JEE Chemistry Electrochemistry MCQs Set 04

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

Explanation:
1. The anode is defined as the electrode where oxidation takes place in every kind of cell.
2. Only its sign differs.

Teacher's Note:
1. An Ox, Red Cat.
2. Galvanic anode is negative, electrolytic anode is positive.

Answer: (d) Na\(^+\)

Explanation:
1. Zn is above Ag, Cu and Fe in the electrochemical series but below Na (E\(^\circ\) of Na+/Na = -2.71 V).
2. So it cannot reduce Na+.

Teacher's Note:
1. A metal displaces ions of metals lower in the series.
2. Na is more reactive than zinc.

Answer: (a) Zn

Explanation:
1. The lower (more negative) the reduction potential, the stronger the reducing agent.
2. Zn has the lowest value, -0.762 V.

Teacher's Note:
1. Strong reductants have very negative E°.
2. Strong oxidants have very positive E°.

Answer: (b) ZYX

Explanation:
1. Z is so reactive that carbon cannot reduce ZO, so Z is the most active.
2. Y takes oxygen from XO, so Y is more active than X.
3. The order is Z > Y > X.

Teacher's Note:
1. A more active metal displaces a less active metal from its oxide.
2. Carbon reduces only oxides of the less reactive metals.

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

Explanation:
1. Fe (E° = -0.44 V) is a stronger reductant than Pb (E° = -0.13 V).
2. Fe reduces Pb2+ to Pb and goes into solution as Fe2+.

Teacher's Note:
1. Fe + Pb2+ gives Fe2+ + Pb.
2. Pb cannot reduce Fe2+.

Answer: (d) No reaction will take place

Explanation:
1. Cu (E° = +0.34 V) is below Fe (-0.44 V) in the series.
2. Cu cannot reduce Fe2+, so no reaction occurs.

Teacher's Note:
1. Only a stronger reducing metal can displace another.
2. Copper is relatively unreactive.

Answer: (c) Cl\(_2\)

Explanation:
1. Chlorine is a stronger oxidising agent than bromine, so Cl2 oxidises Br- to Br2.
2. The others are weaker oxidants.

Teacher's Note:
1. Halogen displacement order: F2 > Cl2 > Br2 > I2.
2. Oxidising power falls down the group.

Answer: (c) II and IV

Explanation:
1. A metal displaces another metal from solution if its reduction potential is lower.
2. Fe displaces Cu and Ag; Cu cannot displace Fe, and Ag cannot displace Cu.
3. So II and IV are correct.

Teacher's Note:
1. Compare reduction potentials.
2. Lower potential metal is the displacer.

Answer: (d) B > C > A

Explanation:
1. Reducing power increases as the reduction potential decreases.
2. So B (-2.50) > C (0.50) > A (0.68).

Teacher's Note:
1. More negative E° means better reducing agent.
2. Order is reverse of E°.

Answer: (b) \( \ce{Zn + 2Ag+ -> 2Ag + Zn^{2+}} \)

Explanation:
1. Emf = oxidation potential of the reductant plus reduction potential of the oxidant.
2. Zn with Ag+: 0.76 + 0.80 = 1.56 V, the largest.
3. Zn with Cu2+ gives 1.10 V, H2 with Cu2+ 0.34 V and H2 with Ni2+ -0.55 V.

Teacher's Note:
1. Take the best reductant with the best oxidant.
2. Add oxidation and reduction potentials.

Answer: (a) \( E^\circ_{W^-/W_2}>E^\circ_{Y^-/Y_2}>E^\circ_{X^-/X_2}>E^\circ_{Z^-/Z_2} \)

Explanation:
1. Y2 oxidises X-, so E(Y) > E(X).
2. Y2 does not oxidise W-, so E(W) > E(Y).
3. X2 oxidises Z-, so E(X) > E(Z).
4. Hence W > Y > X > Z.

Teacher's Note:
1. A reaction occurs when the oxidant has the higher E°.
2. Chain the three results.

Answer: (d) Mg\(^{2+}\)

Explanation:
1. The ion with the highest (least negative) reduction potential is the strongest oxidant.
2. Mg2+ has -2.37 V.

Teacher's Note:
1. Most positive E° means strongest oxidising agent.
2. Li+ is the weakest oxidant.

Answer: (c) Zn can reduce Ag+ and Cu2+

Explanation:
1. Zn has the lowest E°, so it is the strongest reducing agent.
2. It reduces both Ag+ and Cu2+.

Teacher's Note:
1. The metal with the lowest E° reduces ions of the others.
2. Silver cannot reduce Zn2+ or Cu2+.

Answer: (c) D

Explanation:
1. A metal displaces another if its reduction potential is lower (more negative).
2. Only D (-0.402 V) is lower than A (-0.250 V).

Teacher's Note:
1. Check which values are more negative than the target.
2. Lower E° means stronger reductant.

Answer: (d) AgNO\(_3\)

Explanation:
1. Copper displaces Ag+ from AgNO3, forming blue Cu2+ in solution.
2. Zn, Mg and K are more reactive than Cu, so no displacement occurs.

Teacher's Note:
1. Blue colour indicates Cu2+.
2. Cu reduces only ions of metals below it.

Answer: (c) -43.04 kJ

Explanation:
1. \( \Delta G^\circ \) is an extensive property and doubles when the equation is doubled.
2. \( \Delta G^\circ=2\times(-21.52)=-43.04 \) kJ.

Teacher's Note:
1. Free energy is extensive, E° is intensive.
2. Double the equation means double ΔG.

Answer: (a) Increases by 0.03 V

Explanation:
1. \( E=E^\circ-\dfrac{0.059}{2}\log[Zn^{2+}] \) for the reduction potential.
2. Ten times dilution lowers log by 1, so E increases by 0.0295, about 0.03 V.

Teacher's Note:
1. Here E is the reduction potential of the electrode.
2. n = 2 gives 0.0295.

Answer: (d) Decreases by 0.41 V

Explanation:
1. E = -0.059 pH.
2. Going from pH 0 to 7 changes E by -0.059 x 7 = -0.41 V, a decrease.

Teacher's Note:
1. Hydrogen electrode potential decreases as pH rises.
2. 0.059 x 7 = 0.413.

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

Explanation:
1. \( E=E^\circ-\dfrac{0.059}{2}\log\dfrac{[Sn^{2+}]}{[Ag^+]^2} \).
2. Higher [Ag+] lowers Q and raises E; electrode size has no effect.

Teacher's Note:
1. Reactant concentration up, voltage up.
2. Solid size does not matter.

Answer: (b) Not feasible

Explanation:
1. A negative \( E^\circ \) means a positive \( \Delta G^\circ \).
2. So the reaction is not feasible under standard conditions.

Teacher's Note:
1. Feasible if E° is positive.
2. ΔG° = -nFE°.