Electrochemistry and Ionic Equilibria Questions and Answers

Answer: E

The correct match is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).

• Milk of magnesia, a suspension of magnesium hydroxide, is basic: pH about 10.
• Black coffee is mildly acidic: pH about 5.
• Egg white is slightly basic: pH about 7.8.
• Milk is very slightly acidic: pH about 6.8.

Options matching coffee to 7.8 or milk of magnesia to 6.8 put an acidic substance on the basic side or the other way round; option A also uses (ii) twice.

Answer: A

\(\Delta_{r}G^\circ\) is -196.86 kJ mol^-1.

Chromium is oxidised (anode) and cadmium ions are reduced (cathode). Six electrons are transferred in the balanced reaction.

\[ \begin{aligned} E^\circ_{cell} &= -0.40 - (-0.74) = 0.34 \text{ V} \\ \Delta_{r}G^\circ &= -nFE^\circ_{cell} \\ &= -6 \times 96500 \times 0.34 \\ &= -196860 \text{ J mol}^{-1} \end{aligned} \]

The cell EMF is positive, so the reaction is spontaneous and \(\Delta_{r}G^\circ\) must be negative, ruling out all the positive values. -96.50 kJ mol^-1 does not use the 0.34 V EMF with six electrons.

Answer: B

The percentage ionisation is 16.7 %.

By Kohlrausch's law, the limiting molar conductivity is the sum of the ionic contributions, and the degree of dissociation of a weak electrolyte is \(\alpha = \frac{\Lambda_{m}}{\Lambda^\circ_{m}}\).

\[ \begin{aligned} \Lambda^\circ_{m} &= 340 + 80 = 420 \text{ S cm}^{2}\text{ mol}^{-1} \\ \alpha &= \frac{70}{420} = 0.167 \end{aligned} \]

That is 16.7 %. 8.35 % is half of it, and 20 % would need a limiting molar conductivity of 350, which ignores the contribution of one of the ions.

Answer: E

The key gives HF, the strongest of the five acids listed, but the textbook tables do not agree on its margin over nitrous acid.

The other three are far weaker: phenol has \(K_a\) near \(10^{-10}\), hypochlorous acid near \(10^{-8}\), and a pyridinecarboxylic acid, \(\mathrm{C_5H_4NCOOH}\), only of the order of \(10^{-5}\). That leaves HF and \(\mathrm{HNO_2}\), both near \(10^{-4}\).

General chemistry reference tables at 298 K give HF about \(6.4 \times 10^{-4}\) and \(\mathrm{HNO_2}\) about \(4.6 \times 10^{-4}\), so HF is the stronger and option E matches the key. The NCERT table lists HF as \(3.5 \times 10^{-4}\) and \(\mathrm{HNO_2}\) as \(4.5 \times 10^{-4}\), which would put nitrous acid (option A) first. The question rests on which table is meant; the official key follows the first.

Answer: C

The article to be electroplated is the cathode.

In electroplating, the object is connected to the negative terminal. Metal ions in the electrolyte, which are positive, move to it, gain electrons and are deposited as a thin metal layer. The anode is usually a block of the plating metal, which dissolves to replenish the ions.

The electrolyte is the solution of the metal salt, and 'conductor' is too general a term.

Answer: B

In a dry cell, chemical energy is converted into electrical energy.

A dry (Leclanche type) cell has a zinc container as the negative electrode and a carbon rod surrounded by manganese dioxide as the positive electrode, with an ammonium chloride paste as electrolyte. The redox reactions at the electrodes release chemical energy as a flow of electrons.

No mechanical, kinetic or potential (gravitational) energy is involved. Mechanical energy is converted into electrical energy in a generator, not a cell.

Answer: A

The pH is 11.28.

First find pOH from the hydroxide concentration, then use pH + pOH = 14 at 25 °C.

\[ \begin{aligned} \mathrm{pOH} &= -\log(1.9 \times 10^{-3}) \\ &= 3 - 0.28 = 2.72 \\ \mathrm{pH} &= 14 - 2.72 = 11.28 \end{aligned} \]

An ammonia solution is basic, so its pH must exceed 7; 5.29 would be acidic. 9.28 and 7.28 do not follow from this concentration: they would need an hydroxide concentration a hundred or ten thousand times smaller.

Answer: B

The pH is 11.3.

Barium hydroxide is a strong base giving two hydroxide ions per formula unit: \(\text{Ba(OH)}_{2} \rightarrow \text{Ba}^{2+} + 2\text{OH}^{-}\).

\[ \begin{aligned} [\text{OH}^{-}] &= 2 \times 0.001 = 0.002\ \text{M} \\ \text{pOH} &= -\log(2 \times 10^{-3}) = 2.70 \\ \text{pH} &= 14 - 2.70 = 11.30 \end{aligned} \]

11.0 is what you get by forgetting that each Ba(OH)₂ gives two OH^- ions. 12.0 and 12.5 would need ten or more times the hydroxide concentration.

Answer: A

The pH is 3.

HCl is a strong acid and ionises completely, so the hydrogen ion concentration equals the acid concentration:

\[ \begin{aligned} [\text{H}^{+}] &= 0.001 = 10^{-3}\ \text{M} \\ \text{pH} &= -\log 10^{-3} = 3 \end{aligned} \]

Each tenfold dilution raises the pH by one unit, so 4 and 5 would correspond to 0.0001 M and 0.00001 M acid. 'None of the above' is wrong because 3 is offered. The answer is below 7, as it must be for an acid.

Answer: D

Electrons flow from the Zn electrode to the Pb electrode through the Cu wire.

The standard reduction potentials are Zn²⁺/Zn -0.76 V and Pb²⁺/Pb -0.13 V. Zinc, with the more negative potential, is oxidised and is the anode; lead ions are reduced at the cathode. The cell reaction is Zn + Pb²⁺ → Zn²⁺ + Pb, with E°cell = 0.63 V.

Electrons released at the zinc anode travel through the external circuit, the Cu wire, to the lead cathode. Electrons never pass through the salt bridge; it carries ions that keep the two solutions electrically neutral. That rules out both salt-bridge options, and the option with electrons going from Pb to Zn has the direction reversed.