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JEE Advanced · Chemistry · Electrochemistry

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1
The molar conductivities (Λm\Lambda_m, in S cm2^2 mol−1^{-1}) of aqueous NaNO3_3, NaCl and AgNO3_3 were measured at two concentrations: NaNO3\textbf{NaNO}_3: 111111 at 0.01 0.01\,M, 101101 at 0.04 0.04\,M NaCl\textbf{NaCl}: 117117 at 0.01 0.01\,M, 107107 at 0.04 0.04\,M AgNO3\textbf{AgNO}_3: 125125 at 0.01 0.01\,M, 116116 at 0.04 0.04\,M The specific conductance (κ\kappa) of a saturated solution of AgCl in water at the same temperature is 1.40×10−6 1.40\times10^{-6}\,S cm−1^{-1}. If the solubility of AgCl is XX mol L−1^{-1}, the value of log⁡10(X−1)\log_{10}(X^{-1}) is ______.
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Your answer: —
Correct answer: 5
2
In an electrochemical cell, dichromate ions in aqueous acidic medium are reduced to Cr3+Cr^{3+}. The current (in amperes) that flows through the cell for 48.2548.25 minutes to produce 11 mole of Cr3+Cr^{3+} is ______. Use: 11 Faraday =96500 C mol−1= 96500\ C\,mol^{-1}
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Your answer: —
Correct answer: 100
3
An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is XF×103\dfrac{X}{F}\times10^3 volts, where FF is the Faraday constant. The value of XX is ______. Use: Standard Gibbs energies of formation at 298 K are: ΔfGCO2∘=−394 kJ mol−1\Delta_f G^{\circ}_{\text{CO}_2} = -394\ \text{kJ mol}^{-1}; ΔfGwater∘=−237 kJ mol−1\Delta_f G^{\circ}_{\text{water}} = -237\ \text{kJ mol}^{-1}; ΔfGbutane∘=−18 kJ mol−1\Delta_f G^{\circ}_{\text{butane}} = -18\ \text{kJ mol}^{-1}
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Your answer: —
Correct answer: 105.5
4
An aqueous solution of hydrazine (N2H4N_2H_4) is electrochemically oxidized by O2O_2, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is N2(g)N_2(g). Choose the correct statement(s) about the above process:
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A
OH−OH^- ions react with N2H4N_2H_4 at the anode to form N2(g)N_2(g) and water, releasing 4 electrons to the anode.
B
At the cathode, N2H4N_2H_4 breaks to N2(g)N_2(g) and nascent hydrogen released at the electrode reacts with oxygen to form water.
C
At the cathode, molecular oxygen gets converted to OH−OH^-.
D
Oxides of nitrogen are major by-products of the electrochemical process.
5
In a conductometric titration, a small volume of titrant of higher concentration is added stepwise to a much larger volume of titrate of much lower concentration (so dilution effects can be neglected), and the conductance is measured after each addition. The limiting ionic conductivities Λ0\Lambda^0 (in mS m2 mol−1mS\,m^2\,mol^{-1}) are: Ag+=6.2Ag^{+}=6.2, K+=7.4K^{+}=7.4, Na+=5.0Na^{+}=5.0, H+=35.0H^{+}=35.0, NO3−=7.2NO_3^{-}=7.2, Cl−=7.6Cl^{-}=7.6, SO42−=16.0SO_4^{2-}=16.0, OH−=19.9OH^{-}=19.9, CH3COO−=4.1CH_3COO^{-}=4.1. List-I: (P) Titrate KClKCl, Titrant AgNO3AgNO_3; (Q) Titrate AgNO3AgNO_3, Titrant KClKCl; (R) Titrate NaOHNaOH, Titrant HClHCl; (S) Titrate NaOHNaOH, Titrant CH3COOHCH_3COOH. List-II: conductance-vs-volume-of-titrant graphs (1)-(5), shown in the figure. Match List-I with List-II and choose the correct option. (A) P-4, Q-3, R-2, S-5 (B) P-2, Q-4, R-3, S-1 (C) P-3, Q-4, R-2, S-5 (D) P-4, Q-3, R-2, S-1
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(1)ConductanceVolume of titrant (2)ConductanceVolume of titrant (3)ConductanceVolume of titrant (4)ConductanceVolume of titrant (5)ConductanceVolume of titrant
A
P-4, Q-3, R-2, S-5
B
P-2, Q-4, R-3, S-1
C
P-3, Q-4, R-2, S-5
D
P-4, Q-3, R-2, S-1
6
Plotting 1/Λm1/\Lambda_m against cΛmc\Lambda_m for aqueous solutions of a monobasic weak acid (HXHX) resulted in a straight line with yy-axis intercept of PP and slope of SS. The ratio P/SP/S is [Λm\Lambda_m = molar conductivity, Λm∘\Lambda_m^{\circ} = limiting molar conductivity, cc = molar concentration, KaK_a = dissociation constant of HXHX]
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A
KaΛm∘K_a\Lambda_m^{\circ}
B
KaΛm∘/2K_a\Lambda_m^{\circ}/2
C
2KaΛm∘2K_a\Lambda_m^{\circ}
D
1/(KaΛm∘)1/(K_a\Lambda_m^{\circ})
7
Consider the strong electrolytes ZmXnZ_mX_n, UmYpU_mY_p and VmXnV_mX_n. Limiting molar conductivity (Λ∘\Lambda^\circ) of UmYpU_mY_p and VmXnV_mX_n are 250 and 440 S cm2 mol−1S\,cm^2\,mol^{-1}, respectively. The value of (m+n+p)(m+n+p) is ______. Given\textbf{Given} (limiting molar conductivities of ions, in S cm2 mol−1S\,cm^2\,mol^{-1}): λ∘(Zn+)=50.0\lambda^\circ(Z^{n+})=50.0, λ∘(Up+)=25.0\lambda^\circ(U^{p+})=25.0, λ∘(Vn+)=100.0\lambda^\circ(V^{n+})=100.0, λ∘(Xm−)=80.0\lambda^\circ(X^{m-})=80.0, λ∘(Ym−)=100.0\lambda^\circ(Y^{m-})=100.0. The plot of molar conductivity (Λ\Lambda) of ZmXnZ_mX_n vs c1/2c^{1/2} is shown below.
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Your answer: —
Correct answer: 7
8
The reduction potential (E∘E^\circ, in V) of MnO4−(aq)/Mn(s)MnO_4^{-}(aq)/Mn(s) is ______. Given: E∘(MnO4−(aq)/MnO2(s))=1.68 VE^\circ(MnO_4^{-}(aq)/MnO_2(s)) = 1.68\ V; E∘(MnO2(s)/Mn2+(aq))=1.21 VE^\circ(MnO_2(s)/Mn^{2+}(aq)) = 1.21\ V; E∘(Mn2+(aq)/Mn(s))=−1.03 VE^\circ(Mn^{2+}(aq)/Mn(s)) = -1.03\ V
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Your answer: —
Correct answer: 0.77
9
Some standard electrode potentials at 298298 K are given below: Pb2+/Pb:−0.13Pb^{2+}/Pb: -0.13 V Ni2+/Ni:−0.24Ni^{2+}/Ni: -0.24 V Cd2+/Cd:−0.40Cd^{2+}/Cd: -0.40 V Fe2+/Fe:−0.44Fe^{2+}/Fe: -0.44 V To a solution containing 0.0010.001 M of X2+X^{2+} and 0.10.1 M of Y2+Y^{2+}, the metal rods XX and YY are inserted (at 298298 K) and connected by a conducting wire. This resulted in dissolution of XX. The correct combination(s) of XX and YY, respectively, is(are) (Given: Gas constant R=8.314 J K−1mol−1R=8.314\ J\ K^{-1}mol^{-1}, Faraday constant F=96500 C mol−1F=96500\ C\ mol^{-1})
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A
CdCd and NiNi
B
CdCd and FeFe
C
NiNi and PbPb
D
NiNi and FeFe
10
At 298298 K, the limiting molar conductivity of a weak monobasic acid is 4×102 S cm2 mol−14\times10^2\ S\ cm^2\ mol^{-1}. At 298298 K, for an aqueous solution of the acid, the degree of dissociation is α\alpha and the molar conductivity is y×102 S cm2 mol−1y\times10^2\ S\ cm^2\ mol^{-1}. At 298298 K, upon 2020 times dilution with water, the molar conductivity of the solution becomes 3y×102 S cm2 mol−13y\times10^2\ S\ cm^2\ mol^{-1}. The value of α\alpha is ____.
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Your answer: —
Correct answer: 0.22
11
At 298298 K, the limiting molar conductivity of a weak monobasic acid is 4×102 S cm2 mol−14\times10^2\ S\ cm^2\ mol^{-1}. At 298298 K, for an aqueous solution of the acid, the degree of dissociation is α\alpha and the molar conductivity is y×102 S cm2 mol−1y\times10^2\ S\ cm^2\ mol^{-1}. At 298298 K, upon 2020 times dilution with water, the molar conductivity of the solution becomes 3y×102 S cm2 mol−13y\times10^2\ S\ cm^2\ mol^{-1}. The value of yy is ____.
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Your answer: —
Correct answer: 0.88
12
Consider a 70%70\% efficient hydrogen-oxygen fuel cell working under standard conditions at 11 bar and 298298 K. Its cell reaction is H2(g)+12O2(g)→H2O(l)H_2(g) + \frac12 O_2(g) \rightarrow H_2O(l). The work derived from the cell on the consumption of 1.0×10−31.0\times10^{-3} mol of H2(g)H_2(g) is used to compress 1.001.00 mol of a monoatomic ideal gas in a thermally insulated container. What is the change in temperature (in K) of the ideal gas? The standard reduction potentials are: O2(g)+4H+(aq)+4e−→2H2O(l), E∘=1.23O_2(g)+4H^+(aq)+4e^- \rightarrow 2H_2O(l),\ E^\circ=1.23 V; 2H+(aq)+2e−→H2(g), E∘=0.002H^+(aq)+2e^- \rightarrow H_2(g),\ E^\circ=0.00 V. Use F=96500 C mol−1F=96500\ \text{C mol}^{-1}, R=8.314 J mol−1K−1R=8.314\ \text{J mol}^{-1}\text{K}^{-1}.
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Your answer: —
Correct answer: 13.32
13
Consider an electrochemical cell: A(s)∣An+(aq,2 M)∥B2n+(aq,1 M)∣B(s)A(s) \mid A^{n+}(aq, 2\ M) \| B^{2n+}(aq, 1\ M) \mid B(s). The value of ΔH∘\Delta H^{\circ} for the cell reaction is twice that of ΔG∘\Delta G^{\circ} at 300 K. If the emf of the cell is zero, the ΔS∘\Delta S^{\circ} (in J K−1mol−1K^{-1}mol^{-1}) of the cell reaction per mole of BB formed at 300 K is ____. (Given: ln⁡(2)=0.7\ln(2) = 0.7, RR (universal gas constant) = 8.3 J K−1mol−1K^{-1}mol^{-1}. H, S and G are enthalpy, entropy and Gibbs energy, respectively.)
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Your answer: —
Correct answer: -11.62
14
For the electrochemical cell, Mg(s) ∣ Mg2+(aq,1 M) ∣∣ Cu2+(aq,1 M) ∣ Cu(s)Mg(s)\ |\ Mg^{2+}(aq, 1\ M)\ ||\ Cu^{2+}(aq, 1\ M)\ |\ Cu(s) the standard emf of the cell is 2.70 V2.70\ V at 300 K300\ K. When the concentration of Mg2+Mg^{2+} is changed to x Mx\ M, the cell potential changes to 2.67 V2.67\ V at 300 K300\ K. The value of xx is ____. (given, FR=11500 K V−1\dfrac{F}{R}=11500\ K\ V^{-1}, where FF is the Faraday constant and RR is the gas constant, ln⁡(10)=2.30\ln(10)=2.30)
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Your answer: —
Correct answer: 10
15
For the following cell, Zn(s) ∣ ZnSO4(aq) ∥ CuSO4(aq) ∣ Cu(s)Zn(s)\,|\,ZnSO_4(aq)\,\|\,CuSO_4(aq)\,|\,Cu(s) when the concentration of Zn2+Zn^{2+} is 10 times the concentration of Cu2+Cu^{2+}, the expression for ΔG\Delta G (in J mol−1J\ mol^{-1}) is [F is Faraday constant; R is gas constant; T is temperature; Ecello=1.1 VE^o_{cell} = 1.1\ V]
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A
1.1F1.1F
B
2.303RT−2.2F2.303RT - 2.2F
C
2.303RT+1.1F2.303RT + 1.1F
D
−2.2F-2.2F
16
The conductance of a 0.00150.0015 M aqueous solution of a weak monobasic acid was determined using a conductivity cell with platinized Pt electrodes. The distance between the electrodes is 120120 cm with a cross-sectional area of 1 cm21\text{ cm}^2. The conductance of this solution was found to be 5×10−7 S5\times10^{-7}\text{ S}. The pH of the solution is 44. The value of limiting molar conductivity (Λmo)(\Lambda_m^o) of this weak monobasic acid in aqueous solution is Z×102 S cm−1mol−1Z\times10^2\text{ S cm}^{-1}\text{mol}^{-1}. The value of ZZ is:
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Your answer: —
Correct answer: 6
17
For the following electrochemical cell at 298 K: Pt(s)∣H2(g, 1 bar)∣H+(aq, 1 M)∥M4+(aq), M2+(aq)∣Pt(s)\text{Pt}(s) \mid \text{H}_2(g,\ 1\ \text{bar}) \mid \text{H}^+(aq,\ 1\ \text{M}) \parallel \text{M}^{4+}(aq),\ \text{M}^{2+}(aq) \mid \text{Pt}(s) Ecell=0.092 VE_{cell} = 0.092\ \text{V} when [M2+(aq)][M4+(aq)]=10x\dfrac{[\text{M}^{2+}(aq)]}{[\text{M}^{4+}(aq)]} = 10^{x}. Given: EM4+/M2+∘=0.151 VE^{\circ}_{\text{M}^{4+}/\text{M}^{2+}} = 0.151\ \text{V}; 2.303RTF=0.059 V2.303\dfrac{RT}{F} = 0.059\ \text{V} The value of xx is
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Your answer: —
Correct answer: 2
18
The molar conductivity of a solution of a weak acid HXHX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HYHY (0.10 M). If λX−0≈λY−0\lambda^0_{X^-} \approx \lambda^0_{Y^-}, the difference in their pKapK_a values, pKa(HX)−pKa(HY)pK_a(HX) - pK_a(HY), is (consider degree of ionization of both acids to be ≪1\ll 1). Answer as a single-digit integer, 0-9.
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Your answer: —
Correct answer: 3
19
All the energy released from the reaction X→YX\rightarrow Y, ΔrG0=−193 kJ mol−1\Delta_rG^0=-193\ \text{kJ mol}^{-1} is used for oxidizing M+M^+ as M+→M3++2e−M^+\rightarrow M^{3+}+2e^-, E0=−0.25 VE^0=-0.25\ V. Under standard conditions, the number of moles of M+M^+ oxidized when one mole of XX is converted to YY is: [F=96500 C mol−1F=96500\ \text{C mol}^{-1}]
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Your answer: —
Correct answer: 4
20
In a galvanic cell, the salt bridge
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A
does not participate chemically in the cell reaction.
B
stops the diffusion of ions from one electrode to another.
C
is necessary for the occurrence of the cell reaction.
D
ensures mixing of the two electrolytic solutions.
21
An aqueous solution of X is added slowly to an aqueous solution of Y, as shown in List-I. Match the resulting conductivity behaviour in List-II: List-I\textbf{List-I} (P) X=(C2H5)3NX=(C_2H_5)_3N (weak base) added to Y=CH3COOHY=CH_3COOH (weak acid) (Q) X=KI (0.1 M)X=KI\ (0.1\ M) added to Y=AgNO3 (0.01 M)Y=AgNO_3\ (0.01\ M) (R) X=CH3COOHX=CH_3COOH (weak acid) added to Y=KOHY=KOH (strong base) (S) X=NaOHX=NaOH (strong base) added to Y=HIY=HI (strong acid) List-II\textbf{List-II}: (1) decreases then increases (2) decreases then does not change much (3) increases then does not change much (4) does not change much then increases
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A
P→\to3, Q→\to4, R→\to2, S→\to1
B
P→\to4, Q→\to3, R→\to2, S→\to1
C
P→\to2, Q→\to3, R→\to4, S→\to1
D
P→\to1, Q→\to4, R→\to3, S→\to2
22
Standard reduction potential data at 25∘C25^\circ C: E∘(Fe3+,Fe2+)=0.77E^\circ(Fe^{3+},Fe^{2+})=0.77 V; E∘(Fe2+,Fe)=−0.44E^\circ(Fe^{2+},Fe)=-0.44 V; E∘(Cu2+,Cu)=0.34E^\circ(Cu^{2+},Cu)=0.34 V; E∘(Cu+,Cu)=0.52E^\circ(Cu^+,Cu)=0.52 V; E∘(O2(g)+4H++4e−→2H2O)=1.23E^\circ(O_2(g)+4H^++4e^- \to 2H_2O)=1.23 V; E∘(O2(g)+2H2O+4e−→4OH−)=0.40E^\circ(O_2(g)+2H_2O+4e^- \to 4OH^-)=0.40 V; E∘(Cr3+,Cr)=−0.74E^\circ(Cr^{3+},Cr)=-0.74 V; E∘(Cr2+,Cr)=−0.91E^\circ(Cr^{2+},Cr)=-0.91 V. Match E∘E^\circ of the redox pair in List-I with the values in List-II: List-I\textbf{List-I} (P) E∘(Fe3+,Fe)E^\circ(Fe^{3+},Fe) (Q) E∘(4H2O⇌4H++4OH−)E^\circ(4H_2O \rightleftharpoons 4H^+ + 4OH^-) (R) E∘(Cu2++Cu→2Cu+)E^\circ(Cu^{2+}+Cu \to 2Cu^+) (S) E∘(Cr3+,Cr2+)E^\circ(Cr^{3+},Cr^{2+}) List-II\textbf{List-II}: (1) −0.18-0.18 V (2) −0.4-0.4 V (3) −0.04-0.04 V (4) −0.83-0.83 V
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A
P→\to4, Q→\to1, R→\to2, S→\to3
B
P→\to2, Q→\to3, R→\to4, S→\to1
C
P→\to1, Q→\to2, R→\to3, S→\to4
D
P→\to3, Q→\to4, R→\to1, S→\to2
Tip\textbf{Tip}: When combining two given half-reactions with DIFFERENT electron counts nn to get a third, always combine via ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circ (weighted by each reaction's own nn), never by naive averaging of E∘E^\circ values. When the two half-reactions being combined happen to share the SAME nn, their E∘E^\circ values may be added/subtracted directly, since the electrons cancel exactly - that shortcut is only valid in that special case.
Deconstruction\textbf{Deconstruction}: Four composite E∘E^\circ values must each be built from the given elementary half-reactions using Hess's law on ΔG∘=−nFE∘\Delta G^\circ=-nFE^\circ, then matched to List-II. Strategy\textbf{Strategy}: For each target, express it as a combination of given half-reactions with matching electron counts, add/subtract the corresponding ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circ values (not the E° values directly, unless n happens to match), and solve for the target E∘E^\circ. Execution (each sub-part independently)\textbf{Execution (each sub-part independently)}: (P) E∘(Fe3+,Fe)\textbf{(P) }E^\circ(Fe^{3+},Fe): build from Fe3++e−→Fe2+Fe^{3+}+e^-\to Fe^{2+} (n=1n=1, E=0.77E=0.77) and Fe2++2e−→FeFe^{2+}+2e^-\to Fe (n=2n=2, E=−0.44E=-0.44), which sum to Fe3++3e−→FeFe^{3+}+3e^-\to Fe (n=3n=3). −3FEP=−1F(0.77)+(−2F)(−0.44)=−0.77+0.88=0.11F-3F E_P = -1F(0.77) + (-2F)(-0.44) = -0.77+0.88=0.11F, so EP=−0.11/3≈−0.04E_P = -0.11/3 \approx -0.04 V. Matches List-II (3)\textbf{(3)}. (Q) E∘(4H2O→4H++4OH−)\textbf{(Q) }E^\circ(4H_2O\to 4H^++4OH^-): build from O2+2H2O+4e−→4OH−O_2+2H_2O+4e^-\to 4OH^- (n=4n=4, E=0.40E=0.40) plus the REVERSE of O2+4H++4e−→2H2OO_2+4H^++4e^-\to 2H_2O (n=4n=4, E=1.23E=1.23, reversed to 2H2O→O2+4H++4e−2H_2O\to O_2+4H^++4e^-). Since both have the same n=4n=4 and O2O_2/electrons cancel exactly on addition, E∘E^\circ values combine directly: EQ=0.40−1.23=−0.83E_Q = 0.40-1.23=-0.83 V. Matches List-II (4)\textbf{(4)}. (R) E∘(Cu2++Cu→2Cu+)\textbf{(R) }E^\circ(Cu^{2+}+Cu\to 2Cu^+): build from Cu2++2e−→CuCu^{2+}+2e^-\to Cu (n=2n=2, E=0.34E=0.34) plus the REVERSE of 2×[Cu++e−→Cu]2\times[Cu^++e^-\to Cu], i.e. 2Cu→2Cu++2e−2Cu\to 2Cu^++2e^- (n=2n=2, E=0.52E=0.52 for the forward reduction). Both combined half-reactions have the same n=2n=2, so E∘E^\circ values subtract directly: ER=0.34−0.52=−0.18E_R = 0.34-0.52=-0.18 V. Matches List-II (1)\textbf{(1)}. (This negative value correctly signals that the reverse process - disproportionation of Cu+Cu^+ into Cu2+Cu^{2+} and CuCu - is the spontaneous direction, a well-known fact about aqueous Cu+Cu^+ instability.) (S) E∘(Cr3+,Cr2+)\textbf{(S) }E^\circ(Cr^{3+},Cr^{2+}): build from Cr3++3e−→CrCr^{3+}+3e^-\to Cr (n=3n=3, E=−0.74E=-0.74) and Cr2++2e−→CrCr^{2+}+2e^-\to Cr (n=2n=2, E=−0.91E=-0.91), related by Cr3++3e−→Cr=[Cr3++e−→Cr2+]+[Cr2++2e−→Cr]Cr^{3+}+3e^-\to Cr = [Cr^{3+}+e^-\to Cr^{2+}] + [Cr^{2+}+2e^-\to Cr]. −3F(−0.74)=−1F ES+(−2F)(−0.91)⇒2.22F=−F ES+1.82F⇒ES=1.82−2.22=−0.40-3F(-0.74) = -1F\,E_S + (-2F)(-0.91) \Rightarrow 2.22F = -F\,E_S+1.82F \Rightarrow E_S = 1.82-2.22=-0.40 V. Matches List-II (2)\textbf{(2)}. Verification\textbf{Verification}: Each of the four derivations is an independent ΔG∘\Delta G^\circ (electron-weighted) Hess-law calculation, re-checked by re-deriving with the −nFE-nFE bookkeeping shown explicitly above rather than by short-cut averaging; all four land exactly on one of the four distinct List-II values with no leftover ambiguity, a strong self-consistency check. This matches FIITJEE's own worked derivation and the Resonance (Kota) official key (Paper Code 3, Q40 = D). Final Answer\textbf{Final Answer}: (D) P→\to3, Q→\to4, R→\to1, S→\to2.