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Electrochemistry

JEE Chemistry · 118 questions · Page 12 of 12 · Click an option or "Show Solution" to reveal answer

Q111
When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are :
A cathode = pure zinc, anode = pure copper
B cathode = impure sample, anode = pure copper
C cathode = impure zinc, anode = impure sample
D cathode = pure copper, anode = impure sample
Correct Answer
Option D
Solution

Pure metal always deposits at cathode.

Q112
The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.
A ESn4+/Sn2+o=+1.15 VE^o_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +1.15 \text{ V}
B EAl3+/Alo=−1.66 VE^o_{\text{Al}^{3+}/\text{Al}} = -1.66 \text{ V}
C EPb4+/Pb2+o=+1.67 VE^o_{\text{Pb}^{4+}/\text{Pb}^{2+}} = +1.67 \text{ V}
D ETl3+/Tlo=+1.26 VE^o_{\text{Tl}^{3+}/\text{Tl}} = +1.26 \text{ V}
Correct Answer
Option C
Solution

Oxidizing capacity is an element's tendency to donate electrons, or the atmosphere's ability to oxidize compounds.

A higher standard reduction potential indicates a stronger oxidizing capacity.

A substance with a higher standard reduction potential is move likely to gain electrons and act as an oxidizing agent in a redon reaction.

The more positive the reduction potential, the stronger the oxidizing power of a species.

From the given options, the higher standard reduction potential is for lead (Pb).(+1.67)(\mathrm{Pb}) .(+1.67). so, the p-block ion with strongest oxidizing capacity is Pb (lead).

Answer: Option 3) Epb4+/Pb2+0=+1.67 VE_{p_b{ }^{4+} / \mathrm{Pb}^{2+}}^0=+1.67 \mathrm{~V}

Q113
O2\mathrm{O}_2 gas will be evolved as a product of electrolysis of : (A) an aqueous solution of AgNO3\mathrm{AgNO}_3 using silver electrodes. (B) an aqueous solution of AgNO3\mathrm{AgNO}_3 using platinum electrodes. (C) a dilute solution of H2SO4\mathrm{H}_2 \mathrm{SO}_4 using platinum electrodes. (D) a high concentration solution of H2SO4\mathrm{H}_2 \mathrm{SO}_4 using platinum electrodes. Choose the correct answer from the options given below :
A (B) and (C) only
B (B) and (D) only
C (A) and (D) only
D (A)(A) and (C)(C) only
Correct Answer
Option A
Solution

(A) An aqueous solution of AgNO3_3 using silver electrodes. Cathode - Reduction -

Ag(aq)++e−→Ag(s)Ag_{(aq)}^ + + {e^ - } \to Ag(s)

Anode - Oxidation -

Ag(s)→Ag(aq)++e−Ag(s) \to Ag_{(aq)}^ + + {e^ - }

Solid silver will be deposited at the cathode.

Solid anode (silver) will dissolve, releasing silver ions into the solution.

So, there is no formation of O2_2 gas in this electrolysis.

(B) An aqueous solution of AgNO3_3 using platinum electrodes.

Cathode - Reduction -

Ag(aq)++e−→Ag(s)Ag_{(aq)}^ + + {e^ - } \to Ag(s)

Anode - Oxidation -

2H2O→4H++O2+4e−2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - }

When platinum electrodes are used, Ag+^+ from solution is reduced and deposited at cathode whereas O2_2 is produced at the anode.

(C) A dilute soution of H2_2SO4_4 using platinum electrodes.

Cathode - Reduction -

2H++2e−→H22{H^ + } + 2{e^ - } \to {H_2}

Anode - Oxidation -

2H2O→O2+4H++4e−2{H_2}O \to {O_2} + 4{H^ + } + 4{e^ - }

H2_2 gas is produded at the cathode and O2_2 gas is produced at the anode.

(D) a high concentration solution of H2_2SO4_4 using platinum electrodes.

O2_2 gas is not formed in this case.

Cathode - Reduction - The substance formed is H2_2 gas.

Anode - Oxidation - The substance formed is not O2_2 gas.

So, statements (B) and (C) are correct.

Q114
EMF of a cell in terms of reduction potential of its left and right electrodes is :
A E = Eleft - Eright
B E = Eleft + Eright
C E = Eright - Eleft
D E = -(Eright + Eleft)
Correct Answer
Option C
Solution
Ecell=  {E_{cell}} = \,\,

Reduction potential of cathode (right)

                        \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

−- Reduction potential of anode (left)

=Eright−Eleft.= {E_{right}} - {E_{left}}.
Q115
The standard electrode potential Eo{E^o } and its temperature coefficient (dEodT)\left( {{{d{E^o }} \over {dT}}} \right) for a cell are 2V and −- 5 × \times 10−-4 VK−-1 at 300 K respectively. The cell reaction is Zn(s) + Cu2+ (aq) ⟶ \overset{\,}\longrightarrow Zn2+ (aq) + Cu(s) The standard reaction enthalpy (Δ\Delta rHo{^o }) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1]
A −- 412.8
B −- 384.0
C 192.0
D 206.4
Correct Answer
Option A
Solution
Δ\Delta

G = -nFEcell = -2×\times96500×\times2 = -386 kJ

Δ\Delta

S = nF

(dEodT)\left( {{{d{E^o }} \over {dT}}} \right)

= 2×\times96500×\times (−- 5 ×\times 10−-4) = -96.5 kJ At 298 K T

Δ\Delta

S = 298 ×\times (–96.5 J) = – 28.8 kJ at constant T (=248 K) and pressure

Δ\Delta

G =

Δ\Delta

H – T

Δ\Delta

S ⇒\Rightarrow

Δ\Delta

H =

Δ\Delta

G + T

Δ\Delta

S = -386 - 28.8 = -412.8 kJ

Q116
The reaction at cathode in the cells commonly used in clocks involves.
A oxidation of Mn\mathrm{Mn} from +2 to +7
B reduction of Mn\mathrm{Mn} from +4 to +3
C oxidation of Mn\mathrm{Mn} from +3 to +4
D reduction of Mn\mathrm{Mn} from +7 to +2
Correct Answer
Option B
Solution

In the cells commonly used in clocks, specifically alkaline batteries, the reaction at the cathode involves the reduction of manganese dioxide (MnO2).

The manganese in MnO2 is initially in the +4 oxidation state, and it gets reduced to the +3 oxidation state.

Thus, the correct reaction occurring at the cathode can be represented as follows: The correct answer is: Option B reduction of

Mn\mathrm{Mn}

from +4 to +3

Q117
For the electro chemical cell M∣M2+∣∣X∣X2−\mathrm{M}\left|\mathrm{M}^{2+}\right||\mathrm{X}| \mathrm{X}^{2-} If E(M2+/M)0=0.46 V\mathrm{E}_{\left(\mathrm{M}^{2+} / \mathrm{M}\right)}^0=0.46 \mathrm{~V} and E(x/x2−)0=0.34 V\mathrm{E}_{\left(\mathrm{x} / \mathrm{x}^{2-}\right)}^0=0.34 \mathrm{~V}. Which of the following is correct?
A Ecell=0.80 V\mathrm{E}_{\mathrm{cell}}=0.80 \mathrm{~V}
B M+X→M2++X2−\mathrm{M}+\mathrm{X} \rightarrow \mathrm{M}^{2+}+\mathrm{X}^{2-} is a spontaneous reaction
C Ecell =−0.80 V\mathrm{E}_{\text{cell }}=-0.80 \mathrm{~V}
D M2++X2−→M+X\mathrm{M}^{2+}+\mathrm{X}^{2-} \rightarrow \mathrm{M}+\mathrm{X} is a spontaneous reaction
Correct Answer
Option D
Solution
Ecell∘=EX∣X2−∘−EM2+∣M∘=0.34−0.46=−0.12 V (Non-spontaneous) \begin{aligned} & \mathrm{E}_{\mathrm{cell}}^{\circ}=\mathrm{E}_{\mathrm{X} \mid \mathrm{X}^{2-}}^{\circ}-\mathrm{E}_{\mathrm{M}^{2+} \mid \mathrm{M}}^{\circ} \\ & =0.34-0.46 \\ & =-0.12 \mathrm{~V} \text{ (Non-spontaneous) } \end{aligned}

So, reverse reaction will be spontaneous.

M2++X2−→M+X\mathrm{M}^{2+}+\mathrm{X}^{2-} \rightarrow \mathrm{M}+\mathrm{X}
Q118
Which out of the following is a correct equation to show change in molar conductivity with respect to concentration for a weak electrolyte, if the symbols carry their usual meaning :
A Λm−Λm∘+AC12=0\Lambda_{\mathrm{m}}-\Lambda_{\mathrm{m}}^{\circ}+\mathrm{AC}^{\dfrac{1}{2}}=0
B Λm2C+KaΛmo2−KaΛmΛm∘=0\Lambda_{\mathrm{m}}^2 \mathrm{C}+\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}^2}-\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ}=0
C Λm−Λm∘−AC12=0\Lambda_{\mathrm{m}}-\Lambda_{\mathrm{m}}^{\circ}-\mathrm{AC}^{\dfrac{1}{2}}=0
D Λm2C−KaΛm∘2+KaΛmΛm∘=0\Lambda_{\mathrm{m}}^2 \mathrm{C}-\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ 2}+\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ}=0
Correct Answer
Option D
Solution

The dissociation of a weak electrolyte (HA\mathrm{HA}) in solution can be represented as: HA(aq)⇌H+(aq)+A−(aq) \mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{A}^{-}(\mathrm{aq}) The degree of dissociation (α\alpha) for a weak electrolyte is given by: α=ΛmΛm∘ \alpha = \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} Where Λm\Lambda_{\mathrm{m}} is the molar conductivity at a given concentration CC, and Λm∘\Lambda_{\mathrm{m}}^{\circ} is the molar conductivity at infinite dilution.

The dissociation constant (KaK_{\mathrm{a}}) is described by: Ka=α2C1−α K_{\mathrm{a}} = \dfrac{\alpha^2 C}{1 - \alpha} Rewriting this equation in terms of Λm\Lambda_{\mathrm{m}} and Λm∘\Lambda_{\mathrm{m}}^{\circ}: Ka=(ΛmΛm∘)2C1−ΛmΛm∘ K_{\mathrm{a}} = \dfrac{\left(\dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right)^2 C}{1 - \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}} By multiplying out and simplifying the above expression: (ΛmΛm∘)2C+Ka(ΛmΛm∘)=Ka \left( \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \right)^2 C + K_{\mathrm{a}} \left( \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \right) = K_{\mathrm{a}} Substitute α=ΛmΛm∘\alpha = \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}: (ΛmΛm∘)2C+Ka(ΛmΛm∘)−Ka=0 \left(\dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right)^2 C + K_{\mathrm{a}} \left(\dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right) - K_{\mathrm{a}} = 0 Rewriting by multiplying through by (Λm∘)2\left( \Lambda_{\mathrm{m}}^{\circ} \right)^2: Λm2C+KaΛmΛm∘−Ka(Λm∘)2=0 \Lambda_{\mathrm{m}}^2 C + K_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ} - K_{\mathrm{a}} \left( \Lambda_{\mathrm{m}}^{\circ} \right)^2 = 0 Thus, the correct equation that describes the change in molar conductivity with respect to concentration for a weak electrolyte is given by: Option D: Λm2C−KaΛm∘2+KaΛmΛm∘=0 \Lambda_{\mathrm{m}}^2 C - K_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ 2} + K_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ} = 0

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