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Electrochemistry · Question 8

JEE Main 2026 (Online) 2nd April Evening ShiftNumericalHard📈 Topic in 10% papers+4
1

Consider the following two half-cell reactions along with the standard reduction potential given : CO2 + 6H+ + 6e- CH3OH + H2O E red = 0.02 V 12O2 + 2H+ + 2e- H2O E red = 1.23 V A fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with 80% efficiency. If the work derived from the cell using 1 mol of CH3OH is used to compress an ideal gas isothermally against a constant pressure of 1 kPa, then the change in the volume of the gas, V = m3. (nearest integer) Given : F = 96500 C mol-1

Answer

560

Solution

In the fuel cell CH3OH is oxidised (anode) and O2 is reduced (cathode):

CH3OH + 32O2 CO2 + 2H2O, n = 6

E cell = 1.23 - 0.02 = 1.21 V

G = -nFE cell = -6 96500 1.21 = -700590 J

Useful work at 80% efficiency = 0.80 700590 = 560472 J

Work on the gas: W = Pext| V|

⇒ | V| = 5604721000 = 560.47 m3 560 m3 (decrease in volume)

Answer: 560

JEE Main 2026 (Online) 2nd April Evening Shift

Keep practising — Electrochemistry

2

Consider the following redox reaction taking place in acidic medium BH4-(aq) + ClO3-(aq) H2BO3-(aq) + Cl-(aq) If the Nernst equation for the above balanced reaction is Ecell = E cell - RTnF Q, then the value of n is . (Nearest integer)

JEE Main 2026 (Online) 28th January Morning Shift
3

A volume of x mL of 5 M NaHCO3 solution was mixed with 10 mL of 2 M H2CO3 solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = mL (nearest integer). Sn(s) | Sn(OH)62- (0.5 M) | HSnO2- (0.05 M) | OH- | Bi2O3(s) | Bi(s) Consider up to one place of decimal for intermediate calculations [ arrayllGiven: & E Sn(OH)62-|HSnO2- = -0.9 V \ & E Bi2O3|Bi = -0.44 V \ & pKa(H2CO3) = 6.11 \ & 2.303RTF = 0.059 V \ & Antilog (1.29) = 19.5 array ]

JEE Main 2026 (Online) 28th January Evening Shift
4

For strong electrolyte m increases slowly with dilution and can be represented by the equation m = m - Ac1/2 Molar conductivity values of the solutions of strong electrolyte AB at 18°C are given below : array|l|c|c|c|c| c [mol L-1] & 0.04 & 0.09 & 0.16 & 0.25 \ m [S cm2 mol-1] & 96.1 & 95.7 & 95.3 & 94.9 \ array The value of constant A based on the above data [in S cm2 mol-1/(mol/L)1/2] unit is .

JEE Main 2026 (Online) 28th January Evening Shift
5

Electricity is passed through an acidic solution of Cu2+ till all the Cu2+ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is mL. (Nearest integer) [Given: Cu2+(aq) + 2e- Cu(s); E red = +0.34 V O2(g) + 4H+ + 4e- 2H2O; E red = +1.23 V Molar mass of Cu = 63.54 g mol-1 Molar mass of O2 = 32 g mol-1 Faraday Constant = 96500 C mol-1 Molar volume at STP = 22.4 L]

JEE Main 2026 (Online) 24th January Morning Shift
6

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 130 of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If Q- happened to be equal to Z-, then the difference of the pKa values of the two weak acids (pKa(HQ) - pKa(HZ)) is (Nearest integer). [Given: degree of dissociation ( ) 1 for both weak acids, : limiting molar conductivity of ions]

JEE Main 2026 (Online) 24th January Evening Shift
7

In the given electrochemical cell, Ag(s)|AgCl(s)|FeCl2(aq), FeCl3(aq) | Pt(s) at 298 K, the cell potential ( Ecell) will increase when : Choose the correct answer from the options given below :

  1. C, D and E Only
  2. A and B Only
  3. B Only
  4. A and E Only
JEE Main 2026 (Online) 23rd January Morning Shift
8

[figure] Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming M+ (M M+ + e-). The cation M+ is present in two different concentrations c1 and c2 as shown above. Which of the following statement is correct for generating a positive cell potential?

  1. If c1 is present at anode, then c1 > c2.
  2. If c1 is present at cathode, then c1 > c2.
  3. If c1 is present at cathode, then c1 < c2.
  4. If c1 is present at anode, then c1 = c2.
JEE Main 2026 (Online) 23rd January Evening Shift
9

Consider the following electrochemical cell at 298 K Pt | HSnO2-(aq) | Sn(OH)62-(aq) | OH-(aq) | Bi2O3(s) | Bi(s). If the reaction quotient at a given time is 106, then the cell EMF ( Ecell) is 10-1 V (Nearest integer). Given the standard half-cell reduction potential as E Bi2O3/Bi,OH- = -0.44 V and E Sn(OH)62-/HSnO2-,OH- = -0.90 V

JEE Main 2026 (Online) 22nd January Morning Shift
10

Consider the following reduction processes : Al3+ + 3e- Al(s), E0 = -1.66 V Fe3+ + e- Fe2+, E0 = +0.77 V Co3+ + e- Co2+, E0 = +1.81 V Cr3+ + 3e- Cr(s), E0 = -0.74 V The tendency to act as reducing agent decreases in the order :

  1. Al > Fe2+ > Cr > Co2+
  2. Al > Cr > Co2+ > Fe2+
  3. Cr > Fe2+ > Al > Co2+
  4. Al > Cr > Fe2+ > Co2+
JEE Main 2026 (Online) 22nd January Evening Shift
11

Consider the following electrochemical cell : Pt | O2(g)(1 bar) | HCl(aq) || M2+(aq, 1.0 M) | M(s) The pH above which, oxygen gas would start to evolve at anode is (nearest integer). [Given : . arrayl E M2+/M = 0.994 V \ E O2/H2O = 1.23 V array \ standard reduction potential and RTF(2.303) = 0.059 V at the given condition]

JEE Main 2026 (Online) 22nd January Evening Shift

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