to $-2$ in $H_2O$ and $CO_2$ (Decrease = Reduced).

Special Case: Disproportionation Reaction

Definition: Also known as a "self-oxidation-reduction" reaction. This occurs when the same element is both oxidized and reduced within the same reaction.
Example: The decomposition of hydrogen peroxide ($2H_2O_2 \rightarrow 2H_2O + O_2$).
- Oxygen in $H_2O_2$ has an oxidation state of $-1$.
- It is oxidized to $0$ in $O_2$.
- It is reduced to $-2$ in $H_2O$.

Special Redox Reactions & Balancing Methods

Disproportionation Reaction

Definition: A redox reaction where the same element is simultaneously oxidized and reduced.
Alternative Name: Also called a self-oxidation-reduction reaction.
Example Case: The decomposition of hydrogen peroxide ($H_{2}O_{2}$).
- Oxygen in $H_{2}O_{2}$ (oxidation state $-1$) is oxidized to $0$ in $O_{2}$.
- Oxygen in $H_{2}O_{2}$ (oxidation state $-1$) is reduced to $-2$ in $H_{2}O$.

Balancing Redox Equations: Oxidation Number Method

Core Principle: The total number of electrons lost by one element must equal the total number of electrons gained by another element.

Step-by-Step Balancing Process

Assign Oxidation Numbers: Determine the oxidation state for every atom involved in the chemical equation.
Identify Changes: Find which elements are changing their oxidation states to determine the number of electrons lost or gained.
- Example: Phosphorus ($P$) changing from $0$ to $+5$ involves a $5$-electron change.
- Example: Nitrogen ($N$) changing from $+5$ to $+2$ involves a $3$-electron change.
Draw Bridges: Connect the atoms that changed oxidation numbers with a "bridge" and label the number of electrons gained or lost.
Equalize Electrons: Multiply the numbers of electrons gained and lost by small whole numbers to find a common value.
- Note: To balance a $3e^{-}$ gain against a $5e^{-}$ loss, multiply the gain by $5$ and the loss by $3$.
Apply Coefficients: Use these multipliers as the initial coefficients for the respective substances in the equation.

Alternative Balancing Methods

Ion Electron Method: An alternative technique for balancing complex redox equations.

Advanced Redox Concepts and Balancing Methods

Disproportionation Reactions

Definition: A specific type of redox reaction where the same element is oxidized and reduced simultaneously.
Alternative Name: These are also known as self-oxidation-reduction reactions.
Example: The decomposition of hydrogen peroxide ($H_{2}O_{2}$).
- Oxygen starts with an oxidation state of $1-$.
- It is oxidized to $0$ in $O_{2}$.
- It is reduced to $2-$ in $H_{2}O$.

Balancing Equations: Oxidation Number Method

This method is based on the principle that the total electrons lost by one element must equal the total electrons gained by another element.

Assign Oxidation Numbers: Determine the oxidation state for every atom in the equation.
Identify Changes: Find the elements that change their oxidation state and calculate the number of electrons involved in that change.
Draw Bridges: Connect the atoms that changed with a bridge and indicate the number of electrons gained or lost.
Equalize Electrons: Multiply the gain and loss numbers by small whole numbers to reach a common total. Use these multipliers as coefficients for those substances.
Final Inspection: Balance the remaining atoms (other than oxygen and hydrogen first, then oxygen, and finally hydrogen) using the inspection method.

Balancing Equations: Ion-Electron Method

This process involves splitting the reaction into two half-reactions: the oxidation-half and the reduction-half.

Balance Half-Reactions Individually:
- Balance all atoms except Oxygen and Hydrogen.
- Balance Oxygen by adding $H_{2}O$ to the side that needs it.
- Balance Hydrogen by adding $H^{+}$ ions.
- Balance Charge by adding electrons ($e^{-}$) to the side with the higher positive charge.
Equalize Electrons: Multiply one or both half-reactions by integers so that the number of electrons lost equals the number of electrons gained.
Combine and Cancel: Add the two half-reactions together and cancel out any duplicate species (like $e^{-}$, $H_{2}O$, or $H^{+}$) appearing on both sides.
Verification: Ensure the total charges on the left-hand side (LHS) equal the charges on the right-hand side (RHS).

Redox Reaction Study Guide

Special Redox Types: Disproportionation Reaction

Definition: A reaction where the same element is oxidized and reduced at the same time.
Also Known As: Self-oxidation-reduction reactions.
Example: The decomposition of hydrogen peroxide ($H_{2}O_{2}$).
- Oxygen in $H_{2}O_{2}$ (oxidation state $1-$) is oxidized to $0$ in $O_{2}$.
- The same oxygen is reduced to $2-$ in $H_{2}O$.

Oxidizing and Reducing Agents

Oxidizing Agent:
- A substance that accepts electrons and causes oxidation in another substance.
- The oxidation state of the agent decreases because it contains an element being reduced (gaining electrons).
Reducing Agent:
- A substance that causes reduction in another substance by losing electrons.
- The oxidation state of the agent increases because it contains an element being oxidized (losing electrons).
Real-World Example: In the reaction between ferrous sulphate ($FeSO_{4}$) and potassium permanganate ($KMnO_{4}$), $FeSO_{4}$ acts as the reducing agent, while $KMnO_{4}$ is the oxidizing agent that loses its purple color.

Balancing Method 1: Oxidation Number Method

This method is based on the principle that the total electrons lost by one element must equal the total electrons gained by another.

Assign Oxidation Numbers: Identify the oxidation state for every atom in the equation.
Identify Changes: Determine which elements are changing their oxidation state and by how many electrons.
Draw Bridges: Connect atoms that changed states with a bridge, noting the number of electrons gained or lost.
Equalize Electrons: Multiply the gain and loss numbers by small whole numbers to find a common value. Use these multipliers as coefficients for the substances.
Final Balance: Use the inspection method to balance the remaining atoms, starting with atoms other than hydrogen and oxygen, then oxygen, and finally hydrogen.

Balancing Method 2: Ion-Electron Method (Half-Reaction Method)

Redox reactions consist of two parts: an oxidation-half and a reduction-half.

Balance Half-Reactions individually:
- Balance atoms other than Oxygen and Hydrogen.
- Balance Oxygen by adding $H_{2}O$ to the side lacking it.
- Balance Hydrogen by adding $H^{+}$ ions.
- Balance Charge by adding electrons ($e^{-}$) to the side with the higher positive charge.
Equalize Electrons: Multiply the half-reactions by integers so that electrons lost equals electrons gained.
Combine: Add the two half-reactions together and cancel duplicate items (like $e^{-}$, $H_{2}O$, or $H^{+}$) appearing on both sides.
Verify: Ensure the total charges on the left-hand side (LHS) equal the charges on the right-hand side (RHS).

Redox Reactions and Electrochemistry Study Guide

1. Special Redox Reactions

Disproportionation Reaction: A reaction in which the same element is oxidized and reduced simultaneously.
Self-Oxidation-Reduction: Another term for disproportionation reactions.
Example: In the decomposition of hydrogen peroxide ($H_{2}O_{2}$), oxygen is oxidized to $0$ in $O_{2}$ and reduced to $2-$ in $H_{2}O$.

2. Oxidizing and Reducing Agents

Oxidizing Agent: A substance that accepts electrons, causing its oxidation state to decrease while it oxidizes another substance. It contains an element that is being reduced.
Reducing Agent: A substance that causes reduction by losing electrons, resulting in an increase of its own oxidation state. It contains an element that is being oxidized.
Visual Identification: In a reaction between ferrous sulphate ($FeSO_{4}$) and potassium permanganate ($KMnO_{4}$), the purple color of $KMnO_{4}$ is discharged because $FeSO_{4}$ reduces it.

3. Balancing Redox Equations

I. Oxidation Number Method

Based on the principle that total electrons lost must equal total electrons gained.

Assign oxidation numbers to all atoms in the equation.
Identify elements changing oxidation states and determine the electron change per atom.
Draw bridges between the atoms that changed and note the electrons gained or lost.
Equalize electrons by multiplying the gain and loss numbers by small whole numbers to find a common factor.
Apply coefficients from those multipliers to the respective substances.
Balance the rest by the inspection method: balance atoms other than O and H first, then Oxygen, and finally Hydrogen.

II. Ion-Electron Method (Half-Reaction Method)

Redox reactions are viewed as the sum of an oxidation-half and a reduction-half reaction.

Balance half-reactions individually:
- Balance atoms other than O and H.
- Balance O atoms by adding $H_{2}O$.
- Balance H atoms by adding $H^{+}$.
- Balance charges by adding electrons ($e^{-}$).
Equalize electrons transferred in both half-reactions by multiplying the equations by appropriate integers.
Combine and simplify: Add the half-reactions and cancel out duplicate species (like $e^{-}$, $H_{2}O$, or $H^{+}$) appearing on both sides.

4. Electrochemical Cells

Galvanic (Voltaic) Cell: A cell where a spontaneous redox reaction produces an electric current.
Daniel Cell: A specific galvanic cell using a Zinc rod (anode) in $ZnSO_{4}$ and a Copper rod (cathode) in $CuSO_{4}$.
Anode: The electrode where oxidation occurs; it has a more negative reduction potential.
Cathode: The electrode where reduction occurs; it has a more positive reduction potential.
Salt Bridge: A tube filled with an electrolyte (like $KCl$ or $KNO_{3}$) that connects two half-cells. It maintains electrical neutrality and prevents charge buildup by allowing ion movement.

5. Cell Potential and EMF

Electromotive Force (emf): The force that pushes electrons from the anode to the cathode, measured in volts (V).
Cell Potential ($E^{o}_{cell}$): The potential of the cell to do work, calculated as the algebraic difference between the standard reduction potentials of the two half-cells:
$$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$$
Standard Hydrogen Electrode (SHE): Used as a reference electrode with a standard potential of $0.00$ volts to determine relative half-cell potentials.
Standard Electrode Potential: The tendency of a half-cell to undergo reduction relative to the SHE under standard conditions (1 atm, 1M, $25^{\circ}C$).

Study Guide: Electrochemical Potentials and the Standard Hydrogen Electrode

1. Standard Hydrogen Electrode (SHE)

The SHE is the primary reference electrode used to measure the standard electrode potentials of other half-cells.

Components: A platinum foil coated with finely divided platinum, surrounded by hydrogen gas at 1 atm pressure, in contact with 1M HCl solution at 298K.
Standard Potential: Arbitrarily defined as 0.00V at all temperatures.
Versatility: Can act as either an anode or a cathode depending on the electrode it is connected to.

Half-Cell Reactions:

As Cathode (Reduction): $2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}$ ($E^o = 0.00V$)
As Anode (Oxidation): $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^-$ ($E^o = 0.00V$)

2. Determination of Standard Electrode Potentials

To measure the potential of a metal electrode, it is combined with a SHE through a salt bridge to form an electrochemical cell. The voltmeter reading (EMF) determines the electrode potential.

A. Zinc Electrode ($Zn^{2+}/Zn$)

Setup: Zinc rod in $1M$ Zinc Sulphate solution connected to a SHE.
Observations: Electrons move from the Zinc rod to the SHE.
- Zinc Electrode: Acts as the Anode.
- SHE: Acts as the Cathode.
Calculations:
- Voltmeter reading ($E^o_{cell}$) = $0.76V$
- $E^o_{cell} = E^o_{Anode} + E^o_{Cathode}$
- $0.76 = E^o_{Anode} + 0 \rightarrow E^o_{Anode} = +0.76V$
Result: The oxidation potential of Zinc is 0.76V; the reduction potential is -0.76V.

B. Copper Electrode ($Cu^{2+}/Cu$)

Setup: Copper rod in $1M$ Copper Sulphate solution connected to a SHE.
Observations: Electrons move from the SHE to the Copper electrode.
- Copper Electrode: Acts as the Cathode.
- SHE: Acts as the Anode.
Calculations:
- Voltmeter reading ($E^o_{cell}$) = $0.34V$
- $E^o_{cell} = E^o_{Anode} + E^o_{Cathode}$
- $0.34 = 0.0 + E^o_{Cathode} \rightarrow E^o_{Cathode} = +0.34V$
Result: The reduction potential of Copper is 0.34V; the oxidation potential is -0.34V.

Key Definitions

Standard Conditions: $1M$ concentration, $1\text{ atm}$ pressure, and $25^\circ C$ (indicated by the symbol $E^o$).
Salt Bridge: Used to connect two half-cells to complete the electrochemical circuit.

Study Guide: Determination of Cell Potential

Core Concepts of Cell Reactions

An electrochemical cell reaction is composed of two distinct half-reactions determined by their reduction potentials:

Reduction: Occurs in the half-cell with the greater reduction potential.
Oxidation: Occurs in the half-cell with the smaller reduction potential.

Calculating Cell Potential ($E^{\circ}_{cell}$)

The standard cell potential (or electromotive force, emf) is the algebraic difference between the reduction potentials of the two half-cells.

$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$

Process: Determining the Total Cell Reaction

Identify the half-cell with the smaller reduction potential.
Reverse the equation for that half-cell (turning it into an oxidation reaction).
Add the reversed equation to the half-cell equation with the greater reduction potential.
The sum of these two equations represents the final cell reaction.

Example Walkthroughs

1. Zn-Cu Cell

Given Data:
- $Cu^{2+} + 2e^- \rightarrow Cu_{(s)}$ ($E^{\circ} = +0.34V$) → Higher potential (Cathode/Reduction)
- $Zn^{2+} + 2e^- \rightarrow Zn_{(s)}$ ($E^{\circ} = -0.76V$) → Lower potential (Anode/Oxidation)
Cell Reaction: $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Cu_{(s)} + Zn^{2+}_{(aq)}$
Calculation: $E^{\circ}_{cell} = +0.34V - (-0.76V) = \mathbf{+1.10V}$
Electron Flow: Since $E^{\circ}_{cell}$ is positive, electrons flow from the Anode (Zn) to the Cathode (Cu).

2. Ni-Mg Cell

Given Data:
- $Ni^{2+} + 2e^- \rightarrow Ni_{(s)}$ ($E^{\circ} = -0.25V$) → Higher potential (Cathode/Reduction)
- $Mg^{2+} + 2e^- \rightarrow Mg_{(s)}$ ($E^{\circ} = -2.38V$) → Lower potential (Anode/Oxidation)
Action: To find the cell reaction, reverse the Magnesium (Mg) reaction and add it to the Nickel (Ni) reaction.
Anode Identification: The Magnesium electrode serves as the anode because it has the smaller reduction potential.

Study Guide: Feasibility of Chemical Reactions

Feasibility and Spontaneity

The possibility of a chemical reaction occurring on its own can be determined by the standard cell potential ($E^{\circ}$).

Feasible/Spontaneous Reaction: Occurs if the sum of the $E^{\circ}$ values of the two half-cell reactions is positive.
Non-Feasible Reaction: Occurs if the sum of the $E^{\circ}$ values is negative.
Reverse Reaction: If a forward reaction is not feasible (negative $E^{\circ}$), its reverse reaction will be spontaneous.

Determining Reaction Feasibility

To evaluate if a reaction like $Sn + Fe^{2+} \rightarrow Sn^{2+} + Fe$ will occur, follow these steps:

Identify Half-Reactions: Determine which element is being oxidized and which is being reduced based on the equation.
Assign Roles: The oxidized species acts as the anode; the reduced species acts as the cathode.
Apply the Formula: Use the standard reduction potentials to calculate the cell potential:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
Interpret the Result: A positive value means the reaction is feasible; a negative value means it is not.

Example Analysis

Reaction: $Sn + Fe^{2+} \rightarrow Sn^{2+} + Fe$

Data: $E^{\circ}_{Sn} = -0.14V$, $E^{\circ}_{Fe} = -0.44V$.
Oxidation (Anode): $Sn \rightarrow Sn^{2+} + 2e^-$.
Reduction (Cathode): $Fe^{2+} + 2e^- \rightarrow Fe$.
Calculation: $E^{\circ}_{cell} = -0.44V - (-0.14V) = \mathbf{-0.30V}$.
Conclusion: Because the $E^{\circ}_{cell}$ is negative, the reaction is not feasible.

Concept Assessment Practice

Cu-$F_2$ Cell: Given $Cu^{2+} (E = +0.34V)$ and $F_2 (E = +2.87V)$, calculate the $E^{\circ}_{cell}$ and determine electron flow.
Displacement: Use emf data to argue if Magnesium can displace Copper(II) sulphate or if Iodine can displace chlorine.
Al-$Mn^{2+}$ Reaction: Given $E^{\circ}_{Al} = -1.66V$ and $E^{\circ}_{Mn} = -1.18V$, determine if the reaction $Al + Mn^{2+} \rightarrow Al^{3+} + Mn$ is feasible.

Study Guide: Feasibility of Reactions and the Electrochemical Series

Feasibility of Chemical Reactions

The feasibility or spontaneity of a chemical reaction is determined by the standard cell potential ($E^{\circ}$).

Spontaneous/Feasible Reaction: Occurs when the sum of the $E^{\circ}$ values of the half-cell reactions is positive.
Non-Feasible Reaction: Occurs when the sum of the $E^{\circ}$ values is negative.
Reverse Reactions: If a forward reaction is not feasible, the reverse reaction will be spontaneous.

The Electrochemical Series

The electrochemical series is a list of elements arranged according to their standard electrode potentials relative to the standard hydrogen electrode.

Standard Conditions: $E^{\circ}$ values are measured at $1M$ ion concentration, $25^{\circ}C$ ($298K$), and $1 \text{ atm}$ pressure.
Reduction vs. Oxidation: Values are typically given as reduction potentials; the oxidation potential is obtained by reversing the sign.

Activity Series of Metals

Metals are ranked by their ability to replace other metals or hydrogen from compounds, based on their ease of oxidation.

Displacement Reactions: A more active metal (higher on the list) can displace a less active metal (lower on the list) from its compound.
Reaction Vigor: The greater the separation between two species in the series, the more vigorous the reaction will be.
Electron Transfer: Metals higher on the list transfer electrons to metal cations lower on the list.

Examples of Metal Activity:

Zinc and Copper: Zn can replace $Cu^{2+}$ ions because Zn is more easily oxidized; conversely, Cu cannot replace Zn from a solution.
Hydrogen Displacement: Active metals like Na and K can displace $H_2$ from water, while metals like Cu and Ag cannot.
Barium and Lead: Ba replaces Pb in lead (II) oxide because Ba is more active.

Process: Determining Reaction Feasibility

Identify the oxidation and reduction half-reactions from the given equation.
Assign the oxidized species as the anode and the reduced species as the cathode.
Calculate $E^{\circ}_{cell}$ using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Evaluate the result: A negative value (e.g., $-0.30V$ for $Sn + Fe^{2+}$) indicates the reaction is not feasible.

2.7 The Relative Reactivity of Species

Electrode potential values ($E^0$), also known as reduction potentials, indicate how easily a species acts as an oxidizing or reducing agent.

Oxidizing Agents: Species with higher reduction potentials. They have a greater tendency to gain electrons.
Reducing Agents: Species with lower reduction potentials.

Examples of Reactivity

Copper vs. Zinc:
- $Cu^{2+} + 2e^- \longrightarrow Cu$ ($E^0 = +0.34V$)
- $Zn^{2+} + 2e^- \longrightarrow Zn$ ($E^0 = -0.76V$)
- Conclusion: $Cu^{2+}$ is the stronger oxidizing agent because its reduction potential is higher.
Permanganate vs. Iron:
- $MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O$ ($E^0 = +1.52V$)
- $Fe^{2+} \longrightarrow Fe^{3+} + 1e^-$ ($E^0 = +0.77V$)
- Conclusion: $MnO_4^-$ acts as a strong oxidizing agent and $Fe^{2+}$ acts as a strong reducing agent.

2.8 Types of Electro-Chemical Cells

Electrochemical cells are devices that convert electrical energy into chemical energy and vice versa. There are two main types:

Electrolytic cells
Galvanic or voltaic cells

2.8.1 Electrolytic Cells

An electrolytic cell converts electrical energy into chemical energy by a process called electrolysis. This process uses an external electrical current to stimulate a non-spontaneous reaction.

ShutterstockThe Electrolysis Process:

Anode (Positive Electrode): The site where oxidation takes place and electrons are lost.
Cathode (Negative Electrode): The site where reduction takes place and electrons are gained.

Study Guide: Electrolysis

Core Concepts

Definition: A process where an electric current drives a non-spontaneous chemical reaction in a molten or aqueous state.
Apparatus: An electrolytic cell containing an electrolyte and two electrodes connected to a battery.
Conduction:
- Metallic: Current carried by electrons through wires.
- Electrolytic: Current carried by ions (anions and cations) within the electrolyte.
Electrode Reactions:
- Anode: The electrode where oxidation (loss of electrons) occurs.
- Cathode: The electrode where reduction (gain of electrons) occurs.

Units of Measurement

Coulomb (C): The SI unit of charge; the charge on $6.25 \times 10^{18}$ electrons.
Faraday (F): The charge carried by one mole of electrons ($96487 \text{ C}$).
Ampere: The SI unit of current, often defined as "a coulomb per second."

Factors Influencing Products of Electrolysis

The substances liberated depend on three primary factors:

1. State of Electrolyte

Molten State: Only the ions of the electrolyte are discharged.
Aqueous State: Contains ions from both the electrolyte and water; water molecules may produce $H_2$ at the cathode.

2. Position in the Electrochemical/Redox Series

Substances higher in the series tend to be reduced; those lower tend to be oxidized.
Cathode Preference: Cations less reactive than hydrogen (e.g., $Cu^{2+}$) are discharged. Cations more reactive than hydrogen (e.g., $Na^{+}$, $Mg^{2+}$) remain in solution while water is discharged to liberate hydrogen.

3. Concentration of Electrolyte

Higher concentrations can favor the discharge of certain ions.
If an ion is at a low concentration, it may not be discharged; instead, water molecules are discharged.

Case Studies: Electrolysis of NaCl

1. Fused (Molten) NaCl

At Anode: $2Cl^{-} \rightarrow Cl_2 + 2e^{-}$
At Cathode: $2Na^{+} + 2e^{-} \rightarrow 2Na$
Net: $2Na^{+} + 2Cl^{-} \rightarrow 2Na + Cl_2$

2. Concentrated Aqueous NaCl (Brine)

At Anode: $Cl^{-}$ ions are in high concentration and are discharged: $2Cl^{-} \rightarrow Cl_2 + 2e^{-}$
At Cathode: $Na^{+}$ is more reactive than hydrogen, so water is discharged: $2H_2O + 2e^{-} \rightarrow H_2 + 2OH^{-}$
Net: $2H_2O + 2Cl^{-} \rightarrow H_2 + Cl_2 + 2OH^{-}$

3. Dilute Aqueous NaCl

At Anode: $Cl^{-}$ concentration is low, so $OH^{-}$ ions are discharged: $4OH^{-} \rightarrow O_2 + 4e^{-} + 2H_2O$
At Cathode: $Na^{+}$ remains in solution; water is discharged: $4H_2O + 4e^{-} \rightarrow 2H_2 + 4OH^{-}$
Net: $4H_2O \rightarrow 2H_2 + O_2$

Study Guide: Faraday Constant and Electrolysis

1. The Faraday Constant ($F$)

Definition: One faraday is the amount of electric charge carried by one mole of electrons.
Calculation: It is derived by multiplying the charge of a single electron by the number of entities in a mole (Avogadro constant).
- Charge on one electron $\approx 1.602 \times 10^{-19} \text{ C}$
- One mole of electrons $\approx 6.022 \times 10^{23} \text{ per mol}$
- Result: $96485.332 \text{ C/mol}$ (commonly rounded to $96500 \text{ C}$ for calculations).
Key Equation: $$F = N_A \cdot e$$
- $F$ = Faraday constant
- $N_A$ = Avogadro constant
- $e$ = Elementary charge (charge on a single electron)

2. Relationship Between Charge, Current, and Time

The amount of substance produced at an electrode during electrolysis is determined by the total charge passed through the electrolyte.

Formula: $$Q = I \times t$$
Variables:
- $Q$ = Quantity of charge in Coulombs (C)
- $I$ = Current in Ampere (A)
- $t$ = Time of electrolysis in seconds (s)

3. Application in Electrolysis (Examples)

The number of Faradays required depends on the charge of the ion being deposited:

Sodium ($Na^+$):
- Equation: $Na^+ + e^- \rightarrow Na$
- Requirement: To deposit one mole of $Na$, $1F$ ($96500 \text{ C}$) is needed.
Copper ($Cu^{2+}$):
- Equation: $Cu^{2+} + 2e^- \rightarrow Cu$
- Requirement: To deposit one mole of $Cu$, 2 moles of electrons are required, so $2F$ of electricity is needed.

Study Guide: Electrochemistry Calculations

Core Concepts & Constants

Faraday's Constant ($1F$): Represents the charge of one mole of electrons, approximately $96500\text{ C}$.
Charge Calculation ($Q$): Calculated as $\text{Current (Amperes)} \times \text{Time (seconds)}$.
Molar Volume of Gas: Calculated using the Ideal Gas Law: $V = \frac{nRT}{P}$.

Electrolysis of Molten $ZnCl_2$

To determine the mass of Zinc ($Zn$) deposited at the cathode:

Calculate Total Charge: Multiply current by time in seconds (e.g., $0.01\text{ A} \times 3600\text{ s} = 36\text{ C}$).
Convert Charge to Faradays: Divide the coulombs by $96500$.
Identify Stoichiometry: The cathode reaction $Zn^{2+} + 2e^- \rightarrow Zn$ shows that $2F$ of current produces $1$ mole of $Zn$.
Calculate Moles of Product: Multiply the total Faradays by the molar ratio (e.g., $\frac{1}{2}$ mole $Zn$ per Faraday).
Convert Moles to Mass: Multiply moles by the molar mass (for $Zn$, $63.37\text{ g/mol}$).

Electrolysis of $AuCl_4^-$ Solution

1. Cathode Reaction (Gold Deposition)

Reaction: $AuCl_4^- + 3e^- \rightarrow Au + 4Cl^-$
Stoichiometry: $3$ Faradays are required to produce $1$ mole of Gold ($Au$).
Process:
- Determine moles of $Au$ produced: $\text{Mass} / 197\text{ g/mol}$.
- Calculate total Charge in Faradays: $\text{Moles of } Au \times 3\text{ F/mol}$.
- Calculate Current: $\text{Charge (C)} / \text{Time (s)}$.

2. Anode Reaction (Chlorine Evolution)

Reaction: $2Cl^- \rightarrow Cl_2 + 2e^-$
Stoichiometry: $2$ Faradays produce $1$ mole of $Cl_2$ gas.
Process:
- Calculate moles of $Cl_2$ using the total Faradays passed through the system (Molar ratio: $1/2$ mole $Cl_2$ per Faraday).

3. Gas Volume Calculation

To find the volume of $Cl_2$ gas collected at specific conditions ($1\text{ atm}$, $25^{\circ}C$):

Temperature Conversion: $25^{\circ}C + 273 = 298\text{ K}$.
Formula: $V = \frac{nRT}{P}$
- $n = \text{moles of } Cl_2$
- $R = 0.0821\text{ atm}\cdot\text{dm}^3\text{mol}^{-1}\text{K}^{-1}$

Experimental Determination of Avogadro Constant

Electrolysis provides a concrete method to measure the Avogadro Constant experimentally by determining the mass of metal deposited at a cathode using a known current over a specific time.

1. Core Concepts and Variables

Method: Electrolyzing a known electrolyte (e.g., $AgNO_3$ with Ag electrodes).
Current ($I$): Measured in Amperes (A).
Time ($t$): Must be converted into seconds (s) for calculations.
Total Charge ($Q$): The product of current and time ($Q = I \times t$), measured in Coulombs (C).
Cathode Reaction: $Ag^+_{(aq)} + 1e^- \rightarrow Ag_{(s)}$.
Relationship: The reaction equation shows that 1 mole of electrons is required to produce 1 mole of Silver (Ag).

2. The Calculation Process

Calculate Total Charge Used: Multiply the experimental current by the time in seconds (e.g., $0.1 \text{ A} \times 1800 \text{ s} = 180 \text{ C}$).
Find Charge for One Mole: Use the mass of metal deposited to calculate the charge required for 1 mole (molar mass) of that metal.
- Example: If $0.201\text{g}$ of Ag is produced by $180\text{C}$, then 1 mole ($107.868\text{g}$) requires $\approx 96598.209\text{C}$.
Determine Avogadro Constant ($N_A$): Divide the charge on 1 mole of electrons by the charge on a single electron ($1.602 \times 10^{-19}\text{C}$).
- Experimental Result: $96598.209\text{C} / 1.602 \times 10^{-19}\text{C} = 6.0298 \times 10^{23}$.

The experimental value ($6.0298 \times 10^{23}$) is very close to the accepted value of $6.02214 \times 10^{23}$.

3. Concept Assessment Topics

Substance Liberation: Calculating mass based on time, current, and molar mass (e.g., 0.5A for two hours).
Molar Charge: Determining total coulombs for a specific amount of electron flow (e.g., 3 moles).
Deposition Prediction: Calculating the specific mass of Silver deposited under given conditions (e.g., 0.1A for 20 minutes).

Short Question Answers

i. The oxidation potential of Zn is +0.76V and its reduction potential is -0.76V

The oxidation and reduction potentials for the same half-reaction are equal in magnitude but opposite in sign. Oxidation is the reverse process of reduction. Therefore, the potential for one process will be the negative of the potential for the other. This can be represented by the following equations:

$$E_{\text{oxidation}} = -E_{\text{reduction}}$$

For zinc (Zn), the oxidation half-reaction is:

$$Zn \rightarrow Zn^{2+} + 2e^- \quad E^\circ = +0.76V$$

The reduction half-reaction is:

$$Zn^{2+} + 2e^- \rightarrow Zn \quad E^\circ = -0.76V$$

ii. A salt bridge maintains the electrical neutrality in the cell

A salt bridge connects the two half-cells of a galvanic cell and contains an electrolyte solution. As oxidation occurs at the anode, positive ions accumulate in that half-cell. Simultaneously, as reduction occurs at the cathode, negative ions accumulate in that half-cell. The salt bridge allows ions to migrate between the two half-cells to balance these charges, preventing a buildup that would stop the flow of electrons and bring the reaction to a halt.

iii. Na and K can displace hydrogen from acids but Cu and Pt cannot

This is determined by the **activity series** of metals. Metals higher up in the series are more reactive and have a greater tendency to be oxidized (lose electrons) than those below them. Sodium (Na) and Potassium (K) are highly reactive alkali metals, located high in the activity series, meaning their standard reduction potentials are very negative. They are therefore able to reduce the hydrogen ions ($$H^+$$) in acids to hydrogen gas ($$H_2$$).

In contrast, Copper (Cu) and Platinum (Pt) are less reactive and are located below hydrogen in the activity series, with positive standard reduction potentials. They have a lower tendency to be oxidized and therefore cannot displace hydrogen from acids.

$$Na \rightarrow Na^+ + e^-$$

$$K \rightarrow K^+ + e^-$$

$$2H^+ + 2e^- \rightarrow H_2$$

$$Cu \rightarrow Cu^{2+} + 2e^-$$

iv. Define oxidation in terms of electron transfer

Oxidation is the **loss of electrons** by a substance during a chemical reaction. A substance that is oxidized loses one or more electrons, and its oxidation number increases.

v. What is the oxidation number of oxygen in $$H_2O_2$$?

The oxidation number of hydrogen (H) is typically +1. Since the molecule $$H_2O_2$$ is neutral, the sum of the oxidation numbers must be zero. Let 'x' be the oxidation number of oxygen (O).

$$2(+1) + 2(x) = 0$$

$$2 + 2x = 0$$

$$2x = -2$$

$$x = -1$$

Therefore, the oxidation number of oxygen in hydrogen peroxide ($$H_2O_2$$) is **-1**.

vi. Identify the reducing agent in the reaction $$Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$$

A **reducing agent** is the substance that gets oxidized (loses electrons) and causes the reduction of another substance. In the given reaction:

$$Zn$$ goes from an oxidation number of 0 to +2. It loses electrons, so it is oxidized.
$$Cu^{2+}$$ goes from an oxidation number of +2 to 0. It gains electrons, so it is reduced.

Since **Zn** is oxidized, it is the reducing agent.

vii. State the purpose of the Winkler method

The Winkler method is a titration technique used to **determine the concentration of dissolved oxygen (DO)** in water samples. It is a key procedure in environmental and water quality monitoring.

viii. Explain what happens at the anode during the electrolysis of aqueous sodium chloride

In the electrolysis of aqueous sodium chloride ($$NaCl$$ solution), the anode is the positive electrode. During electrolysis, **oxidation** occurs at the anode. The species present are $$Cl^-$$ ions from $$NaCl$$ and $$H_2O$$. Oxidation potentials are compared to see which reaction is more likely to occur:

Oxidation of chloride ions: $$2Cl^-(aq) \rightarrow Cl_2(g) + 2e^- \quad E^\circ = -1.36V$$
Oxidation of water: $$2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- \quad E^\circ = -1.23V$$

Based on the standard potentials, the oxidation of water is more favorable. However, due to the high **overpotential** of oxygen gas formation, the oxidation of chloride ions to form chlorine gas ($$Cl_2$$) is the reaction that predominantly occurs at the anode.

ix. What does a positive standard electrode potential indicate about a substance's tendency to gain electrons?

A positive standard electrode potential ($$E^\circ > 0$$) indicates a **high tendency for a substance to gain electrons** and undergo reduction. The more positive the potential, the stronger the oxidizing agent it is.

x. Calculate the oxidation number of chromium in $$K_2Cr_2O_7$$

The compound $$K_2Cr_2O_7$$ is neutral, so the sum of the oxidation numbers must be zero. The oxidation number of Potassium (K) is +1 (alkali metal), and oxygen (O) is typically -2. Let 'x' be the oxidation number of Chromium (Cr).

$$2(K) + 2(Cr) + 7(O) = 0$$

$$2(+1) + 2(x) + 7(-2) = 0$$

$$2 + 2x - 14 = 0$$

$$2x - 12 = 0$$

$$2x = 12$$

$$x = +6$$

The oxidation number of chromium in $$K_2Cr_2O_7$$ is **+6**.

xiv. Calculation of mass of Al and volume of O₂

First, let's calculate the total charge passed. The current (I) is 15 A, and the time (t) is 10 hours. We need to convert the time to seconds:

$$t = 10 \text{ hours} \times 60 \text{ min/hour} \times 60 \text{ s/min} = 36000 \text{ s}$$

The total charge (Q) is given by the formula $$Q = I \times t$$

$$Q = 15 \text{ A} \times 36000 \text{ s} = 540000 \text{ C}$$

Now, let's find the number of moles of electrons (nₑ⁻) using the Faraday constant ($$F \approx 96485 \text{ C/mol e}^-$$):

$$n_{e^-} = \frac{Q}{F} = \frac{540000 \text{ C}}{96485 \text{ C/mol e}^-} \approx 5.60 \text{ mol e}^-$$

The half-reactions are:

**At the cathode (Reduction):** $$Al^{3+} + 3e^- \rightarrow Al(s)$$
**At the anode (Oxidation):** $$2O^{2-} \rightarrow O_2(g) + 4e^-$$

From the half-reactions, we can see the molar ratios:

1 mol Al requires 3 mol e⁻
1 mol O₂ requires 4 mol e⁻

Using these ratios, let's find the moles of Al and O₂:

$$n_{Al} = 5.60 \text{ mol e}^- \times \frac{1 \text{ mol Al}}{3 \text{ mol e}^-} \approx 1.867 \text{ mol Al}$$
$$n_{O_2} = 5.60 \text{ mol e}^- \times \frac{1 \text{ mol } O_2}{4 \text{ mol e}^-} = 1.4 \text{ mol } O_2$$

Now, let's calculate the mass of Al and the volume of O₂.

Mass of Al:

Molar mass of Al is approximately 26.98 g/mol.

$$\text{Mass of Al} = n_{Al} \times \text{Molar mass}_{Al}$$

$$\text{Mass of Al} = 1.867 \text{ mol} \times 26.98 \text{ g/mol} \approx 50.37 \text{ g}$$

Volume of O₂:

We use the ideal gas law, $$PV = nRT$$. We are given standard conditions (1 atm and 25°C). The gas constant R for these units is $$0.0821 \text{ L} \cdot \text{atm} / (\text{mol} \cdot \text{K})$$.

Temperature in Kelvin: $$T = 25 + 273.15 = 298.15 \text{ K}$$

$$V = \frac{nRT}{P} = \frac{(1.4 \text{ mol})(0.0821 \text{ L} \cdot \text{atm} / (\text{mol} \cdot \text{K}))(298.15 \text{ K})}{1 \text{ atm}}$$

$$V \approx 34.3 \text{ L}$$

The answer provided is 34.20 dm³, and 1 L = 1 dm³, so the volume is approximately 34.3 dm³.

Answer: Mass of Al is approximately 50.37 g and the volume of O₂ is approximately 34.3 dm³.

xv. Comparison of Mass Deposited

The mass of metal deposited is directly proportional to its molar mass and inversely proportional to the number of electrons required for its reduction. The formula is:

$$\text{Mass} = \frac{I \times t \times \text{Molar Mass}}{F \times \text{Charge}}$$

Since the current (I), time (t), and Faraday constant (F) are the same for both cases, we can compare the ratio of Molar Mass to the charge for each metal.

For NaCl, the reaction is $$Na^+ + e^- \rightarrow Na$$. The charge on Na is +1.
Molar Mass of Na is approx. 23 g/mol. Ratio: $$\frac{23}{1} = 23$$
For $$CaCl_2$$, the reaction is $$Ca^{2+} + 2e^- \rightarrow Ca$$. The charge on Ca is +2.
Molar Mass of Ca is approx. 40 g/mol. Ratio: $$\frac{40}{2} = 20$$

Since the ratio for **Na is greater (23 > 20)**, **NaCl will give more mass of metal.**

xvi. Electroplating Time Calculation

We need to find the time (t) to deposit 75 g of copper with a 5 A current. The reaction is: $$Cu^{2+} + 2e^- \rightarrow Cu(s)$$.

First, find the moles of Cu to be deposited:

Molar mass of Cu is approx. 63.55 g/mol.

$$n_{Cu} = \frac{75 \text{ g}}{63.55 \text{ g/mol}} \approx 1.18 \text{ mol Cu}$$

From the reaction, 1 mol of Cu requires 2 mol of electrons. So, moles of electrons needed is:

$$n_{e^-} = 1.18 \text{ mol Cu} \times \frac{2 \text{ mol e}^-}{1 \text{ mol Cu}} = 2.36 \text{ mol e}^-$$

Now, find the total charge required:

$$Q = n_{e^-} \times F = 2.36 \text{ mol} \times 96485 \text{ C/mol} \approx 227600 \text{ C}$$

Finally, calculate the time using $$t = Q/I$$:

$$t = \frac{227600 \text{ C}}{5 \text{ A}} = 45520 \text{ s}$$

Convert seconds to hours:

$$\text{Hours} = \frac{45520 \text{ s}}{3600 \text{ s/hr}} \approx 12.64 \text{ hours}$$

Answer: It would take approximately 12.64 hours.

xvii. Differentiation

(a) Galvanic vs. Electrolytic Cell

**Galvanic Cell (Voltaic Cell):** Converts **chemical energy into electrical energy**. It is a **spontaneous** reaction ($$ \Delta G < 0 $$) that produces a voltage. Examples include batteries.
**Electrolytic Cell:** Converts **electrical energy into chemical energy**. It requires an external power source to drive a **non-spontaneous** reaction ($$ \Delta G > 0 $$). Example: electrolysis of water.

(b) Oxidation Half-Reaction vs. Reduction Half-Reaction

**Oxidation Half-Reaction:** Describes the process of a substance **losing electrons**. The oxidation number of the element **increases**. It occurs at the **anode**.
**Reduction Half-Reaction:** Describes the process of a substance **gaining electrons**. The oxidation number of the element **decreases**. It occurs at the **cathode**.

xviii. Electroplating Gold Calculation

We need to find the mass of gold deposited. The half-reaction is given: $$[Au(CN)_4]^- + 3e^- \rightarrow Au(s) + 4CN^-$$

First, calculate the total charge (Q) passed. Current (I) = 5.0 A, time (t) = 30 minutes. Convert time to seconds:

$$t = 30 \text{ min} \times 60 \text{ s/min} = 1800 \text{ s}$$

$$Q = I \times t = 5.0 \text{ A} \times 1800 \text{ s} = 9000 \text{ C}$$

Next, find the moles of electrons ($$n_{e^-}$$):

$$n_{e^-} = \frac{Q}{F} = \frac{9000 \text{ C}}{96485 \text{ C/mol e}^-} \approx 0.0933 \text{ mol e}^-$$

From the half-reaction, 1 mol of Au is deposited for every 3 mol of electrons. Moles of Au deposited is:

$$n_{Au} = 0.0933 \text{ mol e}^- \times \frac{1 \text{ mol Au}}{3 \text{ mol e}^-} \approx 0.0311 \text{ mol Au}$$

Finally, calculate the mass of Au using its molar mass (approx. 196.97 g/mol):

$$\text{Mass of Au} = n_{Au} \times \text{Molar mass}_{Au}$$

$$\text{Mass of Au} = 0.0311 \text{ mol} \times 196.97 \text{ g/mol} \approx 6.12 \text{ g}$$

Answer: The mass of gold deposited is approximately 6.12 g.

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FBISE Class 12 Chemistry Chapter 2: Electrochemistry Complete Notes | Federal Board FBISE | Read and Download