Chapter Notes

Electrochemistry

Electrochemistry is the branch of chemistry that studies the relationship between chemical reactions and electrical energy. It explores two main processes:

1. How spontaneous chemical reactions can be used to generate electricity.
2. How electrical energy can be used to cause non-spontaneous chemical reactions to occur.

This field is crucial for many technologies, including the production of metals and chemicals, the functioning of batteries and fuel cells, and even biological processes like the transmission of nerve signals.

Electrochemical Cells

An electrochemical cell is a device that can either generate electrical energy from a chemical reaction or use electrical energy to cause one. There are two main types of electrochemical cells.

Galvanic and Electrolytic Cells

1. Galvanic Cell (or Voltaic Cell): This type of cell converts the chemical energy from a spontaneous redox (reduction-oxidation) reaction into electrical energy. A common example is the Daniell cell. In this cell, a zinc strip is dipped in a zinc sulfate solution, and a copper strip is dipped in a copper sulfate solution. The overall spontaneous reaction is: $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$ This reaction produces an electrical potential of $1.1$ V under standard conditions. Devices like batteries are galvanic cells.

2. Electrolytic Cell: This type of cell uses electrical energy from an external source (like a power supply) to drive a non-spontaneous chemical reaction. For instance, if you apply an external voltage greater than $1.1$ V to the Daniell cell, the reaction reverses. Zinc ions are reduced back to zinc metal, and copper metal is oxidized to copper ions. This process is called electrolysis.

How a Galvanic Cell Works

A galvanic cell consists of two half-cells. Each half-cell contains a metal rod, called an electrode, dipped in a solution of its own ions (the electrolyte).

• Oxidation Half-Cell (Anode): The electrode where oxidation (loss of electrons) occurs. In the Daniell cell, this is the zinc electrode. It is considered the negative electrode because electrons are released here. $\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$
• Reduction Half-Cell (Cathode): The electrode where reduction (gain of electrons) occurs. In the Daniell cell, this is the copper electrode. It is the positive electrode because electrons are consumed here. $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})$

The two half-cells are connected externally by a wire, which allows electrons to flow from the anode to the cathode. Internally, they are connected by a salt bridge, which contains an inert electrolyte and allows ions to flow between the half-cells to maintain electrical neutrality.

Electrode Potential

At the interface between an electrode and its electrolyte, a potential difference develops. This is called the electrode potential. It arises from the competing tendencies of metal ions to deposit on the electrode and metal atoms to dissolve into the solution as ions.

• Standard Electrode Potential ($E^\circ$): This is the electrode potential measured under standard conditions: all species have a concentration of $1 \text{ mol dm}^{-3}$ (or $1$ M), the pressure of any gas is $1$ bar, and the temperature is $298$ K. By convention, standard electrode potentials are quoted as standard reduction potentials.

The potential difference between the two electrodes in a galvanic cell is called the cell potential or electromotive force (emf), measured in volts (V). It is calculated as the difference between the reduction potentials of the cathode and the anode.

By convention, the anode is written on the left and the cathode on the right. $E_{\text{cell}} = E_{\text{right}} - E_{\text{left}} = E_{\text{cathode}} - E_{\text{anode}}$

A galvanic cell is represented using a shorthand notation. A single vertical line | represents a phase boundary (e.g., between a solid electrode and a solution), and a double vertical line || represents the salt bridge.

• Anode (Oxidation): $\mathrm{Cu}(\mathrm{s}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$
• Cathode (Reduction): $2 \mathrm{Ag}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Ag}(\mathrm{s})$

The cell is represented as: $\mathrm{Cu}(\mathrm{s})|\mathrm{Cu}^{2+}(\mathrm{aq})||\mathrm{Ag}^{+}(\mathrm{aq})|\mathrm{Ag}(\mathrms)$ The cell potential is: $E_{\text{cell}} = E_{\mathrm{Ag}^{+}|\mathrm{Ag}} - E_{\mathrm{Cu}^{2+}|\mathrm{Cu}}$

Measurement of Electrode Potential

It is impossible to measure the potential of a single half-cell. We can only measure the potential difference between two half-cells. To assign values to individual electrodes, we use a reference electrode.

The universally accepted reference is the Standard Hydrogen Electrode (SHE). It consists of a platinum electrode coated with platinum black, dipped in a $1$ M solution of H$^+$ ions, with pure hydrogen gas at $1$ bar pressure bubbled through it.

The SHE is assigned a potential of exactly $0.00$ V at all temperatures. $\mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \frac{1}{2} \mathrm{H}_{2}(\mathrm{g}) \quad E^\circ = 0.00 \text{ V}$

To find the standard potential of another electrode, we create a galvanic cell with the SHE as one half-cell (the anode, on the left) and the unknown electrode as the other (the cathode, on the right). The measured emf of this cell is the standard electrode potential of the unknown electrode. $E^\circ_{\text{cell}} = E^\circ_{\text{right}} - E^\circ_{\text{left}} = E^\circ_{\text{cathode}} - E^\circ_{\text{SHE}} = E^\circ_{\text{cathode}} - 0 = E^\circ_{\text{cathode}}$

• A positive $E^\circ$ value means the species is more easily reduced than H$^+$ ions.
• A negative $E^\circ$ value means the species is less easily reduced than H$^+$ ions (i.e., its reduced form is a stronger reducing agent than H$_2$ gas).

By measuring various electrodes against the SHE, we can create a list of standard electrode potentials, often called the electrochemical series. This series allows us to predict the spontaneity of redox reactions and the relative strengths of oxidizing and reducing agents.

• Strongest Oxidizing Agent: The species with the highest positive $E^\circ$ value (e.g., F$_2$). It has the greatest tendency to be reduced.
• Strongest Reducing Agent: The species with the most negative $E^\circ$ value (e.g., Li). Its oxidized form has the least tendency to be reduced.

Nernst Equation

Standard electrode potentials are only valid under standard conditions (1 M concentrations). The Nernst Equation allows us to calculate the electrode potential at any concentration.

For a general electrode reaction: $\mathrm{M}^{\mathrm{n}+}(\mathrm{aq})+\mathrm{ne}^{-} \rightarrow \mathrm{M}(\mathrm{s})$ The Nernst equation is: $E_{\left(\mathrm{M}^{\mathrm{n}+} / \mathrm{M}\right)}=E_{\left(\mathrm{M}^{\mathrm{n}+} / \mathrm{M}\right)}^{\mathrm{o}}-\frac{R T}{n F} \ln \frac{[\mathrm{M}]}{\left[\mathrm{M}^{\mathrm{n}+}\right]}$ Since the concentration of a pure solid $[\mathrm{M}]$ is taken as unity (1), the equation simplifies to: $E_{\left(\mathrm{M}^{\mathrm{n}+} / \mathrm{M}\right)}=E_{\left(\mathrm{M}^{\mathrm{n}+} / \mathrm{M}\right)}^{\mathrm{o}}-\frac{R T}{n F} \ln \frac{1}{\left[\mathrm{M}^{\mathrm{n}^{+}}\right]}$ Where:

• $E^\circ$ is the standard electrode potential.
• $R$ is the gas constant ($8.314 \text{ J K}^{-1} \text{ mol}^{-1}$).
• $T$ is the temperature in Kelvin.
• $n$ is the number of moles of electrons transferred.
• $F$ is the Faraday constant ($96487 \text{ C mol}^{-1}$).
• $[\mathrm{M}^{\mathrm{n}+}]$ is the molar concentration of the metal ion.

For a full cell, like the Daniell cell, the Nernst equation for the cell potential is: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln \frac{[\text{Products}]}{[\text{Reactants}]}$ For the Daniell cell reaction $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{2F} \ln \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]}$

At room temperature ($T = 298$ K), converting natural log (ln) to base-10 log and substituting the constants, the equation becomes: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]}$

Given

• Cell reaction: $\mathrm{Mg}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.0001 \mathrm{M}) \rightarrow \mathrm{Mg}^{2+}(0.130 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})$
• $[\mathrm{Mg}^{2+}] = 0.130 \text{ M}$
• $[\mathrm{Ag}^{+}] = 0.0001 \text{ M}$
• $E^\circ_{\text{cell}} = 3.17 \text{ V}$
• Number of electrons transferred, $n=2$

To Find

The cell potential, $E_{\text{cell}}$.

Formula

The cell is represented as $\mathrm{Mg}|\mathrm{Mg}^{2+}(0.130 \mathrm{M})||\mathrm{Ag}^{+}(0.0001 \mathrm{M})|\mathrm{Ag}$. The Nernst equation is: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059 \text{ V}}{n} \log \frac{[\mathrm{Mg}^{2+}]}{[\mathrm{Ag}^{+}]^2}$

Solution

Substitute the given values into the formula: $E_{\text{cell}} = 3.17 \text{ V} - \frac{0.059 \text{ V}}{2} \log \frac{0.130}{(0.0001)^2}$ $E_{\text{cell}} = 3.17 \text{ V} - 0.0295 \text{ V} \times \log(1.3 \times 10^7)$ $E_{\text{cell}} = 3.17 \text{ V} - 0.0295 \text{ V} \times (7.1139)$ $E_{\text{cell}} = 3.17 \text{ V} - 0.21 \text{ V} = 2.96 \text{ V}$

Final Answer The cell potential is $2.96 \text{ V}$.

Equilibrium Constant from Nernst Equation

As a galvanic cell operates, reactant concentrations decrease, product concentrations increase, and the cell potential drops. Eventually, the cell reaches chemical equilibrium, and the potential becomes zero ($E_{\text{cell}} = 0$). At this point, the reaction quotient $Q$ becomes equal to the equilibrium constant ($K_c$).

Setting $E_{\text{cell}} = 0$ in the Nernst equation: $0 = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln K_c$ Rearranging this gives a relationship between the standard cell potential and the equilibrium constant: $E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K_c = \frac{2.303 RT}{nF} \log K_c$ At $T = 298$ K: $E^\circ_{\text{cell}} = \frac{0.059 \text{ V}}{n} \log K_c$

Given

• $E^\circ_{\text{cell}} = 0.46 \text{ V}$
• The reaction involves the transfer of 2 electrons, so $n=2$.

To Find

The equilibrium constant, $K_c$.

Formula

$E^\circ_{\text{cell}} = \frac{0.059 \text{ V}}{n} \log K_c$

Solution

Rearrange the formula to solve for $\log K_c$: $\log K_c = \frac{n \times E^\circ_{\text{cell}}}{0.059 \text{ V}}$ $\log K_c = \frac{2 \times 0.46 \text{ V}}{0.059 \text{ V}} = 15.6$ Now, find $K_c$: $K_c = 10^{15.6} = 3.92 \times 10^{15}$

Final Answer The equilibrium constant for the reaction is $3.92 \times 10^{15}$.

Electrochemical Cell and Gibbs Energy of the Reaction

The maximum electrical work that can be obtained from a galvanic cell is equal to the decrease in Gibbs energy ($\Delta_r G$) of the cell reaction. The relationship is: $\Delta_r G = -nFE_{\text{cell}}$ Under standard conditions, this becomes: $\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$ Here, a positive $E^\circ_{\text{cell}}$ corresponds to a negative $\Delta_r G^\circ$, indicating a spontaneous reaction.

We can also relate the standard Gibbs energy to the equilibrium constant: $\Delta_r G^\circ = -RT \ln K_c$ This means we can calculate important thermodynamic quantities like $\Delta_r G^\circ$ and $K_c$ simply by measuring the standard potential of an electrochemical cell.

Given

• $E^\circ_{\text{cell}} = 1.1 \text{ V}$
• Number of electrons transferred, $n=2$
• Faraday constant, $F = 96487 \text{ C mol}^{-1}$

To Find

The standard Gibbs energy, $\Delta_r G^\circ$.

Formula

$\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$

Solution

Substitute the values into the formula: $\Delta_r G^\circ = -2 \times 96487 \text{ C mol}^{-1} \times 1.1 \text{ V}$ $\Delta_r G^\circ = -212271.4 \text{ J mol}^{-1}$ Converting to kilojoules: $\Delta_r G^\circ = -212.27 \text{ kJ mol}^{-1}$

Final Answer The standard Gibbs energy for the reaction is $-212.27 \text{ kJ mol}^{-1}$.

Conductance of Electrolytic Solutions

While metals conduct electricity via the flow of electrons (electronic conductance), electrolytic solutions conduct electricity via the movement of ions (electrolytic or ionic conductance).

Basic Terms

• Resistance (R): The opposition to the flow of current. Measured in ohms ($\Omega$). It is directly proportional to the length ($l$) of the conductor and inversely proportional to its cross-sectional area ($A$). $R = \rho \frac{l}{A}$
• Resistivity ($\rho$): The resistance of a conductor that is one meter long with a cross-sectional area of one square meter. Its unit is ohm-meter ($\Omega$ m).
• Conductance (G): The inverse of resistance. It measures the ease with which current flows. $G = \frac{1}{R}$ The SI unit of conductance is the siemens (S), where $1 \text{ S} = 1 \Omega^{-1}$.
• Conductivity ($\kappa$): The inverse of resistivity. It is the conductance of a solution that is 1 m long with a cross-sectional area of 1 m$^2$. $\kappa = \frac{1}{\rho} = \frac{l}{RA} = G \frac{l}{A}$ The SI unit is S m$^{-1}$, though S cm$^{-1}$ is also commonly used.

Measurement of the Conductivity of Ionic Solutions

Measuring the resistance of an ionic solution presents two challenges:

1. Passing a direct current (DC) can change the solution's composition through electrolysis.
2. A solution cannot be connected to a circuit like a solid wire.

These problems are solved by:

1. Using an alternating current (AC) source to prevent changes in composition.
2. Using a specially designed conductivity cell containing two platinum electrodes.

The quantity $l/A$ for a specific conductivity cell is constant and is called the cell constant ($G^*$). $G^* = \frac{l}{A}$ The cell constant is determined by measuring the resistance of the cell when it is filled with a solution of known conductivity (e.g., a KCl solution). $G^* = R \times \kappa$ Once $G^*$ is known, the conductivity of any unknown solution can be found by measuring its resistance ($R_{\text{unknown}}$) in the same cell. $\kappa = \frac{G^*}{R_{\text{unknown}}}$

Molar Conductivity

Conductivity ($\kappa$) depends on the concentration of ions. A more useful quantity is molar conductivity ($\Lambda_m$), which is the conductivity of a solution divided by its molar concentration ($c$). $\Lambda_m = \frac{\kappa}{c}$ Molar conductivity is the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution.

• If $\kappa$ is in S m$^{-1}$ and $c$ is in mol m$^{-3}$, then $\Lambda_m$ is in S m$^2$ mol$^{-1}$.
• If $\kappa$ is in S cm$^{-1}$ and $c$ is in mol L$^{-1}$ (molarity), the formula is: $\Lambda_m (\text{S cm}^2 \text{mol}^{-1}) = \frac{\kappa (\text{S cm}^{-1}) \times 1000 (\text{cm}^3/\text{L})}{\text{Molarity (mol/L)}}$

Given

• Concentration, $c = 0.05 \text{ mol L}^{-1}$
• Resistance, $R = 5.55 \times 10^3 \Omega$
• Length, $l = 50 \text{ cm}$
• Diameter = $1$ cm, so radius, $r = 0.5$ cm

To Find

• Resistivity, $\rho$
• Conductivity, $\kappa$
• Molar conductivity, $\Lambda_m$

Formula

• Area, $A = \pi r^2$
• Resistivity, $\rho = \frac{RA}{l}$
• Conductivity, $\kappa = \frac{1}{\rho}$
• Molar conductivity, $\Lambda_m = \frac{\kappa \times 1000}{c}$

Solution

First, calculate the cross-sectional area: $A = \pi (0.5 \text{ cm})^2 = 3.14 \times 0.25 \text{ cm}^2 = 0.785 \text{ cm}^2$ Now, calculate the resistivity: $\rho = \frac{(5.55 \times 10^3 \Omega) \times (0.785 \text{ cm}^2)}{50 \text{ cm}} = 87.135 \Omega \text{ cm}$ Next, calculate the conductivity: $\kappa = \frac{1}{\rho} = \frac{1}{87.135 \Omega \text{ cm}} = 0.01148 \text{ S cm}^{-1}$ Finally, calculate the molar conductivity: $\Lambda_m = \frac{0.01148 \text{ S cm}^{-1} \times 1000 \text{ cm}^3 \text{L}^{-1}}{0.05 \text{ mol L}^{-1}} = 229.6 \text{ S cm}^2 \text{mol}^{-1}$

Final Answer

• Resistivity = $87.135 \Omega \text{ cm}$
• Conductivity = $0.01148 \text{ S cm}^{-1}$
• Molar conductivity = $229.6 \text{ S cm}^2 \text{mol}^{-1}$

Variation of Conductivity and Molar Conductivity with Concentration

• Conductivity ($\kappa$) always decreases as concentration decreases (i.e., on dilution). This is because the number of ions per unit volume of the solution decreases.
• Molar conductivity ($\Lambda_m$) always increases as concentration decreases. This is because on dilution, the total volume ($V$) containing one mole of the electrolyte increases, and this increase in volume more than compensates for the decrease in conductivity.

The way $\Lambda_m$ increases with dilution is different for strong and weak electrolytes.

• Strong Electrolytes: These are substances (like KCl, NaCl) that dissociate completely into ions in solution. For strong electrolytes, $\Lambda_m$ increases slowly with dilution. The relationship can be described by the equation: $\Lambda_m = \Lambda_m^\circ - A c^{1/2}$ Here, $\Lambda_m^\circ$ is the limiting molar conductivity (the molar conductivity at zero concentration or infinite dilution). By plotting $\Lambda_m$ versus $c^{1/2}$, we get a straight line that can be extrapolated to $c=0$ to find $\Lambda_m^\circ$.

• Weak Electrolytes: These are substances (like acetic acid, CH$_3$COOH) that only partially dissociate in solution. For weak electrolytes, $\Lambda_m$ increases very steeply at low concentrations. This is because dilution increases the degree of dissociation ($\alpha$), creating more ions. The plot of $\Lambda_m$ versus $c^{1/2}$ is a curve, and it's impossible to extrapolate to find $\Lambda_m^\circ$.

Kohlrausch's Law of Independent Migration of Ions

Kohlrausch's law states that the limiting molar conductivity of an electrolyte ($\Lambda_m^\circ$) is the sum of the individual contributions of its constituent cations and anions. $\Lambda_m^\circ = v_+ \lambda_+^\circ + v_- \lambda_-^\circ$ Where:

• $\lambda_+^\circ$ and $\lambda_-^\circ$ are the limiting molar conductivities of the cation and anion, respectively.
• $v_+$ and $v_-$ are the number of cations and anions produced per formula unit of the electrolyte.

Applications of Kohlrausch's Law:

1. Calculating $\Lambda_m^\circ$ for Weak Electrolytes: We can calculate the limiting molar conductivity for a weak electrolyte by combining the values for strong electrolytes. For example, for acetic acid (HAc): $\Lambda_{m(\text{HAc})}^\circ = \lambda_{\text{H}^+}^\circ + \lambda_{\text{Ac}^-}^\circ$ This can be found using: $\Lambda_{m(\text{HAc})}^\circ = \Lambda_{m(\text{HCl})}^\circ + \Lambda_{m(\text{NaAc})}^\circ - \Lambda_{m(\text{NaCl})}^\circ$
2. Calculating the Degree of Dissociation ($\alpha$): For a weak electrolyte at concentration $c$: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$
3. Calculating the Dissociation Constant ($K_a$): $K_a = \frac{c \alpha^2}{1-\alpha} = \frac{c (\Lambda_m / \Lambda_m^\circ)^2}{1 - (\Lambda_m / \Lambda_m^\circ)}$

Given

• $\Lambda_{m(\text{NaCl})}^\circ = 126.4 \text{ S cm}^2 \text{mol}^{-1}$
• $\Lambda_{m(\text{HCl})}^\circ = 425.9 \text{ S cm}^2 \text{mol}^{-1}$
• $\Lambda_{m(\text{NaAc})}^\circ = 91.0 \text{ S cm}^2 \text{mol}^{-1}$

To Find

$\Lambda_{m(\text{HAc})}^\circ$

Formula

From Kohlrausch's law, we can combine the ionic conductivities: $\Lambda_{m(\text{HAc})}^\circ = \lambda_{\text{H}^+}^\circ + \lambda_{\text{Ac}^-}^\circ$ We can get these ions from the given strong electrolytes: $\Lambda_{m(\text{HAc})}^\circ = (\lambda_{\text{H}^+}^\circ + \lambda_{\text{Cl}^-}^\circ) + (\lambda_{\text{Na}^+}^\circ + \lambda_{\text{Ac}^-}^\circ) - (\lambda_{\text{Na}^+}^\circ + \lambda_{\text{Cl}^-}^\circ)$ $\Lambda_{m(\text{HAc})}^\circ = \Lambda_{m(\text{HCl})}^\circ + \Lambda_{m(\text{NaAc})}^\circ - \Lambda_{m(\text{NaCl})}^\circ$

Solution

Substitute the given values into the formula: $\Lambda_{m(\text{HAc})}^\circ = (425.9 + 91.0 - 126.4) \text{ S cm}^2 \text{mol}^{-1}$ $\Lambda_{m(\text{HAc})}^\circ = 390.5 \text{ S cm}^2 \text{mol}^{-1}$

Final Answer The limiting molar conductivity for acetic acid is $390.5 \text{ S cm}^2 \text{mol}^{-1}$.

Electrolytic Cells and Electrolysis

In an electrolytic cell, an external voltage is applied to drive a non-spontaneous chemical reaction. This process is called electrolysis. It is used for industrial processes like electroplating, refining metals (like copper), and producing highly reactive metals like sodium, magnesium, and aluminum from their molten salts.

Quantitative Aspects of Electrolysis: Faraday's Laws

Michael Faraday established the quantitative relationship between electricity and chemical change.

• Faraday's First Law of Electrolysis: The amount of chemical substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
• Faraday's Second Law of Electrolysis: When the same quantity of electricity is passed through different electrolytic solutions, the amounts of the substances liberated are proportional to their chemical equivalent weights.

The quantity of electricity ($Q$) is given by: $Q = I \times t$ where $I$ is the current in amperes (A) and $t$ is the time in seconds (s). The unit of charge is the coulomb (C).

The charge on one mole of electrons is called the Faraday constant (F). $F = 96487 \text{ C mol}^{-1} \approx 96500 \text{ C mol}^{-1}$ This means:

• To reduce 1 mole of $\text{Ag}^+$ ions ($\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}$), we need 1 mole of electrons, or $1 F$ of charge.
• To reduce 1 mole of $\text{Mg}^{2+}$ ions ($\text{Mg}^{2+} + 2\text{e}^- \rightarrow \text{Mg}$), we need 2 moles of electrons, or $2 F$ of charge.
• To reduce 1 mole of $\text{Al}^{3+}$ ions ($\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}$), we need 3 moles of electrons, or $3 F$ of charge.

Given

• Current, $I = 1.5 \text{ A}$
• Time, $t = 10 \text{ minutes} = 10 \times 60 \text{ s} = 600 \text{ s}$
• Atomic mass of Cu = $63 \text{ g mol}^{-1}$

To Find

The mass of copper deposited.

Formula

1. Calculate the total charge passed: $Q = I \times t$
2. The reaction is: $\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})$
3. This means $2F$ of charge ($2 \times 96487$ C) deposits 1 mole of Cu (63 g).

Solution

First, find the total charge: $Q = 1.5 \text{ A} \times 600 \text{ s} = 900 \text{ C}$ Now, use the stoichiometry to find the mass of Cu: $\text{Mass of Cu} = \frac{\text{Molar mass of Cu}}{\text{Charge for 1 mole}} \times Q$ $\text{Mass of Cu} = \frac{63 \text{ g mol}^{-1}}{2 \times 96487 \text{ C mol}^{-1}} \times 900 \text{ C}$ $\text{Mass of Cu} = 0.2938 \text{ g}$

Final Answer The mass of copper deposited is $0.2938$ g.

Products of Electrolysis

The products of electrolysis depend on:

1. The nature of the electrolyte (molten vs. aqueous).
2. The type of electrodes used (inert like platinum vs. reactive like copper).

In an aqueous solution, water molecules can also be oxidized or reduced, so there is competition at the electrodes. The reaction with the higher (more positive) reduction potential is preferred at the cathode, and the reaction with the lower reduction potential (i.e., higher oxidation potential) is preferred at the anode.

1. $\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Na}(\mathrm{s}) \quad E^\circ = -2.71 \text{ V}$
2. $2\mathrm{H}_{2}\mathrm{O}(l)+2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(\mathrm{g})+2\mathrm{OH}^{-}(\mathrm{aq}) \quad E^\circ = -0.83 \text{ V}$ (At standard conditions, the reduction of H$^+$ is $0.00$ V, but in neutral water, the potential is different) The reduction of water is preferred because it has a higher reduction potential. Product at Cathode: H$_2$ gas.

At the anode (oxidation), two reactions are possible:

1. $2\mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{g})+2\mathrm{e}^{-} \quad E^\circ_{\text{oxidation}} = -1.36 \text{ V}$
2. $2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{O}_{2}(\mathrm{g})+4\mathrm{H}^{+}(\mathrm{aq})+4\mathrm{e}^{-} \quad E^\circ_{\text{oxidation}} = -1.23 \text{ V}$ Thermodynamically, the oxidation of water should be preferred. However, due to the overpotential of oxygen, the oxidation of chloride ions is favored. Product at Anode: Cl$_2$ gas.

Batteries

A battery is one or more galvanic cells connected in series, used as a practical source of electrical energy. A good battery should be light, compact, and maintain a relatively constant voltage.

Primary Batteries

In primary batteries, the electrode reactions cannot be reversed. Once the reactants are consumed, the battery is "dead" and must be discarded.

• Dry Cell (Leclanché Cell): Used in clocks and transistors.

  • Anode: Zinc container ($\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{e}^{-}$).
  • Cathode: Graphite rod surrounded by a paste of MnO$_2$ and carbon.
  • Electrolyte: Moist paste of NH$_4$Cl and ZnCl$_2$.
  • Voltage: About $1.5$ V.

• Mercury Cell: Used in watches and hearing aids.

  • Anode: Zinc-mercury amalgam.
  • Cathode: Paste of HgO and carbon.
  • Electrolyte: Paste of KOH and ZnO.
  • Voltage: About $1.35$ V. It provides a constant voltage throughout its life because the overall reaction does not involve ions whose concentration can change.

Secondary Batteries

In secondary batteries, the cell reaction can be reversed by passing an external current through it (charging). They are rechargeable.

• Lead Storage Battery: Used in cars and inverters.

  • Anode: Spongy lead (Pb).
  • Cathode: Grid of lead packed with lead dioxide (PbO$_2$).
  • Electrolyte: 38% solution of sulfuric acid (H$_2$SO$_4$).
  • Discharging Reaction: Both electrodes get coated with lead sulfate (PbSO$_4$). $\mathrm{Pb}(\mathrm{s})+\mathrm{PbO}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow 2 \mathrm{PbSO}_{4}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
  • Charging Reaction: The above reaction is reversed, regenerating the original Pb and PbO$_2$ electrodes.

• Nickel-Cadmium (Ni-Cd) Cell: Has a longer life than the lead storage battery but is more expensive.

  • Overall Reaction (Discharge): $\mathrm{Cd}(\mathrm{s})+2 \mathrm{Ni}(\mathrm{OH})_{3}(\mathrm{s}) \rightarrow \mathrm{CdO}(\mathrm{s})+2 \mathrm{Ni}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Fuel Cells

Fuel cells are galvanic cells designed to convert the energy from the combustion of fuels (like hydrogen, methane, or methanol) directly into electrical energy. Reactants are continuously supplied to the electrodes.

• Advantages: They are highly efficient (around 70% efficiency compared to ~40% for thermal power plants) and pollution-free.

• Hydrogen-Oxygen Fuel Cell: This was used in the Apollo space program.

  • Anode: Hydrogen gas is bubbled through a porous carbon electrode. $2 \mathrm{H}_{2}(\mathrm{g})+4 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow 4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+4 \mathrm{e}^{-}$
  • Cathode: Oxygen gas is bubbled through another porous carbon electrode. $\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+4 \mathrm{e}^{-} \longrightarrow 4 \mathrm{OH}^{-}(\mathrm{aq})$
  • Overall Reaction: $2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ The only product is water, which was used as drinking water for the astronauts.

Corrosion

Corrosion is the gradual destruction of metals through chemical or electrochemical reactions with their environment. A common example is the rusting of iron.

Corrosion is essentially an electrochemical process. On the surface of an iron object, some spots act as anodes while others act as cathodes.

• Anode (Oxidation Spot): Iron loses electrons and is oxidized to ferrous ions ($Fe^{2+}$). $2 \mathrm{Fe}(\mathrm{s}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+4 \mathrm{e}^{-} \quad (E^\circ_{\mathrm{Fe}^{2+}/\mathrm{Fe}} = -0.44 \text{ V})$
• Cathode (Reduction Spot): The electrons released travel through the metal to another spot where, in the presence of H$^+$ ions (from dissolved CO$_2$ in water), they reduce oxygen from the air. $\mathrm{O}_{2}(\mathrm{g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad (E^\circ = 1.23 \text{ V})$

The overall reaction has a positive cell potential, indicating it is spontaneous: $2 \mathrm{Fe}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})+4 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad (E^\circ_{\text{cell}} = 1.67 \text{ V})$ The Fe$^{2+}$ ions are further oxidized by atmospheric oxygen to form ferric ions (Fe$^{3+}$), which then combine with water to form hydrated ferric oxide ($\mathrm{Fe}_2\mathrm{O}_3 \cdot x\mathrm{H}_2\mathrm{O}$), which we know as rust.

Prevention of Corrosion:

• Barrier Protection: Covering the surface with paint or chemicals to prevent contact with the atmosphere.
• Galvanization: Coating the surface with a more reactive metal like zinc.
• Sacrificial Protection: Attaching a block of a more reactive metal (like magnesium or zinc) to the iron object. The more reactive metal corrodes instead of the iron, acting as a "sacrificial" anode.

The Hydrogen Economy

The concept of a Hydrogen Economy envisions a future where hydrogen is the primary energy carrier. The combustion of hydrogen produces only water, making it a non-polluting fuel. $2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$ This vision relies on electrochemical principles for both the production of hydrogen (through the electrolysis of water using renewable energy) and its use (in fuel cells to generate electricity).