Electrochemistry and Ionic Equilibria Questions and Answers

Q: The EMF of the following cell at 298K isMg(s)|Mg2+(aq) (0.10M) || Ag+(aq)(0.001M)| Ag(s)(Given: E0cell=3.17V and 2.303RT/F = 0.06 V)

The EMF is 3.02 V.The cell reaction is Mg + 2Ag+ → Mg2+ + 2Ag, with n = 2. Apply the Nernst equation.\[ \begin{aligned} E &= E^{0} - \frac{0.06}{2}\log\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^{+}}]^{2}} \\ &= 3.17 - 0.03\log\frac{0.10}{(10^{-3})^{2}} \\ &= 3.17 - 0.03 \times 5 = 3.02\ \text{V} \end{aligned} \]3.32 V adds the correction (3.17 + 0.15) instead of subtracting it, and 3.17 V is the standard EMF, which would apply only at unit concentrations. Forgetting to square [Ag+] gives log 100 = 2 and 3.11 V, which is not offered.

Q: The electrolyte used in lead storage battery is

The electrolyte is 38% H2SO4 aqueous solution.A lead storage battery has a lead anode and a lead dioxide cathode dipped in sulphuric acid of about 38% by mass (density about 1.3 g mL-1). During discharge both electrodes turn to PbSO4 and the acid is used up, which is why its density is a measure of the state of charge.Hydrochloric acid is not used, because PbCl2 would form and the cell reactions depend on the sulphate. 10% acid is too dilute and 60% too concentrated for the standard cell.

Q: If "S" is the solubility of X3Y2 in pure water, assuming that neither kind of ion reacts with water, then, the solubility product, Ksp is

\(K_{sp}\) is \(108\,S^{5}\).Each formula unit dissolving gives 3 X ions and 2 Y ions, so a solubility S gives [X] = 3S and [Y] = 2S:\[ K_{sp} = (3S)^{3}(2S)^{2} = 27S^{3} \times 4S^{2} = 108S^{5} \]The options with \(S^{2}\), \(S^{4}\) or \(S^{6}\) use the wrong total number of ions (the power of S is 3 + 2 = 5), and 27 drops the factor 4 from \((2S)^{2}\).

Q: In the following cell reaction,Zn(s) + Cu2+ (0.1 M) ⇋ Zn2+ (0.001 M) + Cu(s) , at 298 K,Calculate the Ecell at 298 K if E0cell at this temperature is 1.1V. (2.303 RT/F = 0.059 V at 298 K)

\(E_{cell}\) is 1.159 V.Two electrons are transferred. By the Nernst equation,\[ \begin{aligned} E &= E^{0} - \frac{0.059}{2}\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \\ &= 1.1 - 0.0295\log\frac{0.001}{0.1} \\ &= 1.1 + 0.059 = 1.159\ \text{V} \end{aligned} \]The low Zn2+ concentration pushes the reaction forward, so E is above \(E^{0}\). 1.041 V gets the sign of the log term wrong, and 1.218 V uses n = 1.Note the half in \(\frac{0.059}{n}\): with n = 2 the log term changes E by 0.0295 V per factor of ten in the concentration ratio, and the ratio here is a hundredfold.

Q: For which of the following electrode reactions the standard electrode potential is the highest at 298 K? The ions are present in aqueous solution.

The highest standard potential is for \(\mathrm{F_{2}(g) + 2e^{-} \rightarrow 2F^{-}}\).Fluorine is the strongest oxidising agent, with \(E^{0} = +2.87\) V. The others, in volts, are Co3+/Co2+ +1.81, Cl2/Cl− +1.36, MnO2/Mn2+ +1.23 and AgCl/Ag +0.22.All are positive, but F2 is the largest, which is why fluorine oxidises everything else on the list.

Q: The pH of a solution is 4 then it's OH- ion concentration (in mol dm-3) is

The hydroxide ion concentration is \(10^{-10}\) mol dm-3.At 25 °C, pH + pOH = 14 (from \(K_{w} = [\mathrm{H^{+}}][\mathrm{OH^{-}}] = 10^{-14}\)):\[ \begin{aligned} \mathrm{pOH} &= 14 - 4 = 10 \\ [\mathrm{OH^{-}}] &= 10^{-10}\ \text{mol dm}^{-3} \end{aligned} \]\(10^{-4}\) is the hydrogen ion concentration, not the hydroxide. The solution is acidic, so \([\mathrm{OH^{-}}]\) must be much smaller than \(10^{-7}\), which also rules out \(10^{-2}\) and \(10^{-6}\).

Q: 4 g of NaOH were dissolved in 1 litre of a solution containing 1 mole of CH3COOH and 1 mole of CH3COONa. The [H+] in the resultant solution is(Given: Ka (CH3COOH) = 1.1×10-5)

The hydrogen ion concentration is \(0.9 \times 10^{-5}\) M.4 g of NaOH is 0.1 mol. It neutralises 0.1 mol of the acetic acid, forming 0.1 mol more acetate, so the buffer now holds 0.9 mol acid and 1.1 mol salt in 1 L. For an acidic buffer:\[ \begin{aligned} [\mathrm{H^{+}}] &= K_{a} \times \frac{[\text{acid}]}{[\text{salt}]} \\ &= 1.1 \times 10^{-5} \times \frac{0.9}{1.1} = 0.9 \times 10^{-5}\ \text{M} \end{aligned} \]The added base lowers \([\mathrm{H^{+}}]\) only slightly from \(1.1 \times 10^{-5}\) M, which is the buffer action. Every other option is above \(1.1 \times 10^{-5}\) M, which would mean adding a base made the solution more acidic.

Q: The conductivity \((k)\) of a decinormal solution of KCl is 0.012 ohm-1 cm-1 . The resistance of a cell containing this solution was found to be 50 ohm at 298 K. The cell constant value is

The cell constant is 0.6 cm-1.Conductivity, conductance and cell constant \(G^{*} = \frac{l}{A}\) are related by \(\kappa = \frac{G^{*}}{R}\), so:\[ \begin{aligned} G^{*} &= \kappa R \\ &= 0.012\ \text{ohm}^{-1}\text{cm}^{-1} \times 50\ \text{ohm} \\ &= 0.6\ \text{cm}^{-1} \end{aligned} \]Dividing instead of multiplying gives \(0.012/50 = 2.4 \times 10^{-4}\), not offered. 0.02, 0.5, 0.8 and 0.1 cm-1 do not follow from these data.

Q: Which of the following compound is used to cover the surface of the metallic object to prevent corrosion?

The compound is Bisphenol.One way of preventing corrosion is to cover the metal surface with paint or with certain chemicals, and the standard textbook example of such a chemical is bisphenol (used in epoxy coatings). The coating keeps air and moisture away from the metal.Phenol, benzene, acetone and nitrophenol do not form protective films; acetone and benzene are volatile solvents.The other ways of preventing corrosion in the same account are galvanising (coating with a more active metal such as zinc) and sacrificial protection with magnesium or zinc.

Q: The pH of an aqueous solution of weak mono acidic base is 11. What is the [H+] of the solution? [pKw = 14]

The hydrogen ion concentration is \(1 \times 10^{-11}\) M.pH is defined from the hydrogen ion concentration whatever the solute, so\[ [\text{H}^{+}] = 10^{-\text{pH}} = 10^{-11}\ \text{M} \]That the solute is a weak base only explains why the pH is above 7. \(10^{-3}\) M is the hydroxide ion concentration, from pOH = 14 - 11 = 3, which is the usual trap. \(10^{-14}\) is \(K_{w}\) itself, and the other values do not follow from pH 11.

21.

The EMF of the following cell at 298K is
Mg_(s)|Mg²⁺(aq) (0.10M) || Ag⁺(aq)(0.001M)| Ag_(s)
(Given: E⁰_cell=3.17V and 2.303RT/F = 0.06 V)

A.3.32V
B.2.96V
C.3.02V
D.3.17V
E.3.47V

Exam Year(s)

KEAM - 2025,

Answer: C

The EMF is 3.02 V.

The cell reaction is Mg + 2Ag⁺ → Mg²⁺ + 2Ag, with n = 2. Apply the Nernst equation.

\[ \begin{aligned} E &= E^{0} - \frac{0.06}{2}\log\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^{+}}]^{2}} \\ &= 3.17 - 0.03\log\frac{0.10}{(10^{-3})^{2}} \\ &= 3.17 - 0.03 \times 5 = 3.02\ \text{V} \end{aligned} \]

3.32 V adds the correction (3.17 + 0.15) instead of subtracting it, and 3.17 V is the standard EMF, which would apply only at unit concentrations. Forgetting to square [Ag⁺] gives log 100 = 2 and 3.11 V, which is not offered.

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22.

The electrolyte used in lead storage battery is

A.10% \(\mathrm{H_{2}SO_{4}}\) aqueous solution
B.60% \(\mathrm{H_{2}SO_{4}}\)aqueous solution
C.38% \(\mathrm{H_{2}SO_{4}}\) aqueous solution
D.38% HCl aqueous solution
E.60% HCl aqueous solution

Exam Year(s)

KEAM - 2025,

Answer: C

The electrolyte is 38% H₂SO₄ aqueous solution.

A lead storage battery has a lead anode and a lead dioxide cathode dipped in sulphuric acid of about 38% by mass (density about 1.3 g mL^-1). During discharge both electrodes turn to PbSO₄ and the acid is used up, which is why its density is a measure of the state of charge.

Hydrochloric acid is not used, because PbCl₂ would form and the cell reactions depend on the sulphate. 10% acid is too dilute and 60% too concentrated for the standard cell.

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23.

If "S" is the solubility of X₃Y₂ in pure water, assuming that neither kind of ion reacts with water, then, the solubility product, K_sp is

A.\(27\ S^{4}\)
B.\(108\ S^{5}\)
C.\(108\ S^{2}\)
D.\(27\ S^{6}\)
E.\(27\ S^{2}\)

Exam Year(s)

KEAM - 2025,

Answer: B

\(K_{sp}\) is \(108\,S^{5}\).

Each formula unit dissolving gives 3 X ions and 2 Y ions, so a solubility S gives [X] = 3S and [Y] = 2S:

\[ K_{sp} = (3S)^{3}(2S)^{2} = 27S^{3} \times 4S^{2} = 108S^{5} \]

The options with \(S^{2}\), \(S^{4}\) or \(S^{6}\) use the wrong total number of ions (the power of S is 3 + 2 = 5), and 27 drops the factor 4 from \((2S)^{2}\).

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24.

In the following cell reaction,

Zn_(s) + Cu²⁺ (0.1 M) ⇋ Zn²⁺ (0.001 M) + Cu_(s) , at 298 K,

Calculate the E_cell at 298 K if E⁰_cell at this temperature is 1.1V. (2.303 RT/F = 0.059 V at 298 K)

A.1.218 V
B.1.118 V
C.1.159 V
D.1.041 V
E.0.982 V

Exam Year(s)

KEAM - 2025,

Answer: C

\(E_{cell}\) is 1.159 V.

Two electrons are transferred. By the Nernst equation,

\[ \begin{aligned} E &= E^{0} - \frac{0.059}{2}\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \\ &= 1.1 - 0.0295\log\frac{0.001}{0.1} \\ &= 1.1 + 0.059 = 1.159\ \text{V} \end{aligned} \]

The low Zn²⁺ concentration pushes the reaction forward, so E is above \(E^{0}\). 1.041 V gets the sign of the log term wrong, and 1.218 V uses n = 1.

Note the half in \(\frac{0.059}{n}\): with n = 2 the log term changes E by 0.0295 V per factor of ten in the concentration ratio, and the ratio here is a hundredfold.

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25.

For which of the following electrode reactions the standard electrode potential is the highest at 298 K? The ions are present in aqueous solution.

A.\(\mathrm{Co^{3+} + e^{-} \rightarrow Co^{2+}}\)
B.\(\mathrm{Cl_{2(g)} + 2e^{-} \rightarrow 2Cl^{-}}\)
C.\(\mathrm{MnO_{2(s)} + 4H^{+} + 2e^{-} \rightarrow Mn^{2+} + 2H_{2}O}\)
D.\(\mathrm{F_{2(g)} + 2e^{-} \rightarrow 2F^{-}}\)
E.\(\mathrm{AgCl_{(s)} + e^{-} \rightarrow Ag_{(s)} + Cl^{-}}\)

Exam Year(s)

KEAM - 2025,

Answer: D

The highest standard potential is for \(\mathrm{F_{2}(g) + 2e^{-} \rightarrow 2F^{-}}\).

Fluorine is the strongest oxidising agent, with \(E^{0} = +2.87\) V. The others, in volts, are Co³⁺/Co²⁺ +1.81, Cl₂/Cl⁻ +1.36, MnO₂/Mn²⁺ +1.23 and AgCl/Ag +0.22.

All are positive, but F₂ is the largest, which is why fluorine oxidises everything else on the list.

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26.

The pH of a solution is 4 then it's OH^- ion concentration (in mol dm^-3) is

A.\(10^{-4}\)
B.\(10^{-10}\)
C.\(10^{-2}\)
D.\(10^{-12}\)
E.\(10^{-6}\)

Exam Year(s)

KEAM - 2025,

Answer: B

The hydroxide ion concentration is \(10^{-10}\) mol dm^-3.

At 25 °C, pH + pOH = 14 (from \(K_{w} = [\mathrm{H^{+}}][\mathrm{OH^{-}}] = 10^{-14}\)):

\[ \begin{aligned} \mathrm{pOH} &= 14 - 4 = 10 \\ [\mathrm{OH^{-}}] &= 10^{-10}\ \text{mol dm}^{-3} \end{aligned} \]

\(10^{-4}\) is the hydrogen ion concentration, not the hydroxide. The solution is acidic, so \([\mathrm{OH^{-}}]\) must be much smaller than \(10^{-7}\), which also rules out \(10^{-2}\) and \(10^{-6}\).

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27.

4 g of NaOH were dissolved in 1 litre of a solution containing 1 mole of CH₃COOH and 1 mole of CH₃COONa. The [H⁺] in the resultant solution is

(Given: Ka (CH₃COOH) = 1.1×10^-5)

A.\(1.47 \times 10^{-5}M\)
B.\(2 \times 10^{-5}M\)
C.\(2.5 \times 10^{-5}M\)
D.\(1.5 \times 10^{-5}M\)
E.\(0.9 \times 10^{-5}M\)

Exam Year(s)

KEAM - 2025,

Answer: E

The hydrogen ion concentration is \(0.9 \times 10^{-5}\) M.

4 g of NaOH is 0.1 mol. It neutralises 0.1 mol of the acetic acid, forming 0.1 mol more acetate, so the buffer now holds 0.9 mol acid and 1.1 mol salt in 1 L. For an acidic buffer:

\[ \begin{aligned} [\mathrm{H^{+}}] &= K_{a} \times \frac{[\text{acid}]}{[\text{salt}]} \\ &= 1.1 \times 10^{-5} \times \frac{0.9}{1.1} = 0.9 \times 10^{-5}\ \text{M} \end{aligned} \]

The added base lowers \([\mathrm{H^{+}}]\) only slightly from \(1.1 \times 10^{-5}\) M, which is the buffer action. Every other option is above \(1.1 \times 10^{-5}\) M, which would mean adding a base made the solution more acidic.

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28.

The conductivity \((k)\) of a decinormal solution of KCl is 0.012 ohm^-1 cm^-1 . The resistance of a cell containing this solution was found to be 50 ohm at 298 K. The cell constant value is

A.0.02 cm\(^{-1}\)
B.0.5 cm\(^{-1}\)
C.0.8 cm\(^{-1}\)
D.0.1 cm\(^{-1}\)
E.0.6 cm\(^{-1}\)

Exam Year(s)

KEAM - 2025,

Answer: E

The cell constant is 0.6 cm^-1.

Conductivity, conductance and cell constant \(G^{*} = \frac{l}{A}\) are related by \(\kappa = \frac{G^{*}}{R}\), so:

\[ \begin{aligned} G^{*} &= \kappa R \\ &= 0.012\ \text{ohm}^{-1}\text{cm}^{-1} \times 50\ \text{ohm} \\ &= 0.6\ \text{cm}^{-1} \end{aligned} \]

Dividing instead of multiplying gives \(0.012/50 = 2.4 \times 10^{-4}\), not offered. 0.02, 0.5, 0.8 and 0.1 cm^-1 do not follow from these data.

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29.

Which of the following compound is used to cover the surface of the metallic object to prevent corrosion?

A.Phenol
B.Benzene
C.Acetone
D.Bisphenol
E.Nitrophenol

Exam Year(s)

KEAM - 2025,

Answer: D

The compound is Bisphenol.

One way of preventing corrosion is to cover the metal surface with paint or with certain chemicals, and the standard textbook example of such a chemical is bisphenol (used in epoxy coatings). The coating keeps air and moisture away from the metal.

Phenol, benzene, acetone and nitrophenol do not form protective films; acetone and benzene are volatile solvents.

The other ways of preventing corrosion in the same account are galvanising (coating with a more active metal such as zinc) and sacrificial protection with magnesium or zinc.

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30.

The pH of an aqueous solution of weak mono acidic base is 11. What is the [H⁺] of the solution? [pKw = 14]

A.1 × \(10^{-11}\) M
B.1 × \(10^{-13}\) M
C.1 × \(10^{-14}\) M
D.1 × \(10^{-3}\) M
E.1 × \(10^{-10}\) M

Exam Year(s)

KEAM - 2026,

Answer: A

The hydrogen ion concentration is \(1 \times 10^{-11}\) M.

pH is defined from the hydrogen ion concentration whatever the solute, so

\[ [\text{H}^{+}] = 10^{-\text{pH}} = 10^{-11}\ \text{M} \]

That the solute is a weak base only explains why the pH is above 7. \(10^{-3}\) M is the hydroxide ion concentration, from pOH = 14 - 11 = 3, which is the usual trap. \(10^{-14}\) is \(K_{w}\) itself, and the other values do not follow from pH 11.

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