Electrochemistry - MCQs with Explanations

1. The position of some metals in the electrochemical series in decreasing electropositive character is given as \(Mg > Al > Zn > Cu > Ag\). What will happen if a copper spoon is used to stir a solution of aluminium nitrate?

• (a) The spoon will get coated with aluminium.
• (b) An alloy of copper and aluminium is formed.
• (c) The solution becomes blue.
• (d) There is no reaction.

Correct Answer: (d)

Explanation: Reduction potential of copper is higher than that of aluminium, meaning copper cannot displace aluminium from its nitrate solution.

2. The ionic conductances of \(Al^{3+}\) and \(SO_{4}^{2-}\) at infinite dilution are \(x\) and \(y\text{ ohm}^{-1} \text{ cm}^{2} \text{ mol}^{-1}\) respectively. If Kohlrausch's law is valid, then molar conductance of aluminium sulphate at infinite dilution will be:

• (a) \(3x + 2y\)
• (b) \(2x + 3y\)
• (c) \(2x + 2y\)
• (d) \(3x + 3y\)

Correct Answer: (b)

Explanation: According to Kohlrausch's law, \(\Lambda_{m}^{\circ} [Al_{2}(SO_{4})_{3}] = 2\lambda_{Al^{3+}}^{\circ} + 3\lambda_{SO_{4}^{2-}}^{\circ}\), which simplifies to \(2x + 3y\).

3. A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at the cathode and anode are respectively:

• (a) \(H_{2}, O_{2}\)
• (b) \(O_{2}, H_{2}\)
• (c) \(O_{2}, Na\)
• (d) \(O_{2}, SO_{2}\)

Correct Answer: (a)

Explanation: At the anode, water is oxidised to \(O_{2}\) gas, and at the cathode, water is reduced to \(H_{2}\) gas instead of sodium ions.

4. Saturated solution of \(KNO_{3}\) is used to make 'salt-bridge' because:

• (a) velocity of \(K^{+}\) ions is greater than that of \(NO_{3}^{-}\) ions
• (b) velocity of \(NO_{3}^{-}\) ions is greater than that of \(K^{+}\) ions
• (c) velocities of both \(K^{+}\) and \(NO_{3}^{-}\) ions are nearly the same
• (d) \(KNO_{3}\) is highly soluble in water.

Correct Answer: (c)

Explanation: The nearly identical ionic mobilities of \(K^{+}\) and \(NO_{3}^{-}\) help maintain electrical neutrality in both half-cells.

5. Chlorine gas is passed into a solution of \(KF, KI\) and \(KBr\) and \(CHCl_{3}\) is added. There is a colour in \(CHCl_{3}\) layer. It is due to:

• (a) formation of \(I_{2}\) (violet)
• (b) formation of \(Br_{2}\) (orange)
• (c) formation of both \(I_{2}\) and \(Br_{2}\)
• (d) formation of \(F_{2}\) (colourless).

Correct Answer: (c)

Explanation: Chlorine is a stronger oxidising agent than iodine and bromine, so it oxidises \(I^{-}\) to \(I_{2}\) and \(Br^{-}\) to \(Br_{2}\), both of which colour the chloroform layer.

6. The quantity of electricity needed to electrolyse completely 1 M solution of \(CuSO_{4}, Bi_{2}(SO_{4})_{3}, AlCl_{3}\) and \(AgNO_{3}\) each will be:

• (a) 2 F, 6 F, 3 F and 1 F respectively
• (b) 6 F, 2 F, 3 F and 1 F respectively
• (c) 2 F, 6 F, 1 F and 3 F respectively
• (d) none of the above.

Correct Answer: (a)

Explanation: The moles of electrons (Faradays) required correspond to the charge of the metal ions: \(Cu^{2+}\) (2 F), \(Bi^{3+}\) (\(2 \times 3 = 6\) F per mole of sulphate), \(Al^{3+}\) (3 F), and \(Ag^{+}\) (1 F).

7. How many minutes would be required to deposit copper in 500 mL of 0.25 N \(CuSO_{4}\) by a current of 75 milliampere?

• (a) 1340 min
• (b) 670 min
• (c) 2680 min
• (d) 5360 min

Correct Answer: (c)

Explanation: Using \(Q = It\) and Faraday's laws, the number of equivalents is \(0.25 \times 0.5 = 0.125\). Solving for time gives 2680 minutes.

8. Which of the following expressions correctly represents the equivalent conductance at infinite dilution of \(Al_{2}(SO_{4})_{3}\). Given that \(\Lambda_{Al^{3+}}^{\circ}\) and \(\Lambda_{SO_{4}^{2-}}^{\circ}\) are the equivalent conductances at infinite dilution of the respective ions?

• (a) \(2\Lambda_{Al^{3+}}^{\circ} + 3\Lambda_{SO_{4}^{2-}}^{\circ}\)
• (b) \(\Lambda_{Al^{3+}}^{\circ} + \Lambda_{SO_{4}^{2-}}^{\circ}\)
• (c) \((\Lambda_{Al^{3+}}^{\circ} + \Lambda_{SO_{4}^{2-}}^{\circ}) \times 6\)
• (d) \(\frac{1}{3}\Lambda_{Al^{3+}}^{\circ} + \frac{1}{2}\Lambda_{SO_{4}^{2-}}^{\circ}\)

Correct Answer: (b)

Explanation: For equivalent conductance, the value is simply the sum of the equivalent conductances of the individual ions.

9. The Gibb's energy for the decomposition of \(Al_{2}O_{3}\) at \(500^{\circ}C\) is as follows: \(2/3 Al_{2}O_{3} \rightarrow 4/3 Al + O_{2}\), \(\Delta_{r}G = +966 \text{ kJ mol}^{-1}\). The potential difference needed for electrolytic reduction of \(Al_{2}O_{3}\) at \(500^{\circ}C\) is at least:

• (a) 5.0 V
• (b) 4.5 V
• (c) 3.0 V
• (d) 2.5 V

Correct Answer: (d)

Explanation: Using \(\Delta G = -nFE\), with \(n = 4\) electrons transferred, the potential \(E\) is calculated to be 2.5 V.

10. Standard electrode potentials for Fe electrode are given as: \(Fe^{2+} + 2e^{-} \rightarrow Fe, E^{\circ} = -0.44 \text{ V}\); \(Fe^{3+} + e^{-} \rightarrow Fe^{2+}, E^{\circ} = +0.77 \text{ V}\). If \(Fe^{2+}, Fe^{3+}\) and Fe blocks are kept together then:

• (a) \([Fe^{3+}]\) decreases
• (b) \([Fe^{3+}]\) increases
• (c) \([Fe^{2+}/Fe^{3+}]\) remains unchanged
• (d) \([Fe^{2+}]\) decreases.

Correct Answer: (a)

Explanation: The \(Fe/Fe^{2+}\) system acts as the anode while \(Fe^{3+}/Fe^{2+}\) acts as the cathode. Consequently, \(Fe^{3+}\) is reduced to \(Fe^{2+}\), decreasing its concentration.

11. Which one of the following does not hold good for S.H.E (Standard Hydrogen Electrode)?

• (a) The pressure of hydrogen gas is 1.5 atmosphere.
• (b) The concentration of \(H^{+}\) in solution is 1 M.
• (c) The temperature is 298 K.
• (d) The surface of platinum electrode is coated with platinum black.

Correct Answer: (a)

Explanation: In a Standard Hydrogen Electrode, the pressure of hydrogen gas must be exactly 1 atmosphere.

12. On passing \(C\) ampere of current for time \(t\) sec through 1 litre of 2 M \(CuSO_{4}\) solution (atomic weight of \(Cu = 63.5\)), the amount \(m\) of Cu (in g) deposited on cathode will be:

• (a) \(m = \frac{Ct}{63.5 \times 96500}\)
• (b) \(m = \frac{Ct}{31.25 \times 96500}\)
• (c) \(m = \frac{C \times 96500}{31.25 \times t}\)
• (d) \(m = \frac{31.75 \times C \times t}{96500}\)

Correct Answer: (d)

Explanation: Using \(m = ZIt\), where \(Z = \frac{\text{Eq. wt.}}{96500}\) and Eq. wt. of \(Cu = 31.75\), the formula is \(m = \frac{31.75 \times C \times t}{96500}\).

13. The e.m.f. of the cell, \(Zn | Zn^{2+}(0.01 \text{ M}) || Fe^{2+}(0.001 \text{ M}) | Fe\), at 298 K is 0.2905 V. The value of the equilibrium constant for cell reaction is:

• (a) \(10^{\frac{0.32}{0.0295}}\)
• (b) \(10^{\frac{0.32}{0.0295}}\) (Re-checking source: Option (b) shows the formula)
• (c) \(10^{0.0295}\)
• (d) \(10^{\frac{0.32}{0.0591}}\)

Correct Answer: (b)

Explanation: Using the Nernst equation at equilibrium (\(E_{cell} = 0\)) and calculating \(E^{\circ}\) as 0.32, the equilibrium constant \(K_{eq}\) is \(10^{\frac{0.32}{0.0295}}\).

14. Molar conductance \(\Lambda_{m}\) is plotted against \(\sqrt{C}\text{ (mol L}^{-1}\text{)}\) for three electrolytes (\(NaCl, HCl, NH_{4}OH\)). Identify (1), (2), and (3):

• (a) \(BaCl_{2}, HCl, NH_{4}OH\)
• (b) \(HCl, BaCl_{2}, NH_{4}OH\)
• (c) \(NH_{4}OH, NaCl, HCl\)
• (d) \(NH_{4}OH, HCl, NaCl\)

Correct Answer: (b)

Explanation: \(HCl\) and \(BaCl_{2}\) are strong electrolytes with high conductances (1 and 2), while \(NH_{4}OH\) is a weak electrolyte showing a curve (3).

15. The conductivity of \(0.01 \text{ mol/dm}^{3}\) aqueous acetic acid at 300 K is \(19.5 \times 10^{-5} \text{ ohm}^{-1} \text{ cm}^{-1}\) and the limiting molar conductivity is \(390 \text{ ohm}^{-1} \text{ cm}^{2} \text{ mol}^{-1}\). The degree of dissociation is:

• (a) 0.5
• (b) 0.05
• (c) \(5 \times 10^{-3}\)
• (d) \(5 \times 10^{-7}\)

Correct Answer: (b)

Explanation: \(\Lambda_{m} = 19.5\). \(\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{19.5}{390} = 0.05\).

16. \(\Lambda_{m}^{\circ}\) for \(NaCl, HCl\) and \(NaAc\) are 126.4, 425.9 and \(91.0 \text{ S cm}^{2} \text{ mol}^{-1}\) respectively. Then \(\Lambda_{m}^{\circ}\) for \(HAc\) will be:

• (a) \(380.5 \text{ S cm}^{2} \text{ mol}^{-1}\)
• (b) \(390.5 \text{ cm}^{2} \text{ mol}^{-1}\)
• (c) \(390.5 \text{ S cm}^{2} \text{ mol}^{-1}\)
• (d) \(380.5 \text{ cm}^{2} \text{ mol}^{-1}\)

Correct Answer: (c)

Explanation: \(\Lambda_{HAc}^{\circ} = \Lambda_{HCl}^{\circ} + \Lambda_{NaAc}^{\circ} - \Lambda_{NaCl}^{\circ} = 425.9 + 91.0 - 126.4 = 390.5\).

17. The conductivity of \(0.001028 \text{ mol L}^{-1}\) acetic acid is \(4.95 \times 10^{-5} \text{ S cm}^{-1}\). Calculate its dissociation constant if \(\Lambda_{m}^{\circ}\) for acetic acid is \(390.5 \text{ S cm}^{2} \text{ mol}^{-1}\).

• (a) \(1.78 \times 10^{-5} \text{ mol L}^{-1}\)
• (b) \(1.87 \times 10^{-5} \text{ mol L}^{-1}\)
• (c) \(0.178 \times 10^{-5} \text{ mol L}^{-1}\)
• (d) \(0.0178 \times 10^{-5} \text{ mol L}^{-1}\)

Correct Answer: (a)

Explanation: After calculating \(\alpha = 0.1233\), the dissociation constant \(K = \frac{c\alpha^{2}}{1-\alpha} = 1.78 \times 10^{-5} \text{ mol L}^{-1}\).

18. Consider the following four electrodes: \(P=Cu^{2+}(0.0001\text{ M})/Cu_{(s)}\), \(Q=Cu^{2+}(0.1\text{ M})/Cu_{(s)}\), \(R=Cu^{2+}(0.01\text{ M})/Cu_{(s)}\), \(S=Cu^{2+}(0.001\text{ M})/Cu_{(s)}\). The reduction potentials follow the order:

• (a) \(P > S > R > Q\)
• (b) \(S > R > Q > P\)
• (c) \(R > S > Q > P\)
• (d) \(Q > R > S > P\)

Correct Answer: (d)

Explanation: Electrode potential \(E\) increases as the concentration of \(M^{n+}\) ions increases. Thus, the order is \(Q (0.1) > R (0.01) > S (0.001) > P (0.0001)\).

19. The EMF of a concentration cell consisting of two zinc electrodes, one in \(M/4\text{ ZnSO}_{4}\) and the other in \(M/16\) at \(25^{\circ}C\) is:

• (a) 0.0125 V
• (b) 0.0250 V
• (c) 0.0178 V
• (d) 0.0356 V

Correct Answer: (c)

Explanation: \(E_{cell} = \frac{0.059}{2} \log\left(\frac{16}{4}\right) = 0.0178 \text{ V}\).

20. Which of the following metals cannot be obtained by the electrolysis of the aqueous solution of their salts?

• (a) \(Ag\) and \(Mg\)
• (b) \(Mg\) and \(Al\)
• (c) \(Ag\) and \(Al\)
• (d) \(Cu\) and \(Cr\)

Correct Answer: (b)

Explanation: \(Mg\) and \(Al\) have lower reduction potentials than water, so water is reduced preferentially to produce \(H_{2}\) gas.

21. Which one of the following statements is correct?

• (a) The emf of a cell is the sum of oxidation potentials of two half-cells.
• (b) An element with higher oxidation potential is a stronger oxidizing agent.
• (c) The emf of a cell may increase or decrease with increase of temperature.
• (d) The absolute value of electrode potential can be measured.

Correct Answer: (c)

Explanation: Temperature changes can either increase or decrease cell EMF depending on the sign of \(\Delta S\).

22. Which of the following ions in aqueous solution is the best conductor of electricity?

• (a) \(Li^{+}\)
• (b) \(Na^{+}\)
• (c) \(Mg^{2+}\)
• (d) \(Cs^{+}\)

Correct Answer: (d)

Explanation: \(Cs^{+}\) has the smallest hydrated radius, giving it the highest ionic mobility among the alkali metal ions.

23. Which of the following curve shows the variation of \(\Lambda_{m}\) of a weak acid with \(\sqrt{C}\)?

(Refer to graph in source)

Correct Answer: (c)

Explanation: Weak electrolytes show a sharp increase in molar conductance at very low concentrations.

24. For the electrochemical cell, \(M | M^{+} || X^{-} | X\), \(E^{\circ}(M^{+}/M) = 0.44 \text{ V}\) and \(E^{\circ}(X/X^{-}) = 0.33 \text{ V}\). One can deduce that:

• (a) \(M + X \rightarrow M^{+} + X^{-}\) is spontaneous
• (b) \(M^{+} + X^{-} \rightarrow M + X\) is spontaneous
• (c) \(E_{cell} = 0.77 \text{ V}\)
• (d) \(E_{cell} = -0.77 \text{ V}\)

Correct Answer: (b)

Explanation: The EMF for reaction (b) is positive (\(+0.11\text{ V}\)), indicating it is spontaneous.

25. The molar ionic conductivities of \(NH_{4}^{+}\) and \(OH^{-}\) at infinite dilution are 72 and \(198 \text{ ohm}^{-1} \text{ cm}^{2}\) respectively. \(\Lambda_{m}\) of a centinormal \(NH_{4}OH\) is 9. Percentage dissociation is:

• (a) 3.33%
• (b) 7.14%
• (c) 12.5%
• (d) 4.54%

Correct Answer: (a)

Explanation: \(\Lambda_{m}^{\circ} = 270\). \(\alpha = \frac{9}{270} = 3.33\%\).

26. During the charging of lead storage battery, the reaction at anode is represented by:

• (a) \(Pb^{2+} + SO_{4}^{2-} \rightarrow PbSO_{4}\)
• (b) \(PbSO_{4} + H_{2}O \rightarrow PbO_{2} + SO_{4}^{2-} + 2H^{+}\)
• (c) \(Pb \rightarrow Pb^{2+} + 2e^{-}\)
• (d) \(Pb^{2+} + 2e^{-} \rightarrow Pb\)

Correct Answer: (b)

Explanation: During charging, the anode reaction is the reverse of the discharge cathode reaction, forming \(PbO_{2}\).

27. Salts of metals A, B and C are electrolysed under identical conditions. 4.2 g of A, 5.4 g of B and 19.2 g of C are deposited. Atomic weights are 7, 27 and 64. Their ratio of valencies is:

• (a) 3:2:2
• (b) 1:2:3
• (c) 2:3:1
• (d) 1:3:2

Correct Answer: (d)

Explanation: Molar ratio is \(0.6:0.2:0.3\). Since equivalents are identical, the valency ratio is \(1:3:2\).

28. The time required to coat \(80 \text{ cm}^{2}\) with \(5 \times 10^{-3} \text{ cm}\) thick silver (density \(1.05 \text{ g cm}^{-3}\)) using 3 A current is:

• (a) 115 sec
• (b) 125 sec
• (c) 135 sec
• (d) 145 sec

Correct Answer: (b)

Explanation: Weight of Ag required is 0.42 g. Solving \(W = \frac{EIt}{96500}\) gives 125 seconds.

29. Which statement is not correct?

• (a) Conductivity depends on size of ions.
• (b) Conductivity depends on viscosity.
• (c) Conductivity does not depend on solvation.
• (d) Conductivity increases with temperature.

Correct Answer: (c)

Explanation: In fact, conductivity does depend on solvation; higher solvation decreases ion mobility.

30. In electrochemical corrosion of metals, the metal undergoing corrosion:

• (a) becomes anode
• (b) becomes cathode
• (c) becomes inert
• (d) none of these.

Correct Answer: (a)

Explanation: The metal atom loses electrons (oxidation), which defines the anode in electrochemical systems.

31. Standard potentials for \(Mn^{2+} + 2e^{-} \rightarrow Mn\) and \(Mn^{3+} + e^{-} \rightarrow Mn^{2+}\) are -1.18 V and 1.51 V. Potential for \(Mn^{3+} + 3e^{-} \rightarrow Mn\) is:

• (a) 0.33 V
• (b) 1.69 V
• (c) -0.28 V
• (d) -0.85 V

Correct Answer: (c)

Explanation: Summing \(\Delta G^{\circ}\) values gives \(0.85\text{ F}\) for the 3e- change, resulting in \(E^{\circ} = -0.28\text{ V}\).

32. Correct order of equivalent conductances at infinite dilution for \(H^{+}, K^{+}, CH_{3}COO^{-}\) and \(HO^{-}\) is:

• (a) \(H^{+} > HO^{-} > K^{+} > CH_{3}COO^{-}\)
• (b) \(H^{+} < HO^{-} > K^{+} > CH_{3}COO^{-}\)
• (c) \(H^{+} > K^{+} > HO^{-} > CH_{3}COO^{-}\)
• (d) \(H^{+} > K^{+} > CH_{3}COO^{-} > HO^{-}\)

Correct Answer: (a)

Explanation: \(H^{+}\) and \(OH^{-}\) have exceptionally high conductances due to the Grotthuss mechanism.

33. Equivalent conductivity at infinite dilution for sodium-potassium oxalate will be (given oxalate=148.2, \(K^{+}\)=50.1, \(Na^{+}\)=73.5):

• (a) \(271.8 \text{ S cm}^{2} \text{ eq}^{-1}\)
• (b) \(67.95 \text{ S cm}^{2} \text{ eq}^{-1}\)
• (c) \(543.6 \text{ S cm}^{2} \text{ eq}^{-1}\)
• (d) \(135.9 \text{ S cm}^{2} \text{ eq}^{-1}\)

Correct Answer: (d)

Explanation: \(\lambda_{m}^{\infty} = 271.8\). Equivalent conductivity is \(\lambda_{m}^{\infty} / 2 = 135.9\).

34. At \(25^{\circ}C\), \(\Lambda_{m}\) of 0.1 M \(NH_{4}OH\) is 9.54 and at infinite dilution it is 238. The degree of ionisation is:

• (a) 4.008%
• (b) 40.800%
• (c) 2.080%
• (d) 20.800%

Correct Answer: (a)

Explanation: \(\alpha = \frac{9.54}{238} = 0.04008 = 4.008\%\).

35. Two solutions X and Y are diluted. \(\Lambda_{m}\) of X increases by 1.5 times while Y increases by 20 times. X and Y are:

• (a) \(X \rightarrow NaCl, Y \rightarrow KCl\)
• (b) \(X \rightarrow NaCl, Y \rightarrow CH_{3}COOH\)
• (c) \(X \rightarrow KOH, Y \rightarrow NaOH\)
• (d) \(X \rightarrow CH_{3}COOH, Y \rightarrow NaCl\)

Correct Answer: (b)

Explanation: Strong electrolytes (X) show small changes, while weak electrolytes (Y) show massive increases on dilution.

36. In the electrolysis of dilute \(H_{2}SO_{4}\) using platinum electrode:

• (a) \(H_{2}\) is liberated at cathode
• (b) \(O_{2}\) is produced at cathode
• (c) \(Cl_{2}\) is obtained at cathode
• (d) \(NH_{3}\) is produced at anode.

Correct Answer: (a)

Explanation: \(H^{+}\) ions are reduced to \(H_{2}\) gas at the cathode during water electrolysis.

37. An electrochemical cell has two half cell reactions: \(A^{2+} + 2e^{-} \rightarrow A (0.34\text{ V})\) and \(X \rightarrow X^{2+} + 2e^{-} (-2.37\text{ V})\). The cell voltage will be:

• (a) 2.71 V
• (b) 2.03 V
• (c) -2.71 V
• (d) -2.03 V

Correct Answer: (a)

Explanation: \(E_{cell}^{\circ} = 0.34 - (-2.37) = 2.71 \text{ V}\).

38. Match List I with List II (Ionic mobility, extrapolate graphs, etc.):

• (a) P-4, Q-1, R-2, S-3
• (b) P-3, Q-1, R-2, S-4
• (c) P-1, Q-2, R-3, S-4
• (d) P-2, Q-1, R-4, S-3

Correct Answer: (d)

Explanation: Strong electrolytes (\(NaNO_{3}\)) can be extrapolated (2); weak acids use Kohlrausch law (1); mobility is velocity under 1 V/cm (4); conductance is ion contribution (3).

39. The standard emf of a galvanic cell involving 2 moles of electrons is 0.59 V. The equilibrium constant is:

• (a) \(10^{20}\)
• (b) \(10^{5}\)
• (c) \(10\)
• (d) \(10^{10}\)

Correct Answer: (a)

Explanation: \(\log K_{c} = \frac{0.59 \times 2}{0.059} = 20\). \(K_{c} = 10^{20}\).

40. 9.65 coulombs is passed through fused \(MgCl_{2}\). The magnesium obtained is converted to Grignard reagent. How many moles of reagent?

• (a) \(5 \times 10^{-4}\)
• (b) \(1 \times 10^{-4}\)
• (c) \(5 \times 10^{-5}\)
• (d) \(1 \times 10^{-5}\)

Correct Answer: (c)

Explanation: Magnesium obtained is \(1.2 \times 10^{-3} \text{ g}\), which corresponds to \(5 \times 10^{-5}\) moles.

41. Which one of the following is correct?

• (a) Equivalent conductance decreases with dilution.
• (b) Specific conductance increases with dilution.
• (c) Specific conductance decreases with dilution.
• (d) Equivalent conductance increases with increasing concentration.

Correct Answer: (c)

Explanation: Specific conductance decreases because the number of ions per unit volume decreases upon dilution.

42. Time required to deposit one millimole of aluminium metal by the passage of 9.65 amperes is:

• (a) 30 s
• (b) 10 s
• (c) 30,000 s
• (d) 10,000 s.

Correct Answer: (a)

Explanation: 1 millimole requires 3 millielectrons. Solving \(W = ZIt\) gives 30 seconds.

43. How many coulombs are required for the reduction of 1 mol of \(MnO_{4}^{-}\) to \(Mn^{2+}\)?

• (a) 96500 C
• (b) \(1.93 \times 10^{5}\text{ C}\)
• (c) \(4.83 \times 10^{5}\text{ C}\)
• (d) \(5.62 \times 10^{5}\text{ C}\)

Correct Answer: (c)

Explanation: The change in oxidation state is from +7 to +2, requiring 5 Faradays: \(5 \times 96500 \approx 4.83 \times 10^{5}\text{ C}\).

44. What will be the weight of silver deposited on passing 965 coulombs of electricity in solution of \(AgNO_{3}\)?

• (a) 1.08 g
• (b) 2.16 g
• (c) 0.54 g
• (d) 0.27 g

Correct Answer: (a)

Explanation: 96500 C deposits 108 g. Thus, 965 C deposits 1.08 g.

45. An electrolytic cell contains \(Ag_{2}SO_{4}\). A current is passed until 1.6 g of \(O_{2}\) is liberated at anode. Amount of silver deposited is:

• (a) 107.88 g
• (b) 1.6 g
• (c) 0.8 g
• (d) 21.60 g

Correct Answer: (d)

Explanation: Using \(\frac{W_{O_{2}}}{W_{Ag}} = \frac{E_{O_{2}}}{E_{Ag}}\), with equivalent weights 8 and 108, the amount is \(21.60 \text{ g}\).