Electrochemistry
Complete Revision Guide for CBSE Board Exams
1. Introduction to Electrochemistry
Electrochemistry is the branch of physical chemistry that deals with the relationship between electrical energy and chemical changes. It essentially studies the production of electricity from the energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous chemical transformations.
These reactions involve the transfer of electrons and are known as Oxidation-Reduction (Redox) reactions. The subject is broadly divided into two major parts:
- Galvanic Cells (Voltaic Cells): Devices that convert chemical energy of a spontaneous redox reaction into electrical energy.
- Electrolytic Cells: Devices that use electrical energy to drive a non-spontaneous chemical reaction (Electrolysis).
2. Master Formula Cheat Sheet
A. Electrochemical Cells & Nernst Equation
- Standard EMF of a Cell ($E^\circ_{cell}$):
$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Right} - E^\circ_{Left}$$
- Nernst Equation (General form at 298 K): For a cell reaction $aA + bB \rightarrow cC + dD$:
$$E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log_{10} \frac{[C]^c [D]^d}{[A]^a [B]^b}$$(Where $n$ is the number of moles of electrons transferred).
- Gibbs Free Energy and EMF:
$$\Delta G^\circ = -n F E^\circ_{cell} \quad \text{and} \quad \Delta G = -n F E_{cell}$$(Where $F = 96487 \text{ C mol}^{-1} \approx 96500 \text{ C mol}^{-1}$).
- Equilibrium Constant ($K_c$) from Nernst Equation: At equilibrium, $E_{cell} = 0$.
$$E^\circ_{cell} = \frac{0.0591}{n} \log_{10} K_c$$
B. Conductance of Electrolytic Solutions
- Resistance ($R$) and Resistivity ($\rho$): $$R = \rho \frac{l}{A}$$
- Conductance ($G$) and Specific Conductivity ($\kappa$): $$G = \frac{1}{R} \quad \text{and} \quad \kappa = \frac{1}{\rho} = G \times \frac{l}{A} = G \times G^*$$
- Cell Constant ($G^*$): $$G^* = \frac{l}{A} = R \times \kappa$$
- Molar Conductivity ($\Lambda_m$): $$\Lambda_m = \frac{\kappa \times 1000}{M}$$ (If $\kappa$ is in $S \text{ cm}^{-1}$ and $M$ is in $\text{mol L}^{-1}$).
- Kohlrausch's Law: $$\Lambda^\circ_m = \nu_+ \lambda^\circ_+ + \nu_- \lambda^\circ_-$$
- Degree of Dissociation ($\alpha$): $$\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}$$
- Dissociation Constant ($K_a$): $$K_a = \frac{c \alpha^2}{1 - \alpha}$$
C. Faraday's Laws of Electrolysis
- Faraday's First Law: Mass deposited ($m$) is directly proportional to charge ($Q$). $$m = Z \times Q = Z \times I \times t$$ (Where $Z$ is the electrochemical equivalent, $I$ is current, $t$ is time in seconds).
- Calculation of Z: $$Z = \frac{\text{Equivalent Weight}}{F} = \frac{\text{Molar Mass}}{n \times 96500}$$
- Faraday's Second Law: $$\frac{m_1}{m_2} = \frac{E_1}{E_2}$$
3. Exhaustive Theoretical Concepts
3.1 Galvanic Cells (Daniell Cell)
A Galvanic cell converts chemical energy liberated during a spontaneous redox reaction into electrical energy. The classic example is the Daniell cell, which relies on the reaction between Zinc and Copper sulfate.
Cell Reaction: $$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$$
This reaction is split into two half-cells:
- Anode (Oxidation Half-Cell): Zinc rod dipped in $ZnSO_4$ solution. It loses electrons and dissolves.
$$Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$$ (Polarity: Negative) - Cathode (Reduction Half-Cell): Copper rod dipped in $CuSO_4$ solution. $Cu^{2+}$ ions gain electrons and deposit as solid copper.
$$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$$ (Polarity: Positive)
The Salt Bridge and Its Functions
A salt bridge is a U-shaped tube containing an inert electrolyte (like $KCl$, $KNO_3$, $NH_4NO_3$) in a gel-like substance such as agar-agar. It is an absolute necessity for the sustained operation of a galvanic cell.
Functions:- It completes the electrical circuit by allowing ions to flow from one half-cell to the other without mixing the two solutions directly.
- It maintains the electrical neutrality of the solutions in the two half-cells. As $Zn^{2+}$ accumulates at the anode, $Cl^-$ ions from the salt bridge migrate there to neutralize the positive charge. Similarly, $K^+$ ions migrate to the cathode to replace the depleted $Cu^{2+}$ ions.
Note: The mobility of cations and anions in the salt bridge electrolyte must be almost equal to prevent junction potential.
3.2 Standard Electrode Potential and SHE
Absolute electrode potential cannot be measured because a half-cell reaction cannot occur in isolation. We can only measure the potential difference between two half-cells. To establish a scale, a reference electrode is chosen and arbitrarily assigned a potential of zero. This is the Standard Hydrogen Electrode (SHE).
- Construction of SHE: It consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution containing $H^+$ ions at 1 M concentration, and pure hydrogen gas is bubbled through it at 1 bar pressure and 298 K.
- Reaction: $$H^+(aq) + e^- \rightleftharpoons \frac{1}{2}H_2(g)$$
- Standard Potential: $E^\circ = 0.00 \text{ V}$ at all temperatures.
Electrochemical Series: The arrangement of various elements in the increasing order of their standard reduction potentials. A negative $E^\circ$ means the redox couple is a stronger reducing agent than the $H^+/H_2$ couple. A positive $E^\circ$ means it is a weaker reducing agent (stronger oxidizing agent).
3.3 The Nernst Equation
The standard electrode potential ($E^\circ$) is valid only when the concentration of all species is unity (1 M). For any other concentration, the electrode potential is given by the Nernst equation.
For a general reduction reaction: $M^{n+}(aq) + ne^- \rightarrow M(s)$
Since the concentration of a pure solid $[M(s)]$ is taken as unity, substituting $R$, $T = 298 \text{ K}$, $F$, and converting natural log to base 10 log, we get:
3.4 Conductance of Electrolytic Solutions
| Parameter | Definition | Unit |
|---|---|---|
| Specific Conductivity ($\kappa$) | Conductance of 1 cm³ of the solution. | $S \text{ cm}^{-1}$ or $\Omega^{-1} \text{ cm}^{-1}$ |
| Molar Conductivity ($\Lambda_m$) | Conducting power of all the ions produced by dissolving one mole of an electrolyte in solution. | $S \text{ cm}^2 \text{ mol}^{-1}$ |
Variation of Conductivity and Molar Conductivity with Concentration
- Specific Conductivity ($\kappa$) DECREASES with dilution.
Reason: Specific conductivity is the conductance of 1 cm³ of the solution. Upon dilution, the total volume increases, so the number of ions per unit volume decreases. - Molar Conductivity ($\Lambda_m$) INCREASES with dilution.
Reason: $\Lambda_m = \kappa \times V$. Upon dilution, the decrease in $\kappa$ is more than compensated by the increase in volume $V$.- Strong Electrolytes (e.g., KCl, NaCl): Completely dissociated. Increase in $\Lambda_m$ on dilution is slight due to a decrease in interionic attractions, allowing ions to move faster.
- Weak Electrolytes (e.g., $CH_3COOH$): Low degree of dissociation initially. Upon dilution, dissociation ($\alpha$) increases significantly, producing many more ions. Thus, $\Lambda_m$ increases sharply at lower concentrations.
Kohlrausch's Law of Independent Migration of Ions
Statement: The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
Application: To determine $\Lambda^\circ_m$ for weak electrolytes using data from strong electrolytes. For example, for Acetic Acid ($CH_3COOH$):
3.5 Batteries and Commercial Cells
| Type | Description | Examples |
|---|---|---|
| Primary Batteries | Reaction occurs only once. Becomes dead after use and cannot be recharged. | Dry cell (Leclanche cell), Mercury cell. |
| Secondary Batteries | Can be recharged by passing current through it in the opposite direction. | Lead storage battery, Nickel-Cadmium cell. |
| Fuel Cells | Galvanic cells that convert the energy of combustion of fuels directly into electricity. | $H_2 - O_2$ fuel cell (pollution-free, ~70% efficiency). |
3.6 Corrosion (Rusting of Iron)
Corrosion is an electrochemical phenomenon where a metal is oxidized by loss of electrons to oxygen and formation of oxides.
- Anode spot (Iron is oxidized): $$2Fe(s) \rightarrow 2Fe^{2+} + 4e^- \quad (E^\circ = -0.44 \text{ V})$$
- Cathode spot (Oxygen is reduced): $$O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l) \quad (E^\circ = 1.23 \text{ V})$$
- Overall reaction: $$2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l)$$
The $Fe^{2+}$ ions are further oxidized by atmospheric oxygen to form rust ($Fe_2O_3 \cdot xH_2O$). Prevention includes Galvanization (coating iron with Zinc, which acts as a sacrificial anode).
4. Most Important CBSE Board Questions
Strictly aligned with the latest CBSE sample papers and marking scheme.
Section A: 1 Mark Questions (MCQs & A-R)
Q1. Which of the following statements is true for a fuel cell? 1 Mark
- They are secondary cells.
- They produce electrical energy from combustion of fuels directly.
- They have very low efficiency (around 20%).
- They cause severe environmental pollution.
Explanation: Fuel cells convert the chemical energy of a continuously supplied fuel (like $H_2$, $CH_4$) directly into electricity. They are highly efficient (70-75%) and pollution-free.
Q2. The quantity of charge required to obtain one mole of aluminium from $Al_2O_3$ is: 1 Mark
- 1 F
- 6 F
- 3 F
- 2 F
Explanation: The oxidation state of Al in $Al_2O_3$ is +3. To reduce one mole of $Al^{3+}$ to Al metal, 3 moles of electrons are required.
Reaction: $Al^{3+} + 3e^- \rightarrow Al$
Charge required = $n \times F = 3 \times F = 3F$.
Assertion-Reasoning
Q3. Assertion (A): Conductivity of all electrolytes decreases on dilution.
Reason (R): On dilution, the number of ions per unit volume decreases. 1 Mark
Explanation: Conductivity ($\kappa$) is the conductance of 1 cm³ of solution. When water is added (dilution), the total volume increases, so the number of current-carrying ions in that specific 1 cm³ volume decreases, causing conductivity to drop.
Section B: 2 Mark Questions (Very Short Answer)
Q4. Write the Nernst equation and calculate the emf of the following cell at 298 K:
$Mg(s) | Mg^{2+} (0.001\text{ M}) || Cu^{2+} (0.0001\text{ M}) | Cu(s)$
Given: $E^\circ_{(Mg^{2+}/Mg)} = -2.37\text{ V}$, $E^\circ_{(Cu^{2+}/Cu)} = +0.34\text{ V}$. 2 Marks
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-2.37) = \mathbf{2.71\text{ V}}$.
The cell reaction is: $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$. Here, $n = 2$ electrons transferred.
Step 2: Apply Nernst Equation
$E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}$
$E_{cell} = 2.71 - 0.0295 \log \frac{10^{-3}}{10^{-4}}$
$E_{cell} = 2.71 - 0.0295 \log(10)$
$E_{cell} = 2.71 - 0.0295(1) = \mathbf{2.6805\text{ V}}$.
Section C: 3 Mark Questions (Short Answer)
Q5. The conductivity of 0.001028 M acetic acid is $4.95 \times 10^{-5} \text{ S cm}^{-1}$. Calculate its dissociation constant ($K_a$) if $\Lambda^\circ_m$ for acetic acid is $390.5 \text{ S cm}^2 \text{ mol}^{-1}$. 3 Marks
Concentration $C = 0.001028\text{ mol L}^{-1}$
Conductivity $\kappa = 4.95 \times 10^{-5}\text{ S cm}^{-1}$
Limiting molar conductivity $\Lambda^\circ_m = 390.5\text{ S cm}^2 \text{ mol}^{-1}$
Step 1: Calculate Molar Conductivity ($\Lambda_m$)
$\Lambda_m = \frac{\kappa \times 1000}{C}$
$\Lambda_m = \frac{4.95 \times 10^{-5} \times 10^3}{0.001028} = \frac{0.0495}{0.001028} = \mathbf{48.15\text{ S cm}^2 \text{ mol}^{-1}}$
Step 2: Calculate Degree of Dissociation ($\alpha$)
$\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{48.15}{390.5} = \mathbf{0.1233}$
Step 3: Calculate Dissociation Constant ($K_a$)
$K_a = \frac{C \alpha^2}{1 - \alpha}$
$K_a = \frac{0.001028 \times (0.1233)^2}{1 - 0.1233}$
$K_a = \frac{0.001028 \times 0.0152}{0.8767} = \mathbf{1.78 \times 10^{-5}\text{ mol L}^{-1}}$.
Section D: 4 Mark Case-Based Questions
Q6. Read the passage given below and answer the following questions:
(i) State Kohlrausch's law of independent migration of ions. (1 Mark)
(ii) Why can't the limiting molar conductivity ($\Lambda^\circ_m$) for a weak electrolyte be determined graphically by extrapolating the $\Lambda_m$ vs $\sqrt{c}$ plot? (1 Mark)
(iii) The $\Lambda^\circ_m$ for $NaCl$, $HCl$ and $NaAc$ (Sodium acetate) are 126.4, 425.9 and 91.0 $\text{S cm}^2\text{ mol}^{-1}$ respectively. Calculate $\Lambda^\circ_m$ for $HAc$ (Acetic acid). (2 Marks)
(ii) For weak electrolytes, the degree of dissociation increases rapidly upon dilution, resulting in a very steep rise in molar conductivity near zero concentration. The curve becomes almost parallel to the y-axis, making precise graphical extrapolation to zero concentration impossible.
(iii) Using Kohlrausch's law:
$\Lambda^\circ_m(HAc) = \Lambda^\circ_m(HCl) + \Lambda^\circ_m(NaAc) - \Lambda^\circ_m(NaCl)$
$\Lambda^\circ_m(HAc) = 425.9 + 91.0 - 126.4$
$\Lambda^\circ_m(HAc) = 516.9 - 126.4 = \mathbf{390.5\text{ S cm}^2 \text{ mol}^{-1}}$.
Section E: 5 Mark Long Answer Questions
Q7. (a) Calculate $\Delta_r G^\circ$ and $\log K_c$ for the following cell reaction at 298 K:
$2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)$
Given: $E^\circ_{cell} = +0.34\text{ V}$, $F = 96500\text{ C mol}^{-1}$. (3 Marks)
(b) Predict the products of electrolysis of an aqueous solution of $CuCl_2$ with platinum electrodes. Support your answer with standard reduction potentials. (2 Marks)
Let's determine the number of electrons transferred ($n$).
Cr goes from 0 to +3 (loss of 3e⁻). For 2 Cr atoms, total electrons lost = 6.
Cd goes from +2 to 0 (gain of 2e⁻). For 3 Cd atoms, total electrons gained = 6.
So, $n = 6$.
Step 1: Calculate Standard Gibbs Free Energy ($\Delta_r G^\circ$)
$\Delta_r G^\circ = -n F E^\circ_{cell}$
$\Delta_r G^\circ = -6 \times 96500 \times 0.34$
$\Delta_r G^\circ = -196860\text{ J mol}^{-1} = \mathbf{-196.86\text{ kJ mol}^{-1}}$.
Step 2: Calculate Equilibrium Constant ($\log K_c$)
$E^\circ_{cell} = \frac{0.0591}{n} \log K_c$
$0.34 = \frac{0.0591}{6} \log K_c$
$\log K_c = \frac{0.34 \times 6}{0.0591} = \mathbf{34.51}$
(Optional: $K_c = \text{antilog}(34.51) \approx 3.2 \times 10^{34}$)
(b) Products of Electrolysis of aq. $CuCl_2$:
The solution contains $Cu^{2+}$, $Cl^-$, and $H_2O$.
At Cathode (Reduction): Competition between $Cu^{2+}$ and $H_2O$.
$Cu^{2+}$ has a higher reduction potential ($+0.34\text{ V}$) than water ($-0.83\text{ V}$). Thus, Copper ($Cu$) is deposited at the cathode.
At Anode (Oxidation): Competition between $Cl^-$ and $H_2O$.
Theoretically, water has a lower oxidation potential and should be oxidized to $O_2$. However, due to the kinetic sluggishness of oxygen formation (overpotential of oxygen), oxidation of chloride ions becomes kinetically favorable. Hence, Chlorine ($Cl_2$) gas is liberated at the anode.
Mastering 'Electrochemistry'
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Master the Nernst Equation: Over 90% of CBSE papers feature a 3-mark or 5-mark question involving it. Carefully balance the redox reaction to find '$n$'. Remember, pure solids do not appear in the log term!
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Units are Crucial: The cell constant unit is $\text{cm}^{-1}$. Conductivity $\kappa$ is $S \text{ cm}^{-1}$. Molar conductivity $\Lambda_m$ is $S \text{ cm}^2 \text{ mol}^{-1}$. A 0.5 mark deduction is strictly enforced for missing units.
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Understand Overpotential: When predicting products of electrolysis for aqueous solutions containing halides ($Cl^-$, $Br^-$), always state that halide oxidation is preferred over water oxidation due to the overpotential of oxygen.
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