Electrochemistry and Ionic Equilibria Questions and Answers

Q: General acid (A) Base (B) titration curve is represented as follows :What should be the equation for estimation of pH at point (L) :

At point L the pH is given by pH = ½ pKa - ½ log [A]initial.Point L is the start of the curve, before any titrant is added, so the solution contains only the weak acid at its initial concentration c. For a weak acid, [H+] = √(Kac), and taking negative logarithms:\[ \mathrm{pH} = \tfrac{1}{2}pK_{a} - \tfrac{1}{2}\log c \]pH = pKa holds at M, the half-equivalence point in the buffer region. pH = ½pKa + ½pKw (plus a concentration term) is used at the equivalence point N, where the salt hydrolyses. The expression with pKw and log [B] belongs to the region past equivalence, where excess base controls the pH.

Q: The stoichiometric point of the titration of 25.0 mL, 0.100 M HClO with 0.10 M NaOH(aq) occurs when the molar concentration of NaClO is 0.050 M. The pH of this solution is about (given pKa of HClO = 7.43; log 0.05 = - 1.301)

The pH is about 10.10.At the stoichiometric point the solution is 0.050 M NaClO, and ClO- is a weak base. Its pKb = 14 - 7.43 = 6.57, and for a weak base pOH = ½(pKb - log c):\[ \begin{aligned} \mathrm{pOH} &= \tfrac{1}{2}(6.57 + 1.301) = 3.94 \\ \mathrm{pH} &= 14 - 3.94 = 10.06 \end{aligned} \]That is about 10.1, so 10.10 is the option meant. 3.94 is the pOH, not the pH, and 2.63 and 11.36 do not follow from the data. A titration of a weak acid with a strong base must have an equivalence point above 7, which also rules out the two acidic values.

Q: The pH of 0.2M HCN(aq.) is (pKa of HCN = 9.31; log10 2 = 0.3010 and log10 0.2 = - 0.6990) :

The pH of 0.2 M HCN is 5.0.HCN is a very weak acid, so the weak-acid approximation [H+] = √(Kac) applies:\[ \begin{aligned} \mathrm{pH} &= \tfrac{1}{2}(\mathrm{p}K_{a} - \log c) \\ &= \tfrac{1}{2}(9.31 + 0.699) \\ &= \tfrac{1}{2}(10.009) = 5.0 \end{aligned} \]9.6 and 8.6 are above 7 and cannot be the pH of an acid solution. 4.3 comes from a sign slip in the log term, ½(9.31 - 0.699).

Q: Consider the following electrochemical cell, connected to the potential source (Potentiostat) with Hg as a working electrode :Hg/Cd2+(0.01M), Zn2+ (0.01M), HCl(1.0M) || HCl(1.0M)| H2(g)|Ptwhen the potential is change towards (i) more negative direction and (ii) change towards the more positive direction from the equilibrium potential value, the following observations are expected :Given :E°(Cd2+/Cd°) = − 0.351 V,E°(Zn2+/Zn°) = − 0.762 V, E°(Hg2+/Hg°) = + 0.855 V

The expected observations are (i) Cd deposits first, followed by Zn; (ii) Hg starts oxidising.On the negative sweep, the ion with the less negative reduction potential is reduced first: Cd2+ (E° = −0.351 V) before Zn2+ (−0.762 V), both depositing into the mercury as amalgams. Hydrogen evolution, which would otherwise come first in 1 M HCl, is suppressed by the very high hydrogen overvoltage on mercury, which is why mercury is used as the working electrode in polarography.On the positive sweep, the electrode itself is oxidised long before water: mercury dissolves (in chloride medium as Hg2Cl2), which limits the anodic range of a mercury electrode, so O2 evolution (A, C) is not reached. Option D leaves out the later deposition of zinc.

Q: An electrochemical cell is constructed using half cell for the reduction reaction is given by :Fe(OH2)(s) + 2e → Fe(s) + 2OH−(aq); E° = −0.877 VIt is combined with another half cell for which the cell reaction is given by :Al3+(aq) + 3e → Al(s); E° = −1.66 VActivities of all the components are assumed to be unity.The overall reaction for the cell in the direction of spontaneous change is :

The spontaneous overall reaction is 3Fe(OH)2(s) + 2Al(s) → 3Fe(s) + 6OH−(aq) + 2Al3+(aq).The half-reaction with the more positive reduction potential, Fe(OH)2/Fe at −0.877 V, is the cathode, and the one with the more negative potential, Al3+/Al at −1.66 V, runs as oxidation at the anode. The cell e.m.f. is\[ E^{\circ} = -0.877 - (-1.66) = +0.783\ \text{V} \]Balancing electrons, 3 Fe(OH)2 take up 6 electrons and 2 Al give up 6, which gives option C. Option D is the reverse, non-spontaneous reaction. Options A and B have the right direction but are not balanced in electrons or charge (A transfers 2 electrons against 3, B 4 against 9).

Q: During the electrolysis of an aqueous solution of KCl which susbtance is formed at the ................ cathode.

The substance formed at the cathode is Hydrogen.At the cathode, reduction occurs. Potassium ions are very hard to reduce (E° about -2.93 V), whereas water is reduced much more easily: 2H2O + 2e- → H2 + 2OH-. So hydrogen gas is evolved and the solution around the cathode becomes alkaline (KOH).Potassium metal is deposited only from molten KCl, not from aqueous solution. Chlorine and, at very low chloride concentration, oxygen are oxidation products formed at the anode.

Q: 100 ml of a NaOH solution of pH 12 is mixed with 900 ml of water. What is the pH of the resulting solution ?

The pH becomes 11.A pH of 12 means pOH = 2, so [OH−] = 10−2 M. Adding 900 mL of water to 100 mL makes 1000 mL, a tenfold dilution:\[ \begin{aligned} [\mathrm{OH^{-}}] &= 10^{-2}/10 = 10^{-3}\ \mathrm{M} \\ \mathrm{pOH} &= 3,\quad \mathrm{pH} = 14 - 3 = 11 \end{aligned} \]A tenfold dilution moves the pH one unit toward 7. pH 13 moves it the wrong way, and 10 and 9 correspond to 100- and 1000-fold dilutions.

Q: According to the following half cell reactions :Cu2+(aq) + 2e− → Cu(s), E0 = 0.34 VCr3+(aq) + e− → Cr2+(aq) E0 = −0.41 VThe reaction that would occur in aqueous solutions under standard condition is :

The reaction that occurs is Cu2+ + 2Cr2+ → Cu + 2Cr3+.The couple with the higher reduction potential, Cu2+/Cu (+0.34 V), is reduced, and the couple with the lower potential, Cr3+/Cr2+ (−0.41 V), runs in reverse as the oxidation. So Cr2+ reduces Cu2+:\[ \begin{aligned} E^{\circ}_{\mathrm{cell}} &= 0.34 - (-0.41) \\ &= +0.75\ \mathrm{V} \end{aligned} \]A positive cell potential means the reaction is spontaneous, and the balanced equation needs two Cr2+ for each Cu2+. Options A and D reduce both Cu2+ and Cr3+, so they have no oxidation half, and option C runs the cell backwards.

Q: When 10 mL of 0.1M NaOH is added to 10 mL of 0.1M HCl, pH of the solution will be :

The pH will be 7.Both solutions contain the same amount of a strong monoprotic acid or base:\[ \begin{aligned} n &= 0.1\ \mathrm{mol\,L^{-1}} \times 0.010\ \mathrm{L} \\ &= 1.0 \times 10^{-3}\ \mathrm{mol} \end{aligned} \]So 1.0 mmol of OH− exactly neutralizes 1.0 mmol of H+, leaving 20 mL of sodium chloride solution. Na+ and Cl− do not hydrolyse, so the solution is neutral and, at 25 °C, its pH is 7.pH 2 would need 0.01 M acid left over, pH 0 a 1 M acid, and pH 14 a 1 M base; none is left once the two equal amounts react.

Q: In the electrolysis of an aqueous solution of HBr, of the following products are formed :

The products are H2 at the cathode and Br2 at the anode.At the cathode (reduction) hydrogen ions are reduced:\[ \mathrm{2H^{+} + 2e^{-} \rightarrow H_{2}} \]At the anode (oxidation) bromide ions are oxidized in preference to water, because the Br2/Br− potential (+1.07 V) is lower than that for oxidizing water to oxygen (+1.23 V, with a sizeable overpotential besides):\[ \mathrm{2Br^{-} \rightarrow Br_{2} + 2e^{-}} \]Oxygen is an oxidation product and hydrogen a reduction product, so the first option has them at the wrong electrodes. Producing HOBr or HBrO4 would need much stronger oxidation than occurs, and Br3− cannot form at the cathode, where species are reduced.

51.

General acid (A) Base (B) titration curve is represented as follows :

Sketch of pH against titrant added: the curve starts at point L on the pH axis, rises gently to M, climbs steeply through N and levels off at O

What should be the equation for estimation of pH at point (L) :

A.\(\mathrm{pH = pK_{w} + \log\,[B]}\)
B.\(\mathrm{pH = \frac{1}{2}pK_{a} + \frac{1}{2}pK_{w}}\)
C.\(\mathrm{pH = \frac{1}{2}pK_{a} - \frac{1}{2}\log\,[A]_{initial}}\)
D.\(\mathrm{pH = pK_{a}}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

At point L the pH is given by pH = ½ pK_a - ½ log [A]_initial.

Point L is the start of the curve, before any titrant is added, so the solution contains only the weak acid at its initial concentration c. For a weak acid, [H⁺] = √(K_ac), and taking negative logarithms:

\[ \mathrm{pH} = \tfrac{1}{2}pK_{a} - \tfrac{1}{2}\log c \]

pH = pK_a holds at M, the half-equivalence point in the buffer region. pH = ½pK_a + ½pK_w (plus a concentration term) is used at the equivalence point N, where the salt hydrolyses. The expression with pK_w and log [B] belongs to the region past equivalence, where excess base controls the pH.

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52.

The stoichiometric point of the titration of 25.0 mL, 0.100 M HClO with 0.10 M NaOH_(aq) occurs when the molar concentration of NaClO is 0.050 M. The pH of this solution is about (given pK_a of HClO = 7.43; log 0.05 = - 1.301)

A.2.63
B.3.94
C.10.10
D.11.36

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The pH is about 10.10.

At the stoichiometric point the solution is 0.050 M NaClO, and ClO^- is a weak base. Its pK_b = 14 - 7.43 = 6.57, and for a weak base pOH = ½(pK_b - log c):

\[ \begin{aligned} \mathrm{pOH} &= \tfrac{1}{2}(6.57 + 1.301) = 3.94 \\ \mathrm{pH} &= 14 - 3.94 = 10.06 \end{aligned} \]

That is about 10.1, so 10.10 is the option meant. 3.94 is the pOH, not the pH, and 2.63 and 11.36 do not follow from the data. A titration of a weak acid with a strong base must have an equivalence point above 7, which also rules out the two acidic values.

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53.

The pH of 0.2M HCN_(aq.) is (pK_a of HCN = 9.31; log₁₀ 2 = 0.3010 and log₁₀ 0.2 = - 0.6990) :

A.5.0
B.9.6
C.8.6
D.4.3

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The pH of 0.2 M HCN is 5.0.

HCN is a very weak acid, so the weak-acid approximation [H⁺] = √(K_ac) applies:

\[ \begin{aligned} \mathrm{pH} &= \tfrac{1}{2}(\mathrm{p}K_{a} - \log c) \\ &= \tfrac{1}{2}(9.31 + 0.699) \\ &= \tfrac{1}{2}(10.009) = 5.0 \end{aligned} \]

9.6 and 8.6 are above 7 and cannot be the pH of an acid solution. 4.3 comes from a sign slip in the log term, ½(9.31 - 0.699).

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54.

Consider the following electrochemical cell, connected to the potential source (Potentiostat) with Hg as a working electrode :

Hg/Cd²⁺(0.01M), Zn²⁺ (0.01M), HCl(1.0M) || HCl(1.0M)| H₂(g)|Pt

when the potential is change towards (i) more negative direction and (ii) change towards the more positive direction from the equilibrium potential value, the following observations are expected :

Given :
E°(Cd²⁺/Cd°) = − 0.351 V,
E°(Zn²⁺/Zn°) = − 0.762 V, E°(Hg²⁺/Hg°) = + 0.855 V

A.(i) H\(_{2}\) evolution will start. (ii) O\(_{2}\) evolution will start.
B.(i) Cd will deposit first followed by Zn. (ii) Hg will start oxidizing.
C.(i) Zn will deposit. (ii) O\(_{2}\) evolution will be observed.
D.(i) Cd will deposit. (ii) Hg will start oxidizing.

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The expected observations are (i) Cd deposits first, followed by Zn; (ii) Hg starts oxidising.

On the negative sweep, the ion with the less negative reduction potential is reduced first: Cd²⁺ (E° = −0.351 V) before Zn²⁺ (−0.762 V), both depositing into the mercury as amalgams. Hydrogen evolution, which would otherwise come first in 1 M HCl, is suppressed by the very high hydrogen overvoltage on mercury, which is why mercury is used as the working electrode in polarography.

On the positive sweep, the electrode itself is oxidised long before water: mercury dissolves (in chloride medium as Hg₂Cl₂), which limits the anodic range of a mercury electrode, so O₂ evolution (A, C) is not reached. Option D leaves out the later deposition of zinc.

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55.

An electrochemical cell is constructed using half cell for the reduction reaction is given by :

Fe(OH₂)(s) + 2e → Fe(s) + 2OH⁻(aq); E° = −0.877 V

It is combined with another half cell for which the cell reaction is given by :

Al³⁺(aq) + 3e → Al(s); E° = −1.66 V

Activities of all the components are assumed to be unity.

The overall reaction for the cell in the direction of spontaneous change is :

A.(Fe(OH)\(_{2}\)(s) + Al(s) \(\rightarrow\) Fe(s) + OH\(^{-}\)(aq.) + Al\(^{3+}\) (aq.)
B.2(Fe(OH)\(_{2}\)(s) + 3Al(s) \(\rightarrow\) 2Fe(s) + 6OH\(^{-}\)(aq.) + 3Al\(^{3+}\) (aq.)
C.3(Fe(OH)\(_{2}\)(s) + 2Al(s) \(\rightarrow\) 3Fe(s) + 6OH\(^{-}\)(aq.) + 2Al\(^{3+}\) (aq.)
D.Fe(s) + OH\(^{-}\)(aq.) + Al\(^{3+}\)(aq.) \(\rightarrow\) (Fe(OH)\(_{2}\)(s) + Al(s)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The spontaneous overall reaction is 3Fe(OH)₂(s) + 2Al(s) → 3Fe(s) + 6OH⁻(aq) + 2Al³⁺(aq).

The half-reaction with the more positive reduction potential, Fe(OH)₂/Fe at −0.877 V, is the cathode, and the one with the more negative potential, Al³⁺/Al at −1.66 V, runs as oxidation at the anode. The cell e.m.f. is

\[ E^{\circ} = -0.877 - (-1.66) = +0.783\ \text{V} \]

Balancing electrons, 3 Fe(OH)₂ take up 6 electrons and 2 Al give up 6, which gives option C. Option D is the reverse, non-spontaneous reaction. Options A and B have the right direction but are not balanced in electrons or charge (A transfers 2 electrons against 3, B 4 against 9).

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56.

During the electrolysis of an aqueous solution of KCl which susbtance is formed at the ................ cathode.

A.Chlorine
B.Hydrogen
C.Oxygen
D.Potassium

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The substance formed at the cathode is Hydrogen.

At the cathode, reduction occurs. Potassium ions are very hard to reduce (E° about -2.93 V), whereas water is reduced much more easily: 2H₂O + 2e^- → H₂ + 2OH^-. So hydrogen gas is evolved and the solution around the cathode becomes alkaline (KOH).

Potassium metal is deposited only from molten KCl, not from aqueous solution. Chlorine and, at very low chloride concentration, oxygen are oxidation products formed at the anode.

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57.

100 ml of a NaOH solution of pH 12 is mixed with 900 ml of water. What is the pH of the resulting solution ?

A.10
B.09
C.11
D.13

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The pH becomes 11.

A pH of 12 means pOH = 2, so [OH⁻] = 10⁻² M. Adding 900 mL of water to 100 mL makes 1000 mL, a tenfold dilution:

\[ \begin{aligned} [\mathrm{OH^{-}}] &= 10^{-2}/10 = 10^{-3}\ \mathrm{M} \\ \mathrm{pOH} &= 3,\quad \mathrm{pH} = 14 - 3 = 11 \end{aligned} \]

A tenfold dilution moves the pH one unit toward 7. pH 13 moves it the wrong way, and 10 and 9 correspond to 100- and 1000-fold dilutions.

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58.

According to the following half cell reactions :

Cu²⁺_(aq) + 2e⁻ → Cu_(s), E⁰ = 0.34 V
Cr³⁺_(aq) + e⁻ → Cr²⁺_(aq) E⁰ = −0.41 V

The reaction that would occur in aqueous solutions under standard condition is :

A.\(\mathrm{Cu^{2+}_{(aq)} + Cr^{3+}_{(aq)} \rightarrow Cu_{(s)} + Cr^{2+}_{(aq)}}\)
B.\(\mathrm{Cu^{2+}_{(aq)} + 2Cr^{2+}_{(aq)} \rightarrow Cu_{(s)} + 2Cr^{3+}_{(aq)}}\)
C.\(\mathrm{Cu_{(s)} + 2Cr^{3+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cr^{2+}_{(aq)}}\)
D.\(\mathrm{2Cu^{2+}_{(aq)} + Cr^{3+}_{(aq)} \rightarrow 2Cu_{(s)} + Cr^{2+}_{(aq)}}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The reaction that occurs is Cu²⁺ + 2Cr²⁺ → Cu + 2Cr³⁺.

The couple with the higher reduction potential, Cu²⁺/Cu (+0.34 V), is reduced, and the couple with the lower potential, Cr³⁺/Cr²⁺ (−0.41 V), runs in reverse as the oxidation. So Cr²⁺ reduces Cu²⁺:

\[ \begin{aligned} E^{\circ}_{\mathrm{cell}} &= 0.34 - (-0.41) \\ &= +0.75\ \mathrm{V} \end{aligned} \]

A positive cell potential means the reaction is spontaneous, and the balanced equation needs two Cr²⁺ for each Cu²⁺. Options A and D reduce both Cu²⁺ and Cr³⁺, so they have no oxidation half, and option C runs the cell backwards.

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59.

When 10 mL of 0.1M NaOH is added to 10 mL of 0.1M HCl, pH of the solution will be :

A.2
B.7
C.0
D.14

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The pH will be 7.

Both solutions contain the same amount of a strong monoprotic acid or base:

\[ \begin{aligned} n &= 0.1\ \mathrm{mol\,L^{-1}} \times 0.010\ \mathrm{L} \\ &= 1.0 \times 10^{-3}\ \mathrm{mol} \end{aligned} \]

So 1.0 mmol of OH⁻ exactly neutralizes 1.0 mmol of H⁺, leaving 20 mL of sodium chloride solution. Na⁺ and Cl⁻ do not hydrolyse, so the solution is neutral and, at 25 °C, its pH is 7.

pH 2 would need 0.01 M acid left over, pH 0 a 1 M acid, and pH 14 a 1 M base; none is left once the two equal amounts react.

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60.

In the electrolysis of an aqueous solution of HBr, of the following products are formed :

A.\(\mathrm{O_{2}}\) at the cathode, \(\mathrm{H_{2}}\) at the anode
B.\(\mathrm{H_{2}}\) at the cathode, \(\mathrm{Br_{2}}\) at the anode
C.\(\mathrm{OH^{-}}\) at the cathode, HOBr at the anode
D.\(\mathrm{Br_{3}^{-}}\) at the cathode, \(\mathrm{HBrO_{4}}\) at the anode

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The products are H₂ at the cathode and Br₂ at the anode.

At the cathode (reduction) hydrogen ions are reduced:

\[ \mathrm{2H^{+} + 2e^{-} \rightarrow H_{2}} \]

At the anode (oxidation) bromide ions are oxidized in preference to water, because the Br₂/Br⁻ potential (+1.07 V) is lower than that for oxidizing water to oxygen (+1.23 V, with a sizeable overpotential besides):

\[ \mathrm{2Br^{-} \rightarrow Br_{2} + 2e^{-}} \]

Oxygen is an oxidation product and hydrogen a reduction product, so the first option has them at the wrong electrodes. Producing HOBr or HBrO₄ would need much stronger oxidation than occurs, and Br₃⁻ cannot form at the cathode, where species are reduced.

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