Easy Practice — Set 1

Quick Recap — Oxidation Number & Redox

• Oxidation = loss of electrons (oxidation number rises); reduction = gain of electrons (falls).
• Oxidising agent is itself reduced; reducing agent is itself oxidised.
• Oxidation-number rules: free element $=0$; monatomic ion $=$ its charge; O usually $-2$ ($-1$ in peroxides); H usually $+1$ ($-1$ in metal hydrides); sum $=$ overall charge.
• Disproportionation: the same element is both oxidised and reduced.

Beyond-NCERT JEE Formulae

These push past the NCERT recap into the numerical shortcuts JEE Main actually rewards. Each group says when to use it and flags a [JEE Tip] trap examiners like to set.

1. Cell EMF, Gibbs Energy and Equilibrium

When to use: any galvanic cell where you must judge spontaneity, find the maximum electrical work, or turn a standard EMF into an equilibrium constant.

• Standard EMF from tabulated reduction potentials: $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$.
• Gibbs energy of the cell reaction: $\Delta G^\circ=-nFE^\circ_{cell}$ (F = 96500 C/mol; n = electrons transferred in the balanced equation).
• Equilibrium constant at 298 K: $\log K_c=\dfrac{nE^\circ_{cell}}{0.059}$.

[JEE Tip] A positive $E^\circ_{cell}$ forces $\Delta G^\circ<0$ and $K_c>1$ (spontaneous as written), so checking the sign is your fastest sanity test. Electrode potential is intensive: if you scale or reverse a half-reaction, only n and the sign change, never the per-mole value of $E^\circ$.

2. Nernst Equation and Concentration Cells

When to use: electrode or cell potentials away from the standard state -- non-unit concentrations, gas pressures, pH-dependent couples, or two identical electrodes at different concentrations.

• General form (298 K): $E=E^\circ-\dfrac{0.059}{n}\log Q$, with Q the reaction quotient (products over reactants; solids and pure liquids omitted; gases as partial pressures in bar).
• Concentration cell (same electrodes, so $E^\circ=0$): $E=\dfrac{0.059}{n}\log\dfrac{C_{cathode}}{C_{anode}}$, positive only when the cathode compartment is the more concentrated.

[JEE Tip] The constant 0.059 V equals $\dfrac{2.303RT}{F}$ at 298 K only; at any other temperature recompute it. A tenfold concentration change moves a one-electron electrode by just 0.059 V, so cell EMF is far less concentration-sensitive than students expect.

3. Faraday's Laws of Electrolysis

When to use: mass or volume of a product deposited or liberated, current-time-charge sums, and comparing two substances discharged by the same charge.

• First law: $m=\dfrac{E\,I\,t}{96500}=ZIt$ (E = equivalent weight, Z = electrochemical equivalent).
• In molar terms: $m=\dfrac{M\,I\,t}{nF}$, where n electrons discharge one ion.
• Charge passed as moles of electrons (equivalents) $=\dfrac{It}{96500}$.
• Second law: for a fixed charge, $\dfrac{m_1}{m_2}=\dfrac{E_1}{E_2}$ (products form in the ratio of their equivalent weights).

[JEE Tip] Route every electrolysis through moles of electrons: find $\dfrac{It}{96500}$, then divide by n for moles of metal, or apply gas stoichiometry (2 electrons per H2, 4 per O2). One faraday (96500 C) always deposits exactly one gram-equivalent.

4. Conductance and Kohlrausch's Law

When to use: any data on conductivity or cell constant, strong-versus-weak electrolyte behaviour, or extracting a dissociation constant from conductance.

• Molar conductivity: $\Lambda_m=\dfrac{\kappa\times1000}{M}$ (conductivity in S/cm, molarity M in mol/L; $\Lambda_m$ in S cm^2/mol).
• Kohlrausch's law of independent migration: $\Lambda^\circ_m=\nu_{+}\lambda^\circ_{+}+\nu_{-}\lambda^\circ_{-}$.
• Degree of dissociation of a weak electrolyte: $\alpha=\dfrac{\Lambda_m}{\Lambda^\circ_m}$, then $K_a=\dfrac{C\alpha^2}{1-\alpha}$.

[JEE Tip] A weak electrolyte's $\Lambda^\circ_m$ can NOT be read off by extrapolating to zero concentration -- assemble it from ions instead, e.g. acetic acid = sodium acetate + HCl - NaCl. Only for strong electrolytes is $\Lambda_m=\Lambda^\circ_m-b\sqrt{C}$ (Debye-Huckel-Onsager), so their plot against the square root of concentration is linear.

5. n-Factor, Equivalent Weight and Balancing Redox

When to use: equivalent-weight questions, redox titrations (KMnO4, K2Cr2O7, iodometry) and normality book-keeping.

• Equivalent weight: $E=\dfrac{M}{n}$, where n (the n-factor) = electrons gained or lost per formula unit.
• Equivalents always balance: (equivalents of oxidant) = (equivalents of reductant), i.e. $N_1V_1=N_2V_2$.
• Ion-electron method: balance the redox atom, then O with H2O, H with H+ (acidic) or OH- (basic), then charge with electrons, and finally scale so the electrons cancel.

[JEE Tip] The n-factor is not fixed by the formula; it depends on the product. MnO4- accepts 5 electrons in acid (to Mn2+), 3 in neutral or weakly basic media (to MnO2) and 1 in strongly basic media (to MnO4^2-), so its equivalent weight is 158/5, 158/3 or 158 in turn. Always fix the product before quoting an equivalent weight.

Solved Examples — Beyond-NCERT Formulae

Each solution names the formula, substitutes, and states the answer with units. All values use $\dfrac{2.303RT}{F}=0.059$ V at 298 K.

Example 1 — EMF, Gibbs energy and spontaneity

Q. For the cell Zn | Zn2+ || Cu2+ | Cu, take E(Cu2+/Cu) = +0.34 V and E(Zn2+/Zn) = -0.76 V. Find the standard EMF and Gibbs energy, and decide if the reaction is spontaneous (n = 2, F = 96500 C/mol).

Step 1 — pick electrodes. The higher reduction potential is the cathode, so Cu2+/Cu is the cathode and Zn2+/Zn is the anode.

Step 2 — standard EMF. Using $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$, $$E^\circ_{cell}=0.34-(-0.76)=+1.10$$ giving an EMF of +1.10 V.

Step 3 — Gibbs energy. With $\Delta G^\circ=-nFE^\circ_{cell}$, $$\Delta G^\circ=-(2)(96500)(1.10)=-2.123\times10^{5}$$ that is -212.3 kJ per mole of reaction.

Step 4 — verdict. $E^\circ_{cell}>0$, so $\Delta G^\circ<0$ and the reaction is spontaneous as written.

Example 2 — Equilibrium constant from standard EMF

Q. For the same cell ($E^\circ_{cell}=+1.10$ V, n = 2), find the equilibrium constant $K_c$ at 298 K.

Step 1 — formula. At equilibrium E = 0, so the Nernst equation collapses to $\log K_c=\dfrac{nE^\circ_{cell}}{0.059}$.

Step 2 — substitute. $\log K_c=\dfrac{2\times1.10}{0.059}=\dfrac{2.20}{0.059}=37.29$.

Step 3 — solve. $K_c=10^{37.29}\approx1.9\times10^{37}$.

Read it. The enormous $K_c$ says the reaction runs essentially to completion -- exactly what the large positive $E^\circ_{cell}$ promised.

Example 3 — Nernst equation at non-standard concentrations

Q. For Zn | Zn2+ (0.10 M) || Cu2+ (0.010 M) | Cu with $E^\circ_{cell}=+1.10$ V and n = 2, find the EMF at 298 K.

Step 1 — reaction quotient. The cell reaction is Zn + Cu2+ -> Zn2+ + Cu (solids omitted), so Q = [Zn2+]/[Cu2+]: $$Q=\dfrac{0.10}{0.010}=10$$

Step 2 — Nernst. Using $E=E^\circ_{cell}-\dfrac{0.059}{n}\log Q$, $$E=1.10-\dfrac{0.059}{2}\log 10=1.10-0.0295=1.0705$$ so the EMF is about +1.07 V.

Note. Diluting the cathode ion (Cu2+) drops the EMF below the standard 1.10 V, just as Le Chatelier predicts.

Example 4 — Mass deposited in electrolysis

Q. A 5.0 A current flows for 30 minutes through aqueous CuSO4. Find the mass of copper deposited (M(Cu) = 63.5 g/mol, n = 2).

Step 1 — moles of electrons. With t = 30 x 60 = 1800 s, $$\dfrac{It}{96500}=\dfrac{5.0\times1800}{96500}=0.0933$$ that is 0.0933 mol of electrons.

Step 2 — mass. Each Cu2+ needs 2 electrons, so use $m=\dfrac{M\,I\,t}{nF}$: $$m=\dfrac{63.5\times5.0\times1800}{2\times96500}=2.96$$ so 2.96 g of copper is deposited.

Cross-check. Equivalent weight of Cu = 63.5/2 = 31.75 g/equiv; mass = 0.0933 x 31.75 = 2.96 g. Consistent.

Example 5 — Molar conductivity, dissociation and Ka (weak acid)

Q. For 0.0010 M acetic acid the conductivity is $\kappa=4.95\times10^{-5}$ S/cm and $\Lambda^\circ_m=390.5$ S cm^2/mol. Find the molar conductivity, the degree of dissociation $\alpha$ and $K_a$.

Step 1 — molar conductivity. Using $\Lambda_m=\dfrac{\kappa\times1000}{M}$, $$\Lambda_m=\dfrac{4.95\times10^{-5}\times1000}{0.0010}=49.5$$ so the molar conductivity is 49.5 S cm^2/mol.

Step 2 — degree of dissociation. $\alpha=\dfrac{\Lambda_m}{\Lambda^\circ_m}=\dfrac{49.5}{390.5}=0.127$, about 12.7%.

Step 3 — dissociation constant. With $K_a=\dfrac{C\alpha^2}{1-\alpha}$, $$K_a=\dfrac{0.0010\times(0.127)^2}{1-0.127}\approx1.8\times10^{-5}$$ close to the accepted value for acetic acid.

Why Kohlrausch. A weak acid's $\Lambda^\circ_m$ cannot be found by extrapolation; it is assembled from ions -- acetic acid = sodium acetate + HCl - NaCl.

Example 6 — Concentration cell EMF

Q. Find the EMF of Cu | Cu2+ (0.0010 M) || Cu2+ (0.10 M) | Cu at 298 K (n = 2).

Step 1 — why the standard EMF is zero. Both electrodes are copper, so $E^\circ_{cell}=0$; the only drive is the concentration gap.

Step 2 — formula. The concentration-cell EMF is $$E=\dfrac{0.059}{n}\log\dfrac{C_{cathode}}{C_{anode}}$$

Step 3 — substitute. The concentrated compartment (0.10 M) is the cathode: $$E=\dfrac{0.059}{2}\log\dfrac{0.10}{0.0010}=\dfrac{0.059}{2}\times2=0.059$$ giving an EMF of 0.059 V.

Insight. The EMF is positive only because the cathode is more concentrated; equal concentrations give exactly zero.

Example 7 — n-factor, equivalent weight and a titration

Q. In acidic medium MnO4- is reduced to Mn2+. (a) Find the equivalent weight of KMnO4 (M = 158 g/mol). (b) What volume of 0.020 M KMnO4 exactly oxidises 20.0 mL of 0.10 M FeSO4?

Step 1 — n-factors. MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O has an n-factor of 5; Fe2+ -> Fe3+ + e- has an n-factor of 1.

Step 2 — equivalent weight. $E=\dfrac{M}{n}=\dfrac{158}{5}=31.6$, so the equivalent weight of KMnO4 is 31.6 g/equiv.

Step 3 — equivalents of Fe2+. equivalents = molarity x volume x n-factor, so $0.10\times0.0200\times1=2.0\times10^{-3}$.

Step 4 — volume of KMnO4. Equivalents match at the end point, so $0.020\times V\times5=2.0\times10^{-3}$, giving $V=\dfrac{2.0\times10^{-3}}{0.10}=0.020$ L, i.e. 20 mL.