Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Electrochemistry question

2019 · 9 Jan · Shift 1 · Q9
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Electrochemistry
  5. /2019 · 9 Jan · Shift 1 · Q9

Electrochemistry question

2019 · 9 Jan · Shift 1 · Q9

JEE MainChemistryElectrochemistryMCQ+4 / −1
The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4PbSO_4PbSO4​ electrolyzed in g during the process is : (Molar mass of PbSO4PbSO_4PbSO4​ = 303 g mol −-− 1)
  1. A
    22.8
  2. B
    15.2
  3. C
    7.6
  4. D
    11.4
View written solutionFree

Correct answer: C

  1. Anodic half-cell reaction during recharge

In a lead-acid battery, during recharging, the anodic half-cell undergoes the reverse of discharge:

PbSO4+2H2O→PbO2+SO42−+4H++2e−PbSO_4 + 2H_2O \rightarrow PbO_2 + SO_4^{2-} + 4H^+ + 2e^-PbSO4​+2H2​O→PbO2​+SO42−​+4H++2e−

So, 1 mole of PbSO4PbSO_4PbSO4​ corresponds to 2 moles of electrons.

  1. Relate Faraday to moles of electrons

Given electricity passed = 0.050.050.05 Faraday.

Since,

1 Faraday=1 mole of electrons1 \text{ Faraday} = 1 \text{ mole of electrons}1 Faraday=1 mole of electrons

Therefore, moles of electrons passed:

ne−=0.05n_{e^-} = 0.05ne−​=0.05

  1. Find moles of PbSO4PbSO_4PbSO4​ electrolyzed

From the half-reaction:

1 mol PbSO4↔2 mol e−1 \text{ mol } PbSO_4 \leftrightarrow 2 \text{ mol } e^-1 mol PbSO4​↔2 mol e−

Hence,

n(PbSO4)=0.052=0.025 moln(PbSO_4) = \frac{0.05}{2} = 0.025 \text{ mol}n(PbSO4​)=20.05​=0.025 mol

  1. Convert moles to mass

Molar mass of PbSO4=303 g mol−1PbSO_4 = 303\, g\,mol^{-1}PbSO4​=303gmol−1

m=n×M=0.025×303=7.575 gm = n \times M = 0.025 \times 303 = 7.575\, gm=n×M=0.025×303=7.575g

m≈7.6 gm \approx 7.6\, gm≈7.6g

  1. Check options
  • A: 22.822.822.8 g
  • B: 15.215.215.2 g
  • C: 7.67.67.6 g
  • D: 11.411.411.4 g

Thus, the correct option is:

C\boxed{C}C​

PreviousNext

More from Electrochemistry

  • If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction Zn(s) + Cu2+ (aq) ⇌ Zn2+(aq) + Cu(s) at 300 K is approximately, (R = 8 JK − 1mol − 1, F =…2019 · MCQ
  • Consider the statements S1 and S2 S1 : Conductivity always increases with decrease in the concentration of electrolyte. S2 : Molar conductivity always increases with decrease in the concentration of electrolyte. The correct option among…2019 · MCQ
  • Which one of the following graphs between molar conductivity (Λm​) versus C​ is correct ?2019 · MCQ
  • Consider the following reduction processes : Zn2+ + 2e– → Zn(s) ; Eo = – 0.76 V Ca2+ + 2e– → Ca(s); Eo = –2.87 V Mg2+ + 2e– → Mg(s) ; Eo = – 2.36 V Ni2​ + 2e– → Ni(s) ; Eo = – 0.25 The reducing power of the…2019 · MCQ
  • In the cell Pt ∣(s)∣ H2​(g, 1 bar) ∣HCl(aq)∣ AgCl ∣(s)∣ Ag(s)|Pt(s) the cell potential is 0.92 V when a 10–6 molal HCl solution is used.…2019 · MCQ
  • For the cell Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s), different half cells and their standard electrode potentials are given below : If Ezn2+/zn0​=− 0.76 V, which cathode will give maximum value of Eocell per electron transferred? Includes table2019 · MCQ
  • Given the equilibrium constant: KC of the reaction : Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s) is 10 × 1015, calculate the E cell0​ of this reaciton at 298 K [2.303 FRT​ at 298 K = 0.059V]2019 · MCQ
  • Given CO3+​ + e– → CO2+ ; Eo = + 1.81 V Pb4+ + 2e– → Pb2+ ; Eo = + 1.67 V Ce4+ + e– → Ce3+ ; Eo = + 1.61 V Bi3+ + 3e– → Bi ; Eo = + 0.20 V Oxidizing power of the species will increase in…2019 · MCQ