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Electrochemistry · Question 16

JEE Main 2026 (Online) 22nd January Morning ShiftNumericalMedium📈 Topic in 39% papers+4
1

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

Answer

4

Solution

Anode (oxidation): HSnO2- + H2O + 3OH- Sn(OH)62- + 2e- ] × 3

Cathode (reduction): Bi2O3 + 3H2O + 6e- 2Bi + 6OH-

Overall: Bi2O3 + 3HSnO2- + 6H2O + 3OH- 2Bi + 3Sn(OH)62-

n = 6

E cell = E cathode - E anode = -0.44 - (-0.90) = 0.46 V

Nernst equation at 298 K:

Ecell = E cell - 0.059n Q

= 0.46 - 0.0596 106

= 0.46 - 0.059 = 0.401 V

= 4.01 10-1 V 4 10-1 V

Answer: 4

JEE Main 2026 (Online) 22nd January Morning Shift

Keep practising — Electrochemistry

2

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
3

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
4

The pH and conductance of a weak acid (HX) were found to be 5 and 4 10-5 S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1 cm2 were at a distance of 15 cm apart. The value of the limiting molar conductivity is S m2 mol-1. (nearest integer) (Given : degree of dissociation of the weak acid ( ) 1)

JEE Main 2026 (Online) 21st January Morning Shift
5

For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2026 (Online) 21st January Evening Shift
6

MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K. MX(s) M+(aq) + X-(aq); Ksp = 10-10 If the standard reduction potential for M+(aq) + e- M(s) is ( E M+/M ) = 0.79 V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode E X-/MX(s)/M is mV. (nearest integer) [Given: 2.303RTF = 0.059 V]

JEE Main 2026 (Online) 21st January Evening Shift
7

Consider the following half cell reaction Cr2O72-(aq) + 6e- + 14H+(aq) 2Cr3+(aq) + 7H2O( ) The reaction was conducted with the ratio of [ Cr3+]2[ Cr2O72-] = 10-6. The pH value at which the EMF of the half cell will become zero is . (nearest integer value) [Given: standard half cell reduction potential E Cr2O72-,H+/Cr3+ = 1.33 V, 2.303RTF = 0.059 V.]

JEE Main 2025 (Online) 8th April Evening Shift
8

Given below are two statements : In the light of the given statements, choose the correct answer from the options given below :

  1. Both Statement I and Statement II are true
  2. Statement I is true but Statement II is false
  3. Both Statement I and Statement II are false
  4. Statement I is false but Statement II is true
JEE Main 2025 (Online) 7th April Morning Shift
9

1 Faraday of electricity was passed through Cu2+(1.5 M, 1 L)/Cu and 0.1 Faraday was passed through Ag+(0.2 M, 1 L)/Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is mV (nearest integer) [figure] Given : E Cu2+/Cu = 0.34 V E Ag+/Ag = 0.8 V 2.303RTF = 0.06 V

JEE Main 2025 (Online) 7th April Morning Shift
10

Given below are two statements : 1 M aqueous solutions of each of Cu(NO3)2, AgNO3, Hg2(NO3)2, Mg(NO3)2 are electrolysed using inert electrodes. Given: E0Ag+/Ag = 0.80 V, E0Hg22+/Hg = 0.79 V, E0Cu2+/Cu = 0.34 V and E0Mg2+/Mg = -2.37 V. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Both Statement I and Statement II are incorrect
  2. Statement I is incorrect but Statement II is correct
  3. Statement I is correct but Statement II is incorrect
  4. Both Statement I and Statement II are correct
JEE Main 2025 (Online) 7th April Evening Shift
11

On charging the lead storage battery, the oxidation state of lead changes from x1 to y1 at the anode and from x2 to y2 at the cathode. The values of x1, y1, x2, y2 are respectively :

  1. 0, +2, +4, +2
  2. +2, 0, 0, +4
  3. +4, +2, 0, +2
  4. +2, 0, +2, +4
JEE Main 2025 (Online) 4th April Morning Shift

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