Electrochemical Cells

Electrochemistry is the study of the relationship between electricity and chemical reactions. It involves converting chemical energy from spontaneous reactions into electrical energy and using electrical energy to drive non-spontaneous chemical reactions.

Electrochemical methods are used for industrial production of many chemicals and metals (e.g., sodium hydroxide, chlorine, aluminum). Batteries and fuel cells, which convert chemical energy to electricity, are important applications.

An electrochemical cell is a device that facilitates the interconversion of chemical and electrical energy.

There are two main types of electrochemical cells:

1. Galvanic cell (or Voltaic cell): Converts the energy released from a spontaneous redox reaction into electrical energy. Example: Daniell cell.
2. Electrolytic cell: Uses electrical energy from an external source to drive a non-spontaneous chemical reaction (electrolysis).

The Daniell cell is a classic example of a galvanic cell. It involves the spontaneous redox reaction:

$\textsf{Zn(s)} + \textsf{Cu}^{2+}\text{(aq)} \to \textsf{Zn}^{2+}\text{(aq)} + \textsf{Cu(s)}$

This reaction converts chemical energy into electrical energy, producing an electric potential (voltage) of 1.1 V under standard conditions.

If an external voltage is applied to a Daniell cell, opposing the cell's potential:

• If external voltage ($\textsf{E}_{\text{ext}}$) < 1.1 V, the cell functions as a galvanic cell (spontaneous reaction produces electricity). Electrons flow from Zn to Cu.
• If $\textsf{E}_{\text{ext}}$ = 1.1 V, the reaction stops, and no current flows (equilibrium).
• If $\textsf{E}_{\text{ext}}$ > 1.1 V, the reaction reverses. The cell now functions as an electrolytic cell, using electrical energy to drive the non-spontaneous reaction ($\textsf{Zn}^{2+}\text{(aq)} + \textsf{Cu(s)} \to \textsf{Zn(s)} + \textsf{Cu}^{2+}\text{(aq)}$). Electrons flow from Cu to Zn (opposite direction).

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Galvanic Cells

A galvanic cell converts the Gibbs energy of a spontaneous redox reaction into electrical work. The overall redox reaction is separated into two half-reactions: oxidation and reduction.

In the Daniell cell reaction ($\textsf{Zn(s)} + \textsf{Cu}^{2+}\text{(aq)} \to \textsf{Zn}^{2+}\text{(aq)} + \textsf{Cu(s)}$):

• Reduction half-reaction: $\textsf{Cu}^{2+}\text{(aq)} + 2\textsf{e}^{-} \to \textsf{Cu(s)}$ (Occurs at the cathode)
• Oxidation half-reaction: $\textsf{Zn(s)} \to \textsf{Zn}^{2+}\text{(aq)} + 2\textsf{e}^{-}$ (Occurs at the anode)

A galvanic cell consists of two half-cells. Each half-cell contains a metallic electrode dipped into an electrolyte solution. The half where oxidation occurs is the anode (negative electrode in a galvanic cell), and the half where reduction occurs is the cathode (positive electrode).

The two half-cells are typically connected externally by a metallic wire (allowing electron flow) and internally by a salt bridge (allowing ion flow to maintain electrical neutrality). Sometimes, both electrodes are in the same electrolyte, eliminating the need for a salt bridge.

At the interface between an electrode and its electrolyte, a potential difference develops due to the tendency of metal atoms to form ions and ions to deposit as metal. This is the electrode potential. When the concentrations of species are unity (1 M), the electrode potential is the standard electrode potential (E°).

In a galvanic cell, the potential difference between the cathode and anode is the cell potential (E$_{cell}$). When no current is drawn, this is the cell's electromotive force (emf).

Conventionally, galvanic cells are represented with the anode on the left and cathode on the right. A single vertical line separates the metal from its electrolyte solution, and a double vertical line represents the salt bridge (or internal connection between electrolytes).

Example: Cell reaction: $\textsf{Cu(s)} + 2\textsf{Ag}^{+}\text{(aq)} \to \textsf{Cu}^{2+}\text{(aq)} + 2\textsf{Ag(s)}$.

Half-reactions: Cathode (reduction): $2\textsf{Ag}^{+}\text{(aq)} + 2\textsf{e}^{-} \to 2\textsf{Ag(s)}$; Anode (oxidation): $\textsf{Cu(s)} \to \textsf{Cu}^{2+}\text{(aq)} + 2\textsf{e}^{-}$.

Cell representation: $\textsf{Cu(s) | Cu}^{2+}\text{(aq)} || \textsf{Ag}^{+}\text{(aq) | Ag(s)}$.

Cell emf: $\textsf{E}_{\text{cell}} = \textsf{E}_{\text{right}} - \textsf{E}_{\text{left}} = \textsf{E}^{0}_{\text{Ag}^{+}|\text{Ag}} - \textsf{E}^{0}_{\text{Cu}^{2+}|\text{Cu}}$.

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Measurement Of Electrode Potential

Individual half-cell potentials cannot be measured directly. Only the difference between two half-cell potentials (the cell emf) can be measured.

To determine the standard electrode potential of a half-cell, it is coupled with a reference electrode whose potential is arbitrarily assigned a value. The Standard Hydrogen Electrode (SHE) is the universally accepted reference electrode, assigned a zero potential at all temperatures.

SHE reaction: $\textsf{H}^{+}\text{(aq)} + \textsf{e}^{-} \to 1/2 \textsf{H}_2\text{(g)}$.

SHE construction: A platinum electrode coated with platinum black, immersed in an acidic solution with 1 M $\textsf{H}^{+}$ concentration, with pure hydrogen gas at 1 bar pressure bubbled through it.

When a half-cell with standard conditions (1 M concentration of ions, 1 bar gas pressure) is connected as the cathode to an SHE anode, the measured cell emf is the standard reduction potential of that half-cell ($\textsf{E}^{0}_{\text{cell}} = \textsf{E}^{0}_{\text{right}} - \textsf{E}^{0}_{\text{left}} = \textsf{E}^{0}_{\text{right}} - 0 = \textsf{E}^{0}_{\text{right}}$).

Example: $\textsf{Pt(s)} | \textsf{H}_2\text{(g, 1 bar)} | \textsf{H}^{+}\text{(aq, 1 M)} || \textsf{Cu}^{2+}\text{(aq, 1 M)} | \textsf{Cu(s)}$. Measured emf is +0.34 V. So, $\textsf{E}^{0}_{\text{Cu}^{2+}|\text{Cu}} = +0.34$ V.

Example: $\textsf{Pt(s)} | \textsf{H}_2\text{(g, 1 bar)} | \textsf{H}^{+}\text{(aq, 1 M)} || \textsf{Zn}^{2+}\text{(aq, 1 M)} | \textsf{Zn(s)}$. Measured emf is -0.76 V. So, $\textsf{E}^{0}_{\text{Zn}^{2+}|\text{Zn}} = -0.76$ V.

Standard electrode potentials (standard reduction potentials) indicate the relative tendency of a species to be reduced. Higher positive $\textsf{E}^0$ indicates a stronger tendency for reduction (and weaker reducing agent). More negative $\textsf{E}^0$ indicates a weaker tendency for reduction (and stronger reducing agent).

For the Daniell cell ($\textsf{Zn | Zn}^{2+} || \textsf{Cu}^{2+} | \textsf{Cu}$), $\textsf{E}^{0}_{\text{cell}} = \textsf{E}^{0}_{\text{Cu}^{2+}|\text{Cu}} - \textsf{E}^{0}_{\text{Zn}^{2+}|\text{Zn}} = 0.34 \text{ V} - (-0.76 \text{ V}) = 1.10 \text{ V}$.

Inert electrodes (like Pt or Au) are used when the reaction involves ions or gases, providing a surface for electron transfer without participating chemically (e.g., Pt in $\textsf{H}_2$ electrode, $\textsf{Br}_2$ electrode).

Standard electrode potentials allow us to predict the feasibility of redox reactions and the relative strengths of oxidising and reducing agents (strongest oxidising agent has highest $\textsf{E}^0$, strongest reducing agent has lowest $\textsf{E}^0$).

Reaction (Oxidised form + ne$^-$ $\to$ Reduced form)	E$^\circ$/V
F$_2$(g) + 2e$^-$ $\to$ 2F$^-$	2.87
Co$^{3+}$ + e$^-$ $\to$ Co$^{2+}$	1.81
H$_2$O$_2$ + 2H$^+$ + 2e$^-$ $\to$ 2H$_2$O	1.78
MnO$_4^-$ + 8H$^+$ + 5e$^-$ $\to$ Mn$^{2+}$ + 4H$_2$O	1.51
Au$^{3+}$ + 3e$^-$ $\to$ Au(s)	1.40
Cl$_2$(g) + 2e$^-$ $\to$ 2Cl$^-$	1.36
Cr$_2$O$_7^{2-}$ + 14H$^+$ + 6e$^-$ $\to$ 2Cr$^{3+}$ + 7H$_2$O	1.33
O$_2$(g) + 4H$^+$ + 4e$^-$ $\to$ 2H$_2$O	1.23
MnO$_2$(s) + 4H$^+$ + 2e$^-$ $\to$ Mn$^{2+}$ + 2H$_2$O	1.23
Br$_2$ + 2e$^-$ $\to$ 2Br$^-$	1.09
NO$_3^-$ + 4H$^+$ + 3e$^-$ $\to$ NO(g) + 2H$_2$O	0.97
2Hg$^{2+}$ + 2e$^-$ $\to$ Hg$_2^{2+}$	0.92
Ag$^+$ + e$^-$ $\to$ Ag(s)	0.80
Fe$^{3+}$ + e$^-$ $\to$ Fe$^{2+}$	0.77
O$_2$(g) + 2H$^+$ + 2e$^-$ $\to$ H$_2$O$_2$	0.68
I$_2$ + 2e$^-$ $\to$ 2I$^-$	0.54
Cu$^+$ + e$^-$ $\to$ Cu(s)	0.52
Cu$^{2+}$ + 2e$^-$ $\to$ Cu(s)	0.34
AgCl(s) + e$^-$ $\to$ Ag(s) + Cl$^-$	0.22
AgBr(s) + e$^-$ $\to$ Ag(s) + Br$^-$	0.10
2H$^+$ + 2e$^-$ $\to$ H$_2$(g)	0.00
Pb$^{2+}$ + 2e$^-$ $\to$ Pb(s)	-0.13
Sn$^{2+}$ + 2e$^-$ $\to$ Sn(s)	-0.14
Ni$^{2+}$ + 2e$^-$ $\to$ Ni(s)	-0.25
Fe$^{2+}$ + 2e$^-$ $\to$ Fe(s)	-0.44
Cr$^{3+}$ + 3e$^-$ $\to$ Cr(s)	-0.74
Zn$^{2+}$ + 2e$^-$ $\to$ Zn(s)	-0.76
2H$_2$O + 2e$^-$ $\to$ H$_2$(g) + 2OH$^-$ (aq)	-0.83
Al$^{3+}$ + 3e$^-$ $\to$ Al(s)	-1.66
Mg$^{2+}$ + 2e$^-$ $\to$ Mg(s)	-2.36
Na$^+$ + e$^-$ $\to$ Na(s)	-2.71
Ca$^{2+}$ + 2e$^-$ $\to$ Ca(s)	-2.87
K$^+$ + e$^-$ $\to$ K(s)	-2.93
Li$^+$ + e$^-$ $\to$ Li(s)	-3.05

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Nernst Equation

The standard electrode potentials ($\textsf{E}^0$) are defined for unit concentrations (1 M) and unit pressures (1 bar). However, in real electrochemical cells, concentrations and pressures are often different from standard conditions. The Nernst equation relates the electrode potential or cell potential at any concentration to its standard potential.

For a general electrode reaction: $\textsf{M}^{\text{n}+}\text{(aq)} + \textsf{ne}^{-} \to \textsf{M(s)}$

The electrode potential (E) at any concentration of $\textsf{M}^{\text{n}+}$ is given by:

$\textsf{E}_{\text{M}^{\text{n}+}|\text{M}} = \textsf{E}^{0}_{\text{M}^{\text{n}+}|\text{M}} - \frac{\textsf{RT}}{\textsf{nF}} \ln \frac{\text{[M(s)]}}{[\textsf{M}^{\text{n}+}\text{(aq)}]}$

Since the concentration/activity of a pure solid is taken as unity ([M(s)] = 1):

$\textsf{E}_{\text{M}^{\text{n}+}|\text{M}} = \textsf{E}^{0}_{\text{M}^{\text{n}+}|\text{M}} - \frac{\textsf{RT}}{\textsf{nF}} \ln \frac{1}{[\textsf{M}^{\text{n}+}\text{(aq)}]}$

Where R is the gas constant (8.314 J K$^{-1}$ mol$^{-1}$), T is temperature in Kelvin, n is the number of electrons involved in the half-reaction, F is Faraday's constant (96487 C mol$^{-1}$), and $[\textsf{M}^{\text{n}+}\text{(aq)}]$ is the molar concentration of the ion.

At 298 K (25$^\circ$C), converting the natural logarithm ($\ln$) to base 10 logarithm ($\log_{10}$, where $\ln x = 2.303 \log_{10} x$) and substituting values for R, T, and F ($\frac{2.303 \times 8.314 \times 298}{96487} \approx 0.0591$):

$\textsf{E}_{\text{M}^{\text{n}+}|\text{M}} = \textsf{E}^{0}_{\text{M}^{\text{n}+}|\text{M}} - \frac{0.0591}{\textsf{n}} \log \frac{1}{[\textsf{M}^{\text{n}+}\text{(aq)}]}$

For a complete cell reaction, E$_{cell}$ is the difference between the reduction potentials of the cathode and anode ($\textsf{E}_{\text{cell}} = \textsf{E}_{\text{cathode}} - \textsf{E}_{\text{anode}}$). Applying the Nernst equation to each half-cell potential gives the cell potential:

For a general reaction: $\textsf{aA} + \textsf{bB} \to \textsf{cC} + \textsf{dD}$ (where A and B are reactants, C and D are products, and n electrons are transferred)

The Nernst equation for the cell potential is:

$\textsf{E}_{\text{cell}} = \textsf{E}^{0}_{\text{cell}} - \frac{\textsf{RT}}{\textsf{nF}} \ln \text{Q} = \textsf{E}^{0}_{\text{cell}} - \frac{\textsf{RT}}{\textsf{nF}} \ln \frac{[\textsf{C}]^{\text{c}}[\textsf{D}]^{\text{d}}}{[\textsf{A}]^{\text{a}}[\textsf{B}]^{\text{b}}}$

Where Q is the reaction quotient, using molar concentrations for aqueous species and partial pressures for gases. Activities of pure solids/liquids are taken as unity.

At 298 K:

$\textsf{E}_{\text{cell}} = \textsf{E}^{0}_{\text{cell}} - \frac{0.0591}{\textsf{n}} \log \frac{[\textsf{C}]^{\text{c}}[\textsf{D}]^{\text{d}}}{[\textsf{A}]^{\text{a}}[\textsf{B}]^{\text{b}}}$

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Equilibrium Constant From Nernst Equation

As a galvanic cell operates, the concentrations of reactants decrease, and product concentrations increase. This causes the cell potential (E$_{cell}$) to decrease over time. Eventually, the reaction reaches equilibrium, at which point $\textsf{E}_{\text{cell}} = 0$ V, and the reaction quotient Q becomes equal to the equilibrium constant ($\textsf{K}_\text{c}$).

At equilibrium, the Nernst equation for the cell becomes:

$\textsf{E}_{\text{cell}} = 0 = \textsf{E}^{0}_{\text{cell}} - \frac{\textsf{RT}}{\textsf{nF}} \ln \text{K}_\text{c}$

Rearranging:

$\textsf{E}^{0}_{\text{cell}} = \frac{\textsf{RT}}{\textsf{nF}} \ln \text{K}_\text{c}$

Or, converting to log$_{10}$ at 298 K:

$\textsf{E}^{0}_{\text{cell}} = \frac{0.0591}{\textsf{n}} \log \textsf{K}_\text{c}$

This equation provides a relationship between the standard cell potential ($\textsf{E}^{0}_{\text{cell}}$) and the equilibrium constant ($\textsf{K}_\text{c}$) of the cell reaction. $\textsf{K}_\text{c}$ can be calculated from a measured $\textsf{E}^{0}_{\text{cell}}$.

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Electrochemical Cell And Gibbs Energy Of The Reaction

The electrical work done by a galvanic cell is equal to the decrease in its Gibbs energy ($\Delta_\text{r}$G). For maximum work, the process must be reversible. If E$_{cell}$ is the cell potential and nF is the total charge transferred (n moles of electrons $\times$ Faraday constant), then:

Electrical work = Charge $\times$ Potential = nF $\times$ E$_{cell}$

$\Delta_\text{r}$G = - (Electrical work) = - nF E$_{cell}$

This equation relates Gibbs energy change ($\Delta_\text{r}$G) to the cell potential (E$_{cell}$). $\Delta_\text{r}$G is an extensive property, while E$_{cell}$ is intensive. The value of $\Delta_\text{r}$G depends on the number of electrons transferred (n) for the specific reaction as written.

Under standard conditions (all reactants and products at unit activity/concentration), E$_{cell}$ becomes $\textsf{E}^{0}_{\text{cell}}$, and $\Delta_\text{r}$G becomes $\Delta_\text{r}$G$^\circ$ (standard Gibbs energy change):

$\Delta_\text{r}$G$^\circ$ = - nF $\textsf{E}^{0}_{\text{cell}}$

This equation allows calculating $\Delta_\text{r}$G$^\circ$ from standard cell potentials. $\Delta_\text{r}$G$^\circ$ is also related to the equilibrium constant ($\textsf{K}_\text{c}$) by $\Delta_\text{r}$G$^\circ$ = - RT $\ln$ $\textsf{K}_\text{c}$. Combining these relationships, we get $\textsf{E}^{0}_{\text{cell}} = \frac{\textsf{RT}}{\textsf{nF}} \ln \textsf{K}_\text{c}$, which is the relationship derived from the Nernst equation at equilibrium.

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Conductance Of Electrolytic Solutions

Electrical resistance (R) of a conductor is measured in ohm ($\Omega$). It's proportional to length (l) and inversely proportional to area (A): $\textsf{R} = \rho \frac{l}{A}$, where $\rho$ is resistivity (specific resistance) with units ohm-metre ($\Omega$ m).

Conductance (G) is the inverse of resistance: $\textsf{G} = \frac{1}{R} = \frac{1}{\rho} \frac{A}{l} = \kappa \frac{A}{l}$. Measured in siemens (S), equal to ohm$^{-1}$ ($\Omega^{-1}$).

Conductivity (specific conductance) ($\kappa$) is the inverse of resistivity: $\kappa = \frac{1}{\rho}$. Units are S m$^{-1}$ or S cm$^{-1}$. Conductivity is the conductance of a material 1 m long with 1 m$^2$ area.

Materials are classified by conductivity magnitude: conductors (high $\kappa$, e.g., metals), insulators (very low $\kappa$, e.g., glass), semiconductors (intermediate $\kappa$, e.g., Si, Ge), superconductors (infinite $\kappa$).

Electronic (metallic) conductance: Due to electron movement in metals. Depends on metal nature, number of valence electrons, and temperature (decreases with T).

Electrolytic (ionic) conductance: Due to ion movement in solutions/melts. Pure water has very low conductivity due to self-ionisation. Dissolving electrolytes increases conductivity by adding more ions. Ionic conductance depends on:

• Nature of electrolyte.
• Size and solvation of ions.
• Solvent nature and viscosity.
• Concentration of electrolyte.
• Temperature (increases with T as ion mobility increases).

Electrolysis (passage of DC) changes the composition of electrolytic solutions.

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Measurement Of The Conductivity Of Ionic Solutions

Measuring the resistance of ionic solutions is tricky: DC causes electrolysis, and solutions can't be connected like wires. Solutions:

• Use an alternating current (AC) source to prevent electrolysis.
• Use a conductivity cell (Fig. 3.4). Consists of two platinized platinum electrodes (area A, distance l) confining the solution. Resistance is $\textsf{R} = \rho \frac{l}{A} = \frac{1}{\kappa} \frac{l}{A}$.

The term $\frac{l}{A}$ is the cell constant (G*). Units are m$^{-1}$ or cm$^{-1}$. It's determined by measuring resistance of a solution with known conductivity (e.g., KCl solution): $\textsf{G}^* = \textsf{R} \kappa$. Once $\textsf{G}^*$ is known, conductivity of any solution can be found from its measured resistance in the same cell: $\kappa = \frac{\textsf{G}^*}{\textsf{R}}$.

Resistance is measured using a Wheatstone bridge setup with an AC source (oscillator) and detector.

Unknown resistance R$_2 = \frac{R_1 R_4}{R_3}$. Conductivity $\kappa = \frac{G^*}{R_2}$. Conductivity meters directly display conductivity/resistance.

Different electrolytes have different conductivity. To compare conductivity of solutions with different concentrations, molar conductivity ($\Lambda_\text{m}$) is used. It's defined as the conductivity of the solution containing one mole of electrolyte.

$\Lambda_\text{m} = \frac{\kappa}{\text{c}}$

Where $\kappa$ is conductivity and c is molar concentration. If $\kappa$ is in S m$^{-1}$ and c in mol m$^{-3}$, $\Lambda_\text{m}$ is in S m$^2$ mol$^{-1}$. If $\kappa$ is in S cm$^{-1}$ and c in mol L$^{-1}$, $\Lambda_\text{m} = \frac{\kappa \times 1000}{\text{c}}$ in S cm$^2$ mol$^{-1}$. $1 \text{ S m}^2\text{mol}^{-1} = 10^4 \text{ S cm}^2\text{mol}^{-1}$.

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Variation Of Conductivity And Molar Conductivity With Concentration

Conductivity ($\kappa$):

Conductivity is the conductance of a unit volume of solution. On dilution (decreasing concentration), the number of ions per unit volume decreases. Therefore, conductivity ($\kappa$) always decreases with a decrease in concentration for both strong and weak electrolytes.

Molar conductivity ($\Lambda_\text{m}$):

Molar conductivity is the conductance of the volume of solution containing one mole of electrolyte. On dilution, the total volume (V) containing one mole of electrolyte increases. $\Lambda_\text{m} = \kappa \text{V}$. Although $\kappa$ decreases with dilution, the increase in volume V (containing one mole) compensates for this decrease and leads to an overall increase in $\Lambda_\text{m}$.

Molar conductivity ($\Lambda_\text{m}$) always increases with a decrease in concentration (dilution).

As concentration approaches zero (infinite dilution), molar conductivity approaches a maximum value called limiting molar conductivity ($\Lambda^0_\text{m}$).

The variation of $\Lambda_\text{m}$ with concentration differs for strong and weak electrolytes (Fig. 3.6):

• Strong Electrolytes: $\Lambda_\text{m}$ increases slowly with dilution. The increase is mainly due to increased ionic mobility as interionic attractions decrease at lower concentrations. This variation can be described by the Debye-Hückel-Onsager equation: $\Lambda_\text{m} = \Lambda^0_\text{m} - \text{A} \text{c}^{1/2}$. A plot of $\Lambda_\text{m}$ vs $\text{c}^{1/2}$ is linear, allowing $\Lambda^0_\text{m}$ to be determined by extrapolation to zero concentration.

Example 3.6 The molar conductivity of KCl solutions at different concentrations at 298 K are given below:

c/mol L$^{-1}$ $\Lambda_\text{m}$/S cm$^2$ mol$^{-1}$

0.000198 148.61

0.000309 148.29

0.000521 147.81

0.000989 147.09

Show that a plot between $\Lambda_\text{m}$ and c$^{1/2}$ is a straight line. Determine the values of $\Lambda^0_\text{m}$ and A for KCl.

Answer:

Calculate c$^{1/2}$ for each concentration:

c/mol L$^{-1}$	c$^{1/2}$/(mol L$^{-1}$)$^{1/2}$	$\Lambda_\text{m}$/S cm$^2$ mol$^{-1}$
0.000198	0.01407	148.61
0.000309	0.01758	148.29
0.000521	0.02283	147.81
0.000989	0.03145	147.09

Plot $\Lambda_\text{m}$ on the y-axis and c$^{1/2}$ on the x-axis. The points fall approximately on a straight line, confirming the relationship for a strong electrolyte.

Determine $\Lambda^0_\text{m}$ (y-intercept) and A (-slope) from the graph or by using two points on the line.

Using points (0.01407, 148.61) and (0.03145, 147.09):

Slope = $\frac{147.09 - 148.61}{0.03145 - 0.01407} = \frac{-1.52}{0.01738} \approx -87.46$.

A = -Slope $\approx 87.46$ S cm$^2$ mol$^{-1}$ (mol/L)$^{-1/2}$.

Equation of line: $\Lambda_\text{m} - 148.61 = -87.46 (c^{1/2} - 0.01407)$.

$\Lambda_\text{m} = -87.46 c^{1/2} + 87.46 \times 0.01407 + 148.61$

$\Lambda_\text{m} = -87.46 c^{1/2} + 1.230 + 148.61 = -87.46 c^{1/2} + 149.84$.

Extrapolating to c$^{1/2} = 0$, the y-intercept $\Lambda^0_\text{m} \approx 149.84$ S cm$^2$ mol$^{-1}$. (Textbook answer 150.0 S cm$^2$ mol$^{-1}$ - very close).

• Weak Electrolytes: $\Lambda_\text{m}$ increases steeply with dilution, especially at low concentrations. This is because weak electrolytes dissociate to a very small extent at higher concentrations. Dilution increases the degree of dissociation, increasing the number of ions in solution and thus $\Lambda_\text{m}$. $\Lambda^0_\text{m}$ for weak electrolytes cannot be obtained by extrapolation of the $\Lambda_\text{m}$ vs $\text{c}^{1/2}$ plot, as the graph is not linear and becomes very steep near zero concentration.

Kohlrausch's Law of Independent Migration of Ions: For weak electrolytes, $\Lambda^0_\text{m}$ is determined using this law. It states that limiting molar conductivity ($\Lambda^0_\text{m}$) of an electrolyte is the sum of the limiting ionic conductivities of its constituent cation and anion, each multiplied by the number of ions produced per formula unit of the electrolyte.

If an electrolyte dissociates into n$^+$ cations with limiting ionic conductivity $\lambda^0_+$ and n$^-$ anions with limiting ionic conductivity $\lambda^0_-$, then:

$\Lambda^0_\text{m} = \text{n}^+ \lambda^0_+ + \text{n}^- \lambda^0_-$

This law is applicable at infinite dilution, where interionic interactions are negligible, and ions move independently.

Values of $\lambda^0$ for some ions at 298 K are given in Table 3.4.

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Applications of Kohlrausch law:

• Calculating $\Lambda^0_\text{m}$ for any electrolyte by summing $\lambda^0$ of individual ions.
• Calculating $\Lambda^0_\text{m}$ for weak electrolytes using $\Lambda^0_\text{m}$ values of strong electrolytes (as shown in Example 3.8).
• Determining the degree of dissociation ($\alpha$) of a weak electrolyte at a given concentration c using $\alpha = \frac{\Lambda_\text{m}(\text{c})}{\Lambda^0_\text{m}}$.
• Calculating the dissociation constant ($\textsf{K}_\text{a}$) of a weak electrolyte using its degree of dissociation and concentration (e.g., for a weak acid HA $\rightleftharpoons$ H$^+$ + A$^-$, $\textsf{K}_\text{a} = \frac{[\textsf{H}^+][\textsf{A}^-]}{[\textsf{HA}]} = \frac{\text{c}\alpha^2}{1-\alpha}$ for dilute solutions, where concentrations are approximated by molarity).

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Electrolytic Cells And Electrolysis

An electrolytic cell uses an external voltage source to drive a non-spontaneous chemical reaction. It converts electrical energy into chemical energy. Electrolysis is used for chemical synthesis and production of elements (e.g., Na, Mg, Al from fused salts) where chemical reduction is difficult.

Example: Electrolysis of aqueous copper sulfate ($\textsf{CuSO}_4$) with copper strips (reactive electrodes).

• At the cathode (negative electrode), $\textsf{Cu}^{2+}$ ions from the solution are reduced: $\textsf{Cu}^{2+}\text{(aq)} + 2\textsf{e}^{-} \to \textsf{Cu(s)}$. Copper metal deposits on the cathode.
• At the anode (positive electrode), copper metal from the electrode is oxidised: $\textsf{Cu(s)} \to \textsf{Cu}^{2+}\text{(aq)} + 2\textsf{e}^{-}$. Copper dissolves from the anode.

This process is used to purify impure copper: impure copper is the anode, pure copper is the cathode.

Electrolysis of molten $\textsf{NaCl}$ yields $\textsf{Na}$ metal (at cathode) and $\textsf{Cl}_2$ gas (at anode). Electrolysis of aqueous $\textsf{NaCl}$ yields $\textsf{NaOH}$, $\textsf{H}_2$ gas (at cathode), and $\textsf{Cl}_2$ gas (at anode).

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Quantitative Aspects Of Electrolysis

Michael Faraday formulated two laws describing the quantitative relationships in electrolysis (amount of substance produced/consumed and quantity of electricity passed).

• Faraday's First Law: The mass of a substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte. Mass $\propto$ Charge (Q).
• Faraday's Second Law: The masses of different substances liberated by the same quantity of electricity passing through different electrolytic solutions connected in series are proportional to their chemical equivalent weights (equivalent weight = Molar mass / Number of electrons involved in the half-reaction).

Quantity of electricity (Q) passed is the product of current (I) and time (t): $\textsf{Q} = \textsf{It}$ (Q in Coulombs (C), I in Amperes (A), t in seconds (s)).

One Faraday (F) is the charge on one mole of electrons ($1 \text{ mol e}^{-}$). $1 \text{ F} = 96487 \text{ C mol}^{-1} \approx 96500 \text{ C mol}^{-1}$.

The charge required for electrolysis depends on the number of electrons in the half-reaction. For $\textsf{Ag}^{+} + \textsf{e}^{-} \to \textsf{Ag(s)}$, 1 mol $\textsf{e}^{-}$ (1F) is needed per mole of $\textsf{Ag}$. For $\textsf{Mg}^{2+} + 2\textsf{e}^{-} \to \textsf{Mg(s)}$, 2 mol $\textsf{e}^{-}$ (2F) are needed per mole of $\textsf{Mg}$.

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Products Of Electrolysis

The products formed during electrolysis depend on several factors:

• Nature of the substance being electrolysed (molten salt vs aqueous solution).
• Type of electrodes (inert vs reactive).
• Relative standard electrode potentials of the competing reactions at each electrode.
• Overpotential (extra voltage needed for some reactions to occur at a reasonable rate).

In an aqueous solution, besides the ions from the electrolyte, water molecules are also present and can be oxidised or reduced. Competing reactions occur at cathode (reduction) and anode (oxidation).

Example: Electrolysis of aqueous $\textsf{NaCl}$ solution with inert electrodes (Pt):

• At cathode (reduction): Competition between $\textsf{Na}^{+}$ reduction and $\textsf{H}_2\textsf{O}$ reduction ($\textsf{H}^{+}$ reduction). $\textsf{H}^{+}$/$\textsf{H}_2$ couple has a higher reduction potential (0.00 V) than $\textsf{Na}^{+}$/$\textsf{Na}$ (-2.71 V). So, $\textsf{H}^{+}$ is preferentially reduced: $2\textsf{H}^{+}\text{(aq)} + 2\textsf{e}^{-} \to \textsf{H}_2\text{(g)}$. Since $\textsf{H}^{+}$ is supplied by water dissociation, net cathode reaction is $2\textsf{H}_2\textsf{O(l)} + 2\textsf{e}^{-} \to \textsf{H}_2\textsf{(g)} + 2\textsf{OH}^{-}\text{(aq)}$. Product: $\textsf{H}_2$ gas.
• At anode (oxidation): Competition between $\textsf{Cl}^{-}$ oxidation and $\textsf{H}_2\textsf{O}$ oxidation. $\textsf{Cl}^{-}$/$\textsf{Cl}_2$ has $\textsf{E}^0 = +1.36$ V. $\textsf{H}_2\textsf{O}$/$\textsf{O}_2$ has $\textsf{E}^0 = +1.23$ V. Thermodynamically, water oxidation should be preferred. However, due to overpotential for oxygen formation, $\textsf{Cl}^{-}$ oxidation is often preferred, especially at higher concentrations of $\textsf{Cl}^{-}$: $2\textsf{Cl}^{-}\text{(aq)} \to \textsf{Cl}_2\text{(g)} + 2\textsf{e}^{-}$. Product: $\textsf{Cl}_2$ gas.

Net reaction for aqueous $\textsf{NaCl}$ electrolysis: $2\textsf{NaCl(aq)} + 2\textsf{H}_2\textsf{O(l)} \to 2\textsf{NaOH(aq)} + \textsf{H}_2\textsf{(g)} + \textsf{Cl}_2\text{(g)}$.

Products: $\textsf{NaOH}$ (in solution), $\textsf{H}_2$ gas, $\textsf{Cl}_2$ gas.

For concentrated sulphuric acid electrolysis, at the anode, $\textsf{S}_2\textsf{O}_8^{2-}$ is formed instead of $\textsf{O}_2$ due to overpotential.

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Batteries

A battery (often multiple cells in series) is a source of electrical energy based on galvanic cells. Desirable battery characteristics include being light, compact, and having stable voltage during discharge. Batteries are classified into primary and secondary types.

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Primary Batteries

In primary batteries, the electrochemical reaction occurs only once. They are non-rechargeable and become dead after use.

• Dry cell (Leclanché cell): Common in transistors, clocks. Anode: Zinc container. Cathode: Carbon (graphite) rod surrounded by $\textsf{MnO}_2$ and carbon powder. Electrolyte: Moist paste of $\textsf{NH}_4\text{Cl}$ and $\textsf{ZnCl}_2$. Potential $\approx$ 1.5 V.
• Mercury cell: For low-current devices (hearing aids, watches). Anode: Zinc-mercury amalgam. Cathode: Paste of HgO and carbon. Electrolyte: Paste of KOH and ZnO. Potential $\approx$ 1.35 V, stable during life as overall reaction doesn't involve concentration changes of dissolved ions.

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Secondary Batteries

Secondary batteries are rechargeable. After discharge, current can be passed in the reverse direction to regenerate the reactants, allowing reuse for many cycles.

• Lead storage battery: Common in automobiles, inverters. Anode: Lead. Cathode: Grid of lead packed with $\textsf{PbO}_2$. Electrolyte: 38% $\textsf{H}_2\textsf{SO}_4$ solution.
• Nickel-cadmium cell: Rechargeable, longer life than lead storage but more expensive.

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Fuel Cells

Fuel cells are galvanic cells designed to convert the energy of combustion of fuels (like $\textsf{H}_2$, $\textsf{CH}_4$, $\textsf{CH}_3\text{OH}$) directly into electrical energy. They are highly efficient compared to thermal power plants.

Reactants are fed continuously to the electrodes, and products are removed. One common fuel cell uses $\textsf{H}_2$ and $\textsf{O}_2$ reactants, producing water.

In a typical $\textsf{H}_2$-$\textsf{O}_2$ fuel cell:

• $\textsf{H}_2$ and $\textsf{O}_2$ are bubbled through porous carbon electrodes.
• Electrolyte is often concentrated aqueous $\textsf{NaOH}$ solution.
• Catalysts (finely divided Pt or Pd) are incorporated into electrodes.

Electrode reactions:

• Cathode: $\textsf{O}_2\text{(g)} + 2\textsf{H}_2\textsf{O(l)} + 4\textsf{e}^{-} \to 4\textsf{OH}^{-}\text{(aq)}$
• Anode: $2\textsf{H}_2\text{(g)} + 4\textsf{OH}^{-}\text{(aq)} \to 4\textsf{H}_2\textsf{O(l)} + 4\textsf{e}^{-}$

Overall reaction: $2\textsf{H}_2\text{(g)} + \textsf{O}_2\text{(g)} \to 2\textsf{H}_2\textsf{O(l)}$.

The cell operates continuously as long as fuel ($\textsf{H}_2$) and oxidant ($\textsf{O}_2$) are supplied. Efficiency is high ($\sim 70\%$). Fuel cells are pollution-free (produce water). They are being explored for vehicles and other applications, driving research into better electrode materials and catalysts.

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Corrosion

Corrosion is the slow degradation of metallic surfaces due to reaction with their environment, typically forming oxides or other salts. Examples: rusting of iron, tarnishing of silver, green coating on copper/bronze. Corrosion causes significant economic damage to structures and objects made of metal, especially iron.

Corrosion is essentially an electrochemical process.

Rusting of iron:

• At certain spots on the iron surface, oxidation occurs: $\textsf{Fe(s)} \to \textsf{Fe}^{2+}\text{(aq)} + 2\textsf{e}^{-}$. This spot acts as the anode.
• Electrons move through the metal to another spot, which acts as the cathode. Here, oxygen is reduced in the presence of $\textsf{H}^{+}$ ions (from $\textsf{H}_2\textsf{CO}_3$ formed from $\textsf{CO}_2$ in water or from acidic oxides in the atmosphere): $\textsf{O}_2\text{(g)} + 4\textsf{H}^{+}\text{(aq)} + 4\textsf{e}^{-} \to 2\textsf{H}_2\textsf{O(l)}$.

Overall electrochemical reaction: $2\textsf{Fe(s)} + \textsf{O}_2\text{(g)} + 4\textsf{H}^{+}\text{(aq)} \to 2\textsf{Fe}^{2+}\text{(aq)} + 2\textsf{H}_2\textsf{O(l)}$.

The $\textsf{Fe}^{2+}$ ions are further oxidised by atmospheric oxygen to $\textsf{Fe}^{3+}$ ions, forming rust (hydrated ferric oxide, $\textsf{Fe}_2\textsf{O}_3$.x$\textsf{H}_2\textsf{O}$).

Prevention of corrosion is crucial for safety and economic reasons.

Methods to prevent corrosion:

• Preventing the metal surface from contacting the atmosphere (e.g., painting, coating with chemicals like bisphenol).
• Covering the surface with other metals:
  • Using an inert metal coating (e.g., tin plating) - provides barrier protection.
  • Using a more reactive metal coating (e.g., zinc coating in galvanisation) - the more reactive metal corrodes preferentially, protecting the underlying metal. This is sacrificial protection.
• Electrochemical methods: Providing a sacrificial electrode of a more reactive metal (like Mg or Zn) connected to the object to be protected. The sacrificial electrode corrodes instead of the object.

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