The limiting molar conductance for aqueous solution of CaCl2 at 298K is 271.6 S cm2 mol-1. If the limiting ionic conductance of Ca2+ ion at the same temperature is 119 S cm2 mol-1 what is the limiting ionic conductance of Cl- ion?

The limiting ionic conductance of Cl- is 76.3 S cm2 mol-1.By Kohlrausch's law, the limiting molar conductance of a salt is the sum of the limiting conductances of its ions, each multiplied by the number of those ions in the formula. CaCl2 gives one Ca2+ and two Cl-:\[ \begin{aligned} \Lambda^{\circ}_{m} &= \lambda^{\circ}_{\text{Ca}^{2+}} + 2\lambda^{\circ}_{\text{Cl}^{-}} \\ 271.6 &= 119 + 2\lambda^{\circ}_{\text{Cl}^{-}} \\ \lambda^{\circ}_{\text{Cl}^{-}} &= 152.6 / 2 = 76.3 \end{aligned} \]152.6 is the answer you get by forgetting that there are two chloride ions per formula unit.

Electrochemistry and Ionic Equilibria Questions and Answers

Q: Which of the following half-cell reaction has the most negative standard electrode potential?

The most negative standard electrode potential belongs to Li+(aq) + e- → Li(s).Lithium has the most negative standard reduction potential of all, about -3.05 V, because it loses its electron very readily in water (its small ion is strongly hydrated). It is therefore the strongest reducing agent in aqueous solution.Na+/Na is also strongly negative (about -2.71 V) but less so than lithium. F2/F- is the most positive of all, about +2.87 V, and I2/I- (about +0.54 V) and Cu+/Cu (about +0.52 V) are positive.

Q: What volume of O2(g) is produced when 2F current is passed through a very dilute aqueous solution of H2SO4 at STP?

The volume of oxygen is 11.2 dm3.At the anode, water is oxidised: \(2\text{H}_{2}\text{O} \rightarrow \text{O}_{2} + 4\text{H}^{+} + 4e^{-}\). One mole of \(\text{O}_{2}\) needs 4 F.\[ \begin{aligned} n(\text{O}_{2}) &= \frac{2}{4} = 0.5\ \text{mol} \\ V &= 0.5 \times 22.4 = 11.2\ \text{dm}^{3} \end{aligned} \]22.4 dm3 is the volume of hydrogen released at the cathode by the same 2 F, since each \(\text{H}_{2}\) needs only 2 electrons.

Q: The standard EMF of the following cellMg(s) | Mg+2 (0.01M) || Ag+(0.0001M) | Ag(s) is 3.17 V at 250C. The EMF of the cell at the same temperature is (2.303RT/F at 250C is 0.06V)

The EMF of the cell is 3.0 V (2.99 V).The cell reaction is \(\mathrm{Mg + 2Ag^+ \rightarrow Mg^{2+} + 2Ag}\), so 2 electrons are transferred. Apply the Nernst equation:\[ \begin{aligned} E &= E^\circ - \frac{0.06}{2}\log\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^+}]^{2}} \\ &= 3.17 - 0.03\log\frac{10^{-2}}{(10^{-4})^{2}} \\ &= 3.17 - 0.03 \times 6 \\ &= 3.17 - 0.18 = 2.99 \approx 3.0 \text{ V} \end{aligned} \]3.17 V is the standard value and ignores the concentrations. Forgetting to square [Ag+] gives \(\log 100 = 2\) and 3.11 V, and adding instead of subtracting the correction gives 3.35 V.

Q: The mass of Al deposited when 2 ampere current is passed through a solution of Al2(SO4)3 for 1 hour is (F = 96500 C)

The mass of aluminium deposited is 0.67 g.The charge passed is \(Q = It\), and each Al3+ ion needs 3 electrons, so 3 F deposit one mole (27 g) of Al.\[ \begin{aligned} Q &= 2 \times 3600 = 7200 \text{ C} \\ m &= \frac{27 \times 7200}{3 \times 96500} \\ &= \frac{194400}{289500} \approx 0.67 \text{ g} \end{aligned} \]1.34 g is exactly double the correct mass, and 0.067 g and 0.134 g are factor-of-ten slips; remember that each Al3+ ion takes three electrons.

Q: The concentration of hydrogen ions in a hydrochloric acid solution is 3 x 10-3 M. Its pH value is about( log3 = 0.4771)

The pH is about 2.52.\[ \begin{aligned} \mathrm{pH} &= -\log(3 \times 10^{-3}) \\ &= 3 - \log 3 = 3 - 0.4771 \\ &= 2.5229 \approx 2.52 \end{aligned} \]3.47 comes from adding \(\log 3\) instead of subtracting it, and 2.32 or 2.47 from rounding slips. A solution with \([\mathrm{H^+}]\) between \(10^{-3}\) and \(10^{-2}\) M must have pH between 2 and 3, which rules out 3.47 and 5.52 at once.

Q: What is the emf of the cell at 298 K in which the following reaction takes place?Ni(s) + 2Ag+(0.002M) → Ni2+(0.04M) + 2Ag(s). (E°cell at 298 K = 1.05V)

The emf is 0.93 V.Two electrons are transferred (n = 2). By the Nernst equation at 298 K:\[ \begin{aligned} E &= E^{\circ} - \frac{0.059}{2}\log\frac{[\mathrm{Ni^{2+}}]}{[\mathrm{Ag^{+}}]^{2}} \\ &= 1.05 - 0.0295\log\frac{0.04}{(0.002)^{2}} \\ &= 1.05 - 0.0295 \times 4 = 0.932\ \text{V} \end{aligned} \]The ratio inside the logarithm is \(\frac{0.04}{4 \times 10^{-6}} = 10^{4}\). Forgetting to square \([\mathrm{Ag^+}]\), or adding the correction, leads to the other values.

Q: In the reaction, O2(g) + 4H+(aq) + 4 e- →2H2O(l) , the quantity of electricity needed to reduce one mole of gaseous oxygen is (E°cell = 1.23V)

The quantity of electricity is 386000 Coulombs.The equation shows 4 electrons for each O2, so reducing one mole of oxygen needs 4 moles of electrons, that is 4 faradays:\[ Q = 4F = 4 \times 96500 = 386000\ \text{C} \]The cell potential of 1.23 V is not needed; it would be used to find the energy, not the charge. 96500 C is one faraday, enough for only a quarter of a mole of O2.

Q: Which of the following metal is incorporated into the electrodes of fuel cell to increase the rate of reaction?

The metal is palladium.In a hydrogen-oxygen fuel cell the gases are bubbled through porous carbon electrodes into the electrolyte. Finely divided platinum or palladium is incorporated into the electrodes as a catalyst, to increase the rate of the electrode reactions.Nickel is used as a catalyst in some alkaline cells and in hydrogenation, but the standard answer for these electrodes is platinum or palladium. Zinc, tin and magnesium are reactive metals that would be consumed, not catalysts.

Q: The solubility product, Ksp of a sparingly soluble salt, AX2, is 3.2 × 10-14 mol3lit-3. Its solubility is

The solubility is \(2 \times 10^{-5}\) mol lit-1.For AX2 dissolving to give A2+ and 2X-, a solubility \(s\) gives \([A^{2+}] = s\) and \([X^{-}] = 2s\):\[ \begin{aligned} K_{sp} &= s(2s)^{2} = 4s^{3} \\ s^{3} &= \frac{3.2 \times 10^{-14}}{4} = 8 \times 10^{-15} \\ s &= 2 \times 10^{-5} \text{ mol lit}^{-1} \end{aligned} \]Forgetting the 4 (using \(K_{sp} = s^{3}\)) gives about \(3.2 \times 10^{-5}\), and the \(10^{-3}\) and \(10^{-2}\) options misplace the cube root of the power of ten.

31.

Which of the following half-cell reaction has the most negative standard electrode potential?

A.\(\mathrm{Li^{+}{}_{(aq)} + e^{-} \rightarrow Li_{(s)}}\)
B.\(\mathrm{F_{2(g)} + 2e^{-} \rightarrow 2F^{-}{}_{(aq)}}\)
C.\(\mathrm{Na^{+}{}_{(aq)} + e^{-} \rightarrow Na_{(s)}}\)
D.\(\mathrm{I_{2(aq)} + 2e^{-} \rightarrow 2I^{-}{}_{(aq)}}\)
E.\(\mathrm{Cu^{+}{}_{(aq)} + e^{-} \rightarrow Cu_{(s)}}\)

Exam Year(s)

KEAM - 2026,

Answer: A

The most negative standard electrode potential belongs to Li⁺(aq) + e^- → Li(s).

Lithium has the most negative standard reduction potential of all, about -3.05 V, because it loses its electron very readily in water (its small ion is strongly hydrated). It is therefore the strongest reducing agent in aqueous solution.

Na⁺/Na is also strongly negative (about -2.71 V) but less so than lithium. F2/F^- is the most positive of all, about +2.87 V, and I2/I^- (about +0.54 V) and Cu⁺/Cu (about +0.52 V) are positive.

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32.

The limiting molar conductance for aqueous solution of CaCl₂ at 298K is 271.6 S cm² mol^-1. If the limiting ionic conductance of Ca²⁺ ion at the same temperature is 119 S cm² mol^-1 what is the limiting ionic conductance of Cl^- ion?

A.152.6 S \(\mathrm{cm^{2}\ mol^{-1}}\)
B.76.3 S \(\mathrm{cm^{2}\ mol^{-1}}\)
C.135.8 S \(\mathrm{cm^{2}\ mol^{-1}}\)
D.228.7 S \(\mathrm{cm^{2}\ mol^{-1}}\)
E.114.35 S \(\mathrm{cm^{2}\ mol^{-1}}\)

Exam Year(s)

KEAM - 2026,

Answer: B

The limiting ionic conductance of Cl^- is 76.3 S cm² mol^-1.

By Kohlrausch's law, the limiting molar conductance of a salt is the sum of the limiting conductances of its ions, each multiplied by the number of those ions in the formula. CaCl2 gives one Ca²⁺ and two Cl^-:

\[ \begin{aligned} \Lambda^{\circ}_{m} &= \lambda^{\circ}_{\text{Ca}^{2+}} + 2\lambda^{\circ}_{\text{Cl}^{-}} \\ 271.6 &= 119 + 2\lambda^{\circ}_{\text{Cl}^{-}} \\ \lambda^{\circ}_{\text{Cl}^{-}} &= 152.6 / 2 = 76.3 \end{aligned} \]

152.6 is the answer you get by forgetting that there are two chloride ions per formula unit.

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33.

What volume of O₂(g) is produced when 2F current is passed through a very dilute aqueous solution of H₂SO₄ at STP?

A.36 \(\text{dm}^{3}\)
B.48.6 \(\text{dm}^{3}\)
C.22.4 \(\text{dm}^{3}\)
D.11.2 \(\text{dm}^{3}\)
E.6 \(\text{dm}^{3}\)

Exam Year(s)

KEAM - 2026,

Answer: D

The volume of oxygen is 11.2 dm³.

At the anode, water is oxidised: \(2\text{H}_{2}\text{O} \rightarrow \text{O}_{2} + 4\text{H}^{+} + 4e^{-}\). One mole of \(\text{O}_{2}\) needs 4 F.

\[ \begin{aligned} n(\text{O}_{2}) &= \frac{2}{4} = 0.5\ \text{mol} \\ V &= 0.5 \times 22.4 = 11.2\ \text{dm}^{3} \end{aligned} \]

22.4 dm³ is the volume of hydrogen released at the cathode by the same 2 F, since each \(\text{H}_{2}\) needs only 2 electrons.

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34.

The standard EMF of the following cell

Mg(s) | Mg⁺² (0.01M) || Ag⁺(0.0001M) | Ag(s) is 3.17 V at 25⁰C. The EMF of the cell at the same temperature is (2.303RT/F at 25⁰C is 0.06V)

A.3.17 V
B.2.81 V
C.3.35 V
D.6.70 V
E.3.0 V

Exam Year(s)

KEAM - 2026,

Answer: E

The EMF of the cell is 3.0 V (2.99 V).

The cell reaction is \(\mathrm{Mg + 2Ag^+ \rightarrow Mg^{2+} + 2Ag}\), so 2 electrons are transferred. Apply the Nernst equation:

\[ \begin{aligned} E &= E^\circ - \frac{0.06}{2}\log\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^+}]^{2}} \\ &= 3.17 - 0.03\log\frac{10^{-2}}{(10^{-4})^{2}} \\ &= 3.17 - 0.03 \times 6 \\ &= 3.17 - 0.18 = 2.99 \approx 3.0 \text{ V} \end{aligned} \]

3.17 V is the standard value and ignores the concentrations. Forgetting to square [Ag⁺] gives \(\log 100 = 2\) and 3.11 V, and adding instead of subtracting the correction gives 3.35 V.

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35.

The mass of Al deposited when 2 ampere current is passed through a solution of Al₂(SO₄)₃ for 1 hour is (F = 96500 C)

A.0.67 g
B.1.34 g
C.0.134 g
D.0.355 g
E.0.067 g

Exam Year(s)

KEAM - 2026,

Answer: A

The mass of aluminium deposited is 0.67 g.

The charge passed is \(Q = It\), and each Al³⁺ ion needs 3 electrons, so 3 F deposit one mole (27 g) of Al.

\[ \begin{aligned} Q &= 2 \times 3600 = 7200 \text{ C} \\ m &= \frac{27 \times 7200}{3 \times 96500} \\ &= \frac{194400}{289500} \approx 0.67 \text{ g} \end{aligned} \]

1.34 g is exactly double the correct mass, and 0.067 g and 0.134 g are factor-of-ten slips; remember that each Al³⁺ ion takes three electrons.

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36.

The concentration of hydrogen ions in a hydrochloric acid solution is 3 x 10^-3 M. Its pH value is about( log3 = 0.4771)

A.2.32
B.2.52
C.2.47
D.3.47
E.5.52

Exam Year(s)

KEAM - 2026,

Answer: B

The pH is about 2.52.

\[ \begin{aligned} \mathrm{pH} &= -\log(3 \times 10^{-3}) \\ &= 3 - \log 3 = 3 - 0.4771 \\ &= 2.5229 \approx 2.52 \end{aligned} \]

3.47 comes from adding \(\log 3\) instead of subtracting it, and 2.32 or 2.47 from rounding slips. A solution with \([\mathrm{H^+}]\) between \(10^{-3}\) and \(10^{-2}\) M must have pH between 2 and 3, which rules out 3.47 and 5.52 at once.

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37.

What is the emf of the cell at 298 K in which the following reaction takes place?
Ni_(s) + 2Ag⁺_(0.002M) → Ni²⁺_(0.04M) + 2Ag_(s). (E°cell at 298 K = 1.05V)

A.1.16 V
B.0.93 V
C.0.73 V
D.0.83 V
E.1.32 V

Exam Year(s)

KEAM - 2026,

Answer: B

The emf is 0.93 V.

Two electrons are transferred (n = 2). By the Nernst equation at 298 K:

\[ \begin{aligned} E &= E^{\circ} - \frac{0.059}{2}\log\frac{[\mathrm{Ni^{2+}}]}{[\mathrm{Ag^{+}}]^{2}} \\ &= 1.05 - 0.0295\log\frac{0.04}{(0.002)^{2}} \\ &= 1.05 - 0.0295 \times 4 = 0.932\ \text{V} \end{aligned} \]

The ratio inside the logarithm is \(\frac{0.04}{4 \times 10^{-6}} = 10^{4}\). Forgetting to square \([\mathrm{Ag^+}]\), or adding the correction, leads to the other values.

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38.

In the reaction, O_2(g) + 4H⁺_(aq) + 4 e^- →2H₂O_(l) , the quantity of electricity needed to reduce one mole of gaseous oxygen is (E°cell = 1.23V)

A.386000 Coulombs
B.96500 Coulombs
C.24125 Coulombs
D.19300 Coulombs
E.579000 Coulombs

Exam Year(s)

KEAM - 2026,

Answer: A

The quantity of electricity is 386000 Coulombs.

The equation shows 4 electrons for each O₂, so reducing one mole of oxygen needs 4 moles of electrons, that is 4 faradays:

\[ Q = 4F = 4 \times 96500 = 386000\ \text{C} \]

The cell potential of 1.23 V is not needed; it would be used to find the energy, not the charge. 96500 C is one faraday, enough for only a quarter of a mole of O₂.

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39.

Which of the following metal is incorporated into the electrodes of fuel cell to increase the rate of reaction?

A.Nickel
B.Zinc
C.Palladium
D.Tin
E.Magnesium

Exam Year(s)

KEAM - 2026,

Answer: C

The metal is palladium.

In a hydrogen-oxygen fuel cell the gases are bubbled through porous carbon electrodes into the electrolyte. Finely divided platinum or palladium is incorporated into the electrodes as a catalyst, to increase the rate of the electrode reactions.

Nickel is used as a catalyst in some alkaline cells and in hydrogenation, but the standard answer for these electrodes is platinum or palladium. Zinc, tin and magnesium are reactive metals that would be consumed, not catalysts.

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40.

The solubility product, Ksp of a sparingly soluble salt, AX₂, is 3.2 × 10^-14 mol³lit^-3. Its solubility is

A.4 × \(10^{-3}\)mol \(lit^{-1}\)
B.2 × \(10^{-2}\)mol \(lit^{-1}\)
C.2 × \(10^{-5}\)mol \(lit^{-1}\)
D.8 × \(10^{-4}\)mol \(lit^{-1}\)
E.2 × \(10^{-3}\)mol \(lit^{-1}\)

Exam Year(s)

KEAM - 2026,

Answer: C

The solubility is \(2 \times 10^{-5}\) mol lit^-1.

For AX₂ dissolving to give A²⁺ and 2X^-, a solubility \(s\) gives \([A^{2+}] = s\) and \([X^{-}] = 2s\):

\[ \begin{aligned} K_{sp} &= s(2s)^{2} = 4s^{3} \\ s^{3} &= \frac{3.2 \times 10^{-14}}{4} = 8 \times 10^{-15} \\ s &= 2 \times 10^{-5} \text{ mol lit}^{-1} \end{aligned} \]

Forgetting the 4 (using \(K_{sp} = s^{3}\)) gives about \(3.2 \times 10^{-5}\), and the \(10^{-3}\) and \(10^{-2}\) options misplace the cube root of the power of ten.

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