Top 50 Most Important
Subjective Questions
Electrochemistry. Master Nernst Equation, Kohlrausch's Law, Faraday's Laws, and Battery mechanisms.
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Electrochemical Cells & Potentials
Galvanic Cell: Converts chemical energy into electrical energy via a spontaneous redox reaction ($\Delta G < 0$). Anode is negative, cathode is positive.
Electrolytic Cell: Converts electrical energy into chemical energy to drive a non-spontaneous redox reaction ($\Delta G > 0$). Anode is positive, cathode is negative.
- It completes the inner electrical circuit by allowing ions to flow from one half-cell to the other.
- It maintains the electrical neutrality of the solutions in the two half-cells, preventing the accumulation of positive or negative charges which would otherwise stop the flow of current.
It is the potential difference developed between a metal electrode and its ions in solution when the concentration of the ions is exactly 1 mol $L^{-1}$ (1 M), the pressure of any gas involved is 1 bar, and the temperature is strictly 298 K.
SHE is a reference electrode consisting of a platinum wire coated with platinum black, dipped in a 1 M $H^+$ (acid) solution, with pure hydrogen gas bubbled through it at 1 bar pressure and 298 K.
Reaction: $H^+_{(aq)} + e^- \rightarrow \frac{1}{2} H_{2(g)}$.
By IUPAC convention, its standard electrode potential is arbitrarily assigned as exactly $0.00 \text{ V}$ at all temperatures.
Oxidation or reduction cannot occur independently; they are halves of a full reaction. Electrons produced at an anode must have a cathode to flow towards. Therefore, we can only measure a potential difference between two half-cells connected together, which is why a reference electrode (like SHE) is required.
A negative standard reduction potential implies that the metal is a stronger reducing agent than hydrogen gas. It means the species prefers to undergo oxidation (lose electrons) rather than reduction when coupled with the Standard Hydrogen Electrode (SHE).
EMF (Electromotive Force): The potential difference between two electrodes when no current is being drawn from the cell (open circuit). It is the maximum voltage a cell can deliver.
Potential Difference: The potential difference when current is flowing through the circuit (closed circuit). It is always less than the EMF due to internal resistance.
No. Zinc is placed above Copper in the electrochemical series (it has a more negative $E^\circ$ value). Thus, Zinc is a stronger reducing agent and will spontaneously displace copper from its solution:
$Zn_{(s)} + CuSO_{4(aq)} \rightarrow ZnSO_{4(aq)} + Cu_{(s)}$.
This reaction will dissolve the zinc vessel, causing it to leak.
When $E_{ext} > 1.1 \text{ V}$, the reaction reverses direction. The cell behaves as an Electrolytic Cell instead of a Galvanic cell. Electrons flow from Cu to Zn, Zinc is deposited at the zinc electrode, and Copper dissolves at the copper electrode.
The Electrochemical Series is the arrangement of various chemical elements/ions in increasing order of their Standard Reduction Potentials ($E^\circ$).
Applications: It helps to compare the relative oxidizing and reducing powers of elements, predict the feasibility (spontaneity) of a redox reaction, and determine whether a metal can liberate $H_2$ gas from dilute acids.
Nernst Eq & Gibbs Free Energy
The Nernst equation relates the electrode potential at non-standard conditions to its standard potential.
$E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln \frac{[M]}{[M^{n+}]}$
Since the concentration of a pure solid $[M] = 1$, it simplifies to: $E = E^\circ - \frac{0.0591}{n} \log \frac{1}{[M^{n+}]}$ at 298 K.
Reaction: $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$. Here, $n = 2$ electrons transferred.
$E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
At chemical equilibrium, the cell potential $E_{cell} = 0$, and the reaction quotient $Q = K_c$. Plugging this into the Nernst equation at 298 K yields:
$E^\circ_{cell} = \frac{0.0591}{n} \log K_c$
Electrical work done by a galvanic cell is equal to the decrease in its Gibbs free energy. The standard relationship is:
$\Delta G^\circ = -n F E^\circ_{cell}$
Where $n$ is moles of electrons and $F$ is the Faraday constant ($96487 \text{ C/mol}$).
As the cell reaction progresses, the concentration of reactants decreases and products increases, causing the potential of the two half-cells to shift. At equilibrium, the rate of the forward reaction perfectly matches the rate of the reverse reaction, meaning there is no longer any driving force (net potential difference) pushing electrons through the external circuit. Thus, $E_{cell} = 0$.
In the Nernst equation for the Daniel cell: $E_{cell} = E^\circ - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
$Zn^{2+}$ is in the numerator of the log term (it is a product). Increasing $[Zn^{2+}]$ increases the value being subtracted from $E^\circ$. Therefore, the overall cell potential ($E_{cell}$) decreases.
$E_{cell}$ is an intensive property. It depends only on the nature and concentration of the substances, not on their physical size or amount. (If you double the stoichiometry of a reaction, $E_{cell}$ remains unchanged).
$\Delta G$ is an extensive property. It depends directly on the amount of substance (moles of electrons, $n$) undergoing the reaction.
For a spontaneous reaction, the system must do work, meaning the change in Gibbs free energy must be negative.
Therefore, $\Delta G$ must be negative ($< 0$), which mathematically requires that the cell potential $E_{cell}$ must be positive ($> 0$).
In a galvanic cell, the chemical reaction occurs spontaneously, releasing free energy ($\Delta G < 0$) which is harnessed as electrical work.
In an electrolytic cell, the internal chemical reaction is inherently non-spontaneous ($\Delta G > 0$). It only occurs because external electrical energy is forcefully pumped in to overcome this positive energy barrier.
The original term in the equation is $\frac{RT}{F} \ln Q$. By converting the natural logarithm ($\ln$) to base-10 logarithm ($\log$) by multiplying by $2.303$, and substituting the standard temperature $T = 298 \text{ K}$, Gas constant $R = 8.314 \text{ J/(K mol)}$, and Faraday constant $F = 96487 \text{ C/mol}$, the entire constant block $\frac{2.303 \times R \times T}{F}$ computes exactly to $0.0591 \text{ V}$ (or $0.059 \text{ V}$).
Conductance & Kohlrausch's Law
Specific Conductivity (or just conductivity, $\kappa$) is defined as the conductance of $1 \text{ cm}^3$ (or $1 \text{ m}^3$) of an electrolytic solution. It is the reciprocal of resistivity ($\rho$).
$\kappa = \frac{1}{\rho} = G \times \frac{l}{A}$.
SI Unit: Siemens per meter ($S \text{ m}^{-1}$) or standard laboratory unit $S \text{ cm}^{-1}$.
Molar conductivity is the conducting power of all the ions produced by dissolving exactly 1 mole of an electrolyte in a solution, when the electrodes are placed 1 cm apart and the area is large enough to contain the entire solution.
Formula: $\Lambda_m = \frac{\kappa \times 1000}{M}$ (where $\kappa$ is in $S \text{ cm}^{-1}$ and $M$ is Molarity).
Unit: $S \text{ cm}^2 \text{ mol}^{-1}$.
Specific conductivity ($\kappa$) always decreases with dilution (i.e., decreasing concentration) for both strong and weak electrolytes.
Reason: Conductivity is the conductance of 1 unit volume ($1 \text{ cm}^3$) of solution. Upon dilution, the total volume increases significantly, so the number of current-carrying ions per unit volume decreases, causing a drop in $\kappa$.
For strong electrolytes (which are 100% dissociated), $\Lambda_m$ increases slowly with dilution.
Reason: Since they are already fully dissociated, dilution does not increase the number of ions. However, dilution pushes the tightly packed ions further apart, decreasing interionic repulsions. This allows the ions to move faster (higher mobility), causing a gradual increase in $\Lambda_m$.
For weak electrolytes, $\Lambda_m$ increases sharply/steeply with dilution, especially at lower concentrations.
Reason: Weak electrolytes are only partially dissociated. According to Ostwald's dilution law, adding more solvent shifts the equilibrium to promote ionization. Thus, dilution drastically increases the total number of ions produced by the 1 mole of electrolyte, causing a steep rise in conductance.
Kohlrausch's Law states that at infinite dilution (when dissociation is complete and interionic effects are zero), each ion migrates independently of its co-ion. The limiting molar conductivity of an electrolyte ($\Lambda^\circ_m$) can be represented as the exact mathematical sum of the individual limiting molar conductivities of its constituent anions and cations.
$\Lambda^\circ_m = v_+ \lambda^\circ_+ + v_- \lambda^\circ_-$
For a weak electrolyte, the graph of $\Lambda_m$ vs $\sqrt{c}$ curves sharply upward as concentration approaches zero because dissociation increases rapidly. Because the curve becomes nearly parallel to the y-axis (asymptotic), it never truly intersects it, making direct graphical extrapolation impossible. Kohlrausch's law is used indirectly to find it instead.
The degree of dissociation ($\alpha$) at any given concentration is the ratio of its molar conductivity at that concentration ($\Lambda_m^c$) to its limiting molar conductivity at infinite dilution ($\Lambda^\circ_m$).
$\alpha = \frac{\Lambda_m^c}{\Lambda^\circ_m}$
Cell constant ($G^*$) is the ratio of the distance between the two electrodes ($l$) to the area of cross-section of the electrodes ($A$).
$G^* = \frac{l}{A}$.
No. It is a fixed geometrical property of a specific conductivity cell. It remains constant regardless of the type or concentration of the electrolyte placed inside it.
They exhibit abnormally high mobility in water due to the Grotthuss mechanism (or proton-jumping). Instead of physically swimming through the liquid, $H^+$ and $OH^-$ ions rapidly pass protons down a chain of hydrogen-bonded water molecules, making charge transfer nearly instantaneous.
Electrolytic Cells & Faraday's Laws
Electrolysis is the process of chemical decomposition of an electrolyte (in molten state or in aqueous solution) into its constituent elements by the passage of direct electric current through it.
It states that the mass of any substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (charge, $Q$) passed through the electrolyte.
$m \propto Q \implies m = Z \cdot Q = Z \cdot I \cdot t$
Where $Z$ is the electrochemical equivalent, $I$ is current, and $t$ is time.
It states that when the same quantity of electricity is passed through different electrolytic solutions connected in series, the masses of the substances produced at the electrodes are directly proportional to their respective Equivalent Weights.
$\frac{m_1}{m_2} = \frac{E_1}{E_2}$
One Faraday ($1 \text{ F}$) is the absolute amount of electrical charge carried by exactly one mole of electrons.
Value: $1 \text{ F} = N_A \times e^- = (6.022 \times 10^{23}) \times (1.602 \times 10^{-19} \text{ C}) \approx \mathbf{96,487 \text{ C/mol}}$ (often rounded to $96500 \text{ C/mol}$ for calculations).
The reduction half-reaction is: $Al^{3+} + 3e^- \rightarrow Al$.
Because 3 moles of electrons are required to reduce 1 mole of $Al^{3+}$, the charge required is $3 \text{ Faradays}$ ($3\text{F}$).
$3 \times 96500 \text{ C} = \mathbf{2,89,500 \text{ C}}$.
- Molten NaCl: Only $Na^+$ and $Cl^-$ are present. Sodium metal ($Na$) forms at the cathode, and Chlorine gas ($Cl_2$) at the anode.
- Aqueous NaCl: Water competes. Water is easier to reduce than $Na^+$, so Hydrogen gas ($H_2$) forms at the cathode. Due to overpotential, $Cl_2$ gas forms at the anode instead of $O_2$.
Overpotential is the extra voltage (beyond the theoretical standard electrode potential) that must be applied to an electrode to overcome the kinetic activation barrier of a specific electrochemical reaction to make it proceed at a measurable rate. For example, oxidizing water to $O_2$ requires a high overpotential, which allows $Cl^-$ to be oxidized to $Cl_2$ first in brine solutions.
Because copper electrodes are active, the anode itself participates.
At Anode: The copper metal dissolves into the solution ($Cu \rightarrow Cu^{2+} + 2e^-$) rather than water oxidizing.
At Cathode: Pure copper ions from the solution are deposited as solid copper ($Cu^{2+} + 2e^- \rightarrow Cu$). This is the principle behind electro-refining of metals.
Electronic: Current flows due to the movement of free electrons. It involves no chemical change or transfer of matter. Conductance decreases with increasing temperature (due to metal lattice vibrations).
Electrolytic: Current flows due to the physical movement of ions in solution. It involves a chemical change (electrolysis) and transfer of matter. Conductance increases with increasing temperature (as viscosity decreases and ion mobility increases).
Yes. By determining the charge ($Q$) required to deposit exactly 1 equivalent weight of a metal during electrolysis, we experimentally find the Faraday constant ($F \approx 96487 \text{ C}$). Since $F$ is the charge of one mole of electrons, dividing $F$ by the known charge of a single electron ($e = 1.602 \times 10^{-19} \text{ C}$) gives Avogadro's number ($N_A$).
Batteries, Fuel Cells & Corrosion
A primary battery is a non-rechargeable galvanic cell where the internal redox reaction occurs only once. When the chemical reactants are consumed, the battery becomes dead and cannot be reused or recharged.
Example: Dry cell (Leclanché cell), Mercury cell.
A secondary battery is a rechargeable cell where the internal redox reaction can be reversed by passing a direct current through it in the opposite direction. It can be discharged and recharged multiple times.
Example: Lead Storage Battery (car battery), Lithium-ion battery.
The overall cell reaction in a mercury cell involves only solid substances: $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$. Because there are no ions in solution whose concentration can change over time during discharging, the Nernst equation term remains constant, giving a steady voltage of $1.35 \text{ V}$.
Anode (Oxidation): $Pb_{(s)} + SO_4^{2-}_{(aq)} \rightarrow PbSO_{4(s)} + 2e^-$
Cathode (Reduction): $PbO_{2(s)} + SO_4^{2-}_{(aq)} + 4H^+_{(aq)} + 2e^- \rightarrow PbSO_{4(s)} + 2H_2O_{(l)}$
Note: Sulphuric acid ($H_2SO_4$) is heavily consumed during this process, decreasing its density.
A fuel cell is a unique type of galvanic cell that continuously converts the chemical energy derived from the combustion of fuels (like $H_2, CH_4$, or methanol) directly into electrical energy. Unlike batteries, fuel cells never go "dead" as long as the fuel and oxidant are continuously supplied.
- High Efficiency: Fuel cells can achieve an efficiency of around 70%, completely bypassing the Carnot cycle limitations of thermal engines (which max out around 40%).
- Zero Pollution: The only chemical byproduct of a $H_2-O_2$ fuel cell is pure water ($H_2O$), making it highly environmentally friendly.
Corrosion is the slow, spontaneous, and destructive oxidation of metals by atmospheric gases (like $O_2, CO_2$) and moisture, transforming the metal into its oxides, carbonates, or sulfides.
Formula of Rust (Hydrated Iron(III) Oxide): $Fe_2O_3 \cdot xH_2O$.
Rusting is essentially the formation of miniature galvanic cells on the iron surface. A water drop loaded with dissolved $O_2$ and $CO_2$ acts as the electrolyte.
Anode (Iron oxidation): $Fe_{(s)} \rightarrow Fe^{2+}_{(aq)} + 2e^-$
Cathode (Oxygen reduction): $O_{2(g)} + 4H^+_{(aq)} + 4e^- \rightarrow 2H_2O_{(l)}$
The $Fe^{2+}$ ions are further oxidized by atmospheric oxygen to form $Fe^{3+}$ (rust).
Galvanization is the process of coating an iron object with a thin layer of Zinc metal. Zinc is more reactive (stronger reducing agent, lower $E^\circ$) than iron. If the coating is scratched, Zinc preferentially undergoes oxidation instead of iron, acting as a sacrificial anode. This is called sacrificial protection.
Corrosion requires an electrolyte to conduct ions between the miniature anode and cathode sites on the metal. Pure water is a poor conductor. Seawater contains a high concentration of dissolved salts (ions like $Na^+, Cl^-$), which drastically increases the electrical conductivity of the water film, massively accelerating the electrochemical rusting process.
Chapter 2 Mastered!
You have just conquered the 50 most critical subjective questions for Class 12 Chemistry, Chapter 2: Electrochemistry. Your Exam Arsenal is fully charged.
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