JEE Chemistry Electrochemistry MCQs Set 22

Multiple Choice Questions (MCQs) for JEE Chemistry: Electrochemistry

Review structured MCQ sets for JEE Chemistry Electrochemistry. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Electrochemistry MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: Which of the following statements is correct?
(a) Oxidation occurs at the anode in both galvanic and electrolytic cells
(b) Reduction occurs at the anode in both galvanic and electrolytic cells
(c) Reduction occurs at the anode in an electrolytic cell whereas oxidation occurs at the cathode in a galvanic cell
(d) Oxidation occurs at the anode in an electrolytic cell whereas reduction occurs at the anode in a galvanic cell
Show Answer & Explanation

Answer: (a) Oxidation occurs at the anode in both galvanic and electrolytic cells

Explanation:
1. By definition oxidation always takes place at the anode and reduction at the cathode, in both types of cell.
2. Answer: (a)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 2: Other things being equal, the life of a Daniell cell may be increased by:
(a) Keeping low temperature
(b) Using a large copper electrode
(c) Decreasing the concentration of copper ions
(d) Using large zinc electrodes
Show Answer & Explanation

Answer: (d) Using large zinc electrodes

Explanation:
1. Zinc is the electrode that is consumed, so a larger zinc electrode supplies more material and lasts longer.
2. Answer: (d)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 3: Zn cannot displace the following ions from their aqueous solution:
(a) Ag+
(b) Cu2+
(c) Fe2+
(d) Na+
Show Answer & Explanation

Answer: (d) Na+

Explanation:
1. Zn (E° = −0.76 V) displaces ions whose E° is higher (Ag+, Cu2+, Fe2+).
2. Na+ (E° = −2.71 V) is more difficult to reduce, so Zn cannot displace it.
3. Answer: (d)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 4: Red hot carbon will remove oxygen from the oxides XO and YO but not from ZO. Y will remove oxygen from XO. The order of activity of the three metals X, Y and Z, putting the most active first, is:
(a) XYZ
(b) ZYX
(c) YXZ
(d) ZXY
Show Answer & Explanation

Answer: (b) ZYX

Explanation:
1. Carbon does not reduce ZO, so Z is the most active.
2. Y removes oxygen from XO, so Y is more active than X.
3. Order: Z > Y > X.
4. Answer: (b)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 5: Adding powdered Pb and Fe to a solution containing 1.0 M of Pb2+ and Fe2+ ions would result in the formation of:
(a) More of Fe and Pb2+ ions
(b) More of Fe2+ and Pb2+ ions
(c) More of Pb and Fe2+ ions
(d) More of Fe and Pb
Show Answer & Explanation

Answer: (c) More of Pb and Fe2+ ions

Explanation:
1. Fe (E° = −0.44 V) is more reactive than Pb (−0.13 V).
2. Fe + Pb2+ → Fe2+ + Pb occurs, giving more Pb and Fe2+.
3. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 6: Which one will liberate Br2 from KBr?
(a) HI
(b) I2
(c) Cl2
(d) SO2
Show Answer & Explanation

Answer: (c) Cl2

Explanation:
1. A stronger oxidising agent is needed to oxidise Br−.
2. Cl2 (E° = 1.36 V) is stronger than Br2 (1.09 V); I2 is weaker.
3. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 7: Using the standard electrode potentials Fe2+ + 2e− ⇌ Fe (−0.44 V), Cu2+ + 2e− ⇌ Cu (+0.34 V), Ag+ + e− ⇌ Ag (+0.80 V), decide which statements are correct. I. Copper can displace iron from FeSO4 solution. II. Iron can displace copper from CuSO4 solution. III. Silver can displace copper from CuSO4 solution. IV. Iron can displace silver from AgNO3 solution. The correct choice is:
(a) I and II
(b) II and III
(c) II and IV
(d) I and IV
Show Answer & Explanation

Answer: (c) II and IV

Explanation:
1. A metal displaces another from solution if its E° is lower.
2. Fe displaces Cu2+ (II) and Ag+ (IV).
3. Cu cannot displace Fe (I) and Ag cannot displace Cu (III).
4. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 8: The standard electrode potentials of the elements A, B and C are 0.68, −2.50 and 0.50 V respectively. The order of their reducing power is:
(a) A > B > C
(b) A > C > B
(c) C > B > A
(d) B > C > A
Show Answer & Explanation

Answer: (d) B > C > A

Explanation:
1. The lower the standard reduction potential, the stronger the reducing power.
2. −2.50 < 0.50 < 0.68, so B > C > A.
3. Answer: (d)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 9: The oxidation potentials of Zn, Cu, Ag, H2 and Ni are 0.76, −0.34, −0.80, 0 and 0.55 V respectively. Which of the following reactions will provide the maximum voltage?
(a) Zn + Cu2+ → Cu + Zn2+
(b) Zn + 2Ag+ → 2Ag + Zn2+
(c) H2 + Cu2+ → 2H+ + Cu
(d) H2 + Ni2+ → 2H+ + Ni
Show Answer & Explanation

Answer: (b) Zn + 2Ag+ → 2Ag + Zn2+

Explanation:
1. Ecell = oxidation potential of the anode metal + reduction potential of the cation.
2. (a) 0.76 + 0.34 = 1.10; (b) 0.76 + 0.80 = 1.56; (c) 0 + 0.34 = 0.34; (d) 0 − 0.55 = −0.55 (nonspontaneous).
3. (b) is maximum.
4. Answer: (b)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 10: The standard reduction potentials at 25°C of Li+/Li, Ba2+/Ba, Na+/Na and Mg2+/Mg are −3.05, −2.73, −2.71 and −2.37 V respectively. Which one is the strongest oxidising agent?
(a) Na+
(b) Li+
(c) Ba2+
(d) Mg2+
Show Answer & Explanation

Answer: (d) Mg2+

Explanation:
1. The species with the highest (least negative) reduction potential is the strongest oxidising agent.
2. Mg2+ (−2.37 V) is highest.
3. Answer: (d)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 11: Each of three metals x, y and z was put in turn into an aqueous solution of the salts of the other two. Observations: y + salt of x gives no action; y + salt of z gives z + salt of y; z + salt of x gives x + salt of z; z + salt of y gives no action. Which of these observations is incorrect?
(a) y + salt of x: no action observed
(b) y + salt of z gives z + salt of y
(c) z + salt of x gives x + salt of z
(d) z + salt of y: no action observed
Show Answer & Explanation

Answer: (c) z + salt of x gives x + salt of z

Explanation:
1. From (b), y is more reactive than z; (d) agrees. From (a), x is more reactive than y, so x > y > z.
2. For z to displace x from its salt, z would have to be more reactive than x, which is false.
3. So (c) is incorrect.
4. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 12: The standard reduction electrode potentials of four elements are A = −0.250 V, B = −0.140 V, C = −0.126 V, D = −0.402 V. The element that displaces A from its compounds is:
(a) B
(b) C
(c) D
(d) None of the above
Show Answer & Explanation

Answer: (c) D

Explanation:
1. A metal displaces another if its reduction potential is lower.
2. Only D (−0.402 V) is lower than A (−0.250 V).
3. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 13: The following four colourless salt solutions are placed in separate test tubes and a strip of Cu is placed in each. Which solution finally turns blue?
(a) Zn(NO3)2
(b) Mg(NO3)2
(c) KNO3
(d) AgNO3
Show Answer & Explanation

Answer: (d) AgNO3

Explanation:
1. Cu displaces only metals with a higher E°: Cu + 2Ag+ → Cu2+ + 2Ag.
2. The Cu2+ formed makes the solution blue.
3. Answer: (d)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 14: Which of the following displacements does NOT occur?
(a) Zn + 2H+ → Zn2+ + H2
(b) Fe + 2Ag+ → Fe2+ + 2Ag
(c) Cu + Fe2+ → Cu2+ + Fe
(d) Zn + Pb2+ → Zn2+ + Pb
Show Answer & Explanation

Answer: (c) Cu + Fe2+ → Cu2+ + Fe

Explanation:
1. Cu (E° = +0.34) is less reactive than Fe (−0.44), so Cu cannot displace Fe from Fe2+.
2. The other three are spontaneous.
3. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 15: E°(Ni2+/Ni) = −0.25 V and E°(Au3+/Au) = 1.50 V. The emf of the voltaic cell Ni | Ni2+ (1.0 M) || Au3+ (1.0 M) | Au is:
(a) 1.25 V
(b) −1.75 V
(c) 1.75 V
(d) 4.0 V
Show Answer & Explanation

Answer: (c) 1.75 V

Explanation:
1. E°cell = 1.50 − (−0.25) = 1.75 V.
2. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 16: The emf of the cell Tl/Tl+(0.0001 M) || Cu2+(0.01 M)/Cu is 0.83 V. The emf of this cell will be increased by:
(a) Increasing the concentration of Cu2+ ions
(b) Decreasing the concentration of Tl+
(c) Increasing the concentration of both
(d) Both (a) and (b)
Show Answer & Explanation

Answer: (d) Both (a) and (b)

Explanation:
1. Reaction: 2Tl + Cu2+ → 2Tl+ + Cu, with E = E° − (0.0591/2) log([Tl+]2/[Cu2+]).
2. E rises when [Cu2+] increases or [Tl+] decreases, so (a) and (b) both work.
3. Answer: (d)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 17: The potential of a hydrogen electrode (PH2 = 1 atm, [H+] = 0.1 M) at 25°C will be:
(a) 0.00 V
(b) −0.059 V
(c) 0.118 V
(d) 0.059 V
Show Answer & Explanation

Answer: (b) −0.059 V

Explanation:
1. E = 0.0591 log[H+] = 0.0591 × (−1) = −0.059 V (reduction potential).
2. Answer: (b)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 18: Which of the following represents the potential of a silver wire dipped into 0.1 M AgNO3 solution at 25°C?
(a) E°red
(b) E°red + 0.059
(c) E°ox − 0.059
(d) E°red − 0.059
Show Answer & Explanation

Answer: (d) E°red − 0.059

Explanation:
1. Ag+ + e− → Ag: E = E° + 0.059 log[Ag+].
2. With [Ag+] = 0.1, E = E°red − 0.059.
3. Answer: (d)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 19: E° for F2 + 2e− → 2F− is 2.8 V. E° for ½F2 + e− → F− is:
(a) 2.8 V
(b) 1.4 V
(c) −2.8 V
(d) −1.4 V
Show Answer & Explanation

Answer: (a) 2.8 V

Explanation:
1. Electrode potential is an intensive property and does not change when the equation is multiplied.
2. It stays 2.8 V.
3. Answer: (a)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 20: ΔG° of the cell reaction AgCl + ½H2 → Ag + H+ + Cl− is −21.52 kJ. ΔG° of 2AgCl + H2 → 2Ag + 2H+ + 2Cl− is:
(a) −21.52 kJ
(b) −10.76 kJ
(c) −43.04 kJ
(d) 43.04 kJ
Show Answer & Explanation

Answer: (c) −43.04 kJ

Explanation:
1. ΔG° is extensive; doubling the equation doubles it: 2 × (−21.52) = −43.04 kJ.
2. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 21: How much will the potential of a hydrogen electrode change when its solution, initially at pH = 0, is neutralised to pH = 7?
(a) Increases by 0.059 V
(b) Decreases by 0.059 V
(c) Increases by 0.41 V
(d) Decreases by 0.41 V
Show Answer & Explanation

Answer: (d) Decreases by 0.41 V

Explanation:
1. E = −0.0591 pH.
2. At pH 0, E = 0; at pH 7, E = −0.414 V.
3. So the potential decreases by about 0.41 V.
4. Answer: (d)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 22: Which of the following will increase the voltage of the cell with the cell reaction Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s)?
(a) Increase in the size of the silver rod
(b) Increase in the concentration of Sn2+ ions
(c) Increase in the concentration of Ag+ ions
(d) Decrease in the concentration of Ag+ ions
Show Answer & Explanation

Answer: (c) Increase in the concentration of Ag+ ions

Explanation:
1. E = E° − (0.0591/2) log([Sn2+]/[Ag+]2).
2. E rises when [Ag+] increases or [Sn2+] decreases.
3. Electrode size does not matter.
4. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

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