Electrolysis Notes
Electrolysis
Electrolytic vs. Galvanic Cells
Electrolytic Cell
Thermodynamically Unfavorable Reaction: Requires energy input to occur.
Anode and Cathode: Often in the same chamber.
Salt Bridge: Not needed.
Power Source: Required (e.g., a battery).
Electrical Energy: Uses electrical energy, resulting in a negative voltage value.
Galvanic Cell
Thermodynamically Favorable Reaction: Occurs spontaneously.
Anode and Cathode: In separate chambers (half-cells).
Salt Bridge: Needed to maintain charge balance.
Electrical Energy: Produces electrical energy (positive voltage value).
Both Types of Cells
Oxidation/Reduction: Oxidation occurs at the anode, and reduction occurs at the cathode.
Ion Flow: Require ion flow in the cell for the reaction to occur (cations flow to the cathode, and anions flow to the anode).
Thermodynamic Favorability
Voltaic Cells: Thermodynamically favorable reactions have positive overall cell potentials (E°_{cell} > 0).
Electrolytic Cells: Thermodynamically unfavorable reactions have negative overall cell potentials (E°_{cell} < 0).
Electrolysis and Electrolytic Cells
Electrolysis: The process in which electrical energy is used to cause a non-spontaneous reaction to occur.
Electrolytic Cell: Apparatus for carrying out electrolysis.
Examples:
Electrolysis of molten sodium chloride.
Electrolysis of water.
Electrolysis of aqueous sodium chloride solution.
Electrolysis of Water
Setup: A pair of electrodes made of a non-reactive metal (platinum or graphite) immersed in water.
Reaction Medium: The reaction occurs readily in a 0.1 M H2SO4 solution due to a sufficient number of ions to conduct electricity.
Observations: Gas bubbles appear on both electrodes immediately.
Reactions in Acidic Medium (0.1 M H2SO4)
Anode (Oxidation): 2H2O(l) \rightarrow O2(g) + 4H^+(aq) + 4e^- with E^0 = 1.23 V
Cathode (Reduction): 4H^+(aq) + 4e^- \rightarrow 2H_2(g) with E^0 = 0 V
Overall Reaction: 2H2O(l) \rightarrow O2(g) + 2H_2(g)
Cell Potential: E^0{cell} = E^0{cathode} - E^0_{anode} = 0 - 1.23 = -1.23 V (negative cell potential)
Reactions in Basic Medium (OH- ions)
Anode: Water is oxidized to oxygen gas and hydrogen ions.
Cathode: Water is reduced to hydrogen gas and hydroxide ions.
Example Problem
Given Half-Reactions:
O2(g) + 4H^+(aq) + 4e^- \rightarrow 2H2O(l), E^0 = 0.70 V
2H2O(l) + 2e^- \rightarrow H2(g) + 2OH^-(aq), E^0 = -0.83 V
Balanced Equation for Water Electrolysis:
Reverse the anode reaction:
2H2O(l) \rightarrow O2(g) + 4H^+(ag) + 4e^-2H2O(l) + 2e^- \rightarrow H2(g) + 2OH^-(aq)
Overall:
H2O(l) \rightarrow 1/2 O2(g) + H_2(g)
Standard Cell Potential Calculation:
E^0{cell} = E^0{cathode} - E^0_{anode} = -0.83 V - (0.70 V) = -1.53 V
Conceptual Question
Given Electrolytic Cell Reaction:
2H2O(l) \rightarrow 2H2(g) + O_2(g)
Question A: Where is hydrogen gas produced (anode or cathode)?
Answer: Hydrogen gas is produced at the cathode because electrons flow into the electrode where hydrogen bubbles appear, and electrons are consumed in reduction, which occurs at the cathode.
Question B: Why is the liquid level lower below the hydrogen bubble than the oxygen bubble?
Answer: The balanced equation shows that twice as much hydrogen as oxygen is produced. Thus, there are more molecules of hydrogen than oxygen, and they occupy more space.
Electroplating
Definition: A process where a thin layer of one metal is deposited onto the surface of another metal.
Key Components and Principles:
Anode:
Made from the pure metal to be coated.
Connected to the positive end of the battery (+).
Oxidation reaction takes place at the anode.
Cathode:
The object intended to be electroplated.
Connected to the negative end of the battery (-).
Reduction reaction takes place at the cathode.
Electrolyte:
Both electrodes are immersed in an aqueous solution containing a soluble salt of the pure metal used as the anode.
Example: Silver-Plated Tableware
The anode is a silver electrode.
The metal spoon is the cathode, connected to the negative terminal of the voltage source.
Both electrodes are immersed in a solution of silver nitrate (AgNO_3).
Reactions
Anode (Oxidation): Ag(s) \rightarrow Ag^+(aq) + e^-
Cathode (Reduction): Ag^+(aq) + e^- \rightarrow Ag(s)
Quantitative Aspects of Electrolysis: Faraday’s Law
Faraday's Observation: The mass formed or consumed at any electrode is proportional to the amount of electricity transferred and the molar mass of the substance.
Charge, Current, and Time Relationship
The amount of charge (q) that flows in an electrolytic cell is a function of the current (I) and the time (t) that the cell operates.
Equation: I = q/t
q is related to changes in the amounts of reactants and products in the cell.
Stoichiometry is used to analyze these relationships.
Units
Current (I) is in amperes (A), which is equivalent to coulombs/second (C/s).
Time (t) needs to be in seconds when using Faraday's law.
Faraday's constant (F) is the charge per mole of electrons: 96,485 C/mol e-
Steps to Solve Electrolysis Problems
Multiply current (amperes) and time (seconds) to get charge in coulombs.
Divide the charge by the Faraday constant to get the number of moles of electrons.
Use the mole ratio in the half-cell reaction to find the moles of substance reduced or oxidized.
Use molar mass or the ideal gas equation to convert moles to grams or liters of product.
Summary of Formulas
Q(C) = I(A) \times t(sec)
mole \space e^- = Q / 96485
mole(element) = (mole \space e^-) / (coefficient \space e^-)
m(g) = n(mol) \times M_m(g/mol)
To Find the Time Required for Electroplating
n = m / M_m
mole \space e^- = mole(element) \times coefficient \space e^-
Q = 96485 \times mole \space e^-
t = Q / I
Practice Problem 1: Plating Copper
Given: Molar mass of Cu = 63.55 g/mol, current = 1.2 A, time = 10 minutes.
Practice Problem 2: Plating Gold
Given: Molar mass of Au = 196.9 g/mol, mass of gold = 0.125 g, current = 0.800 A.