Integer Answer Type Questions for JEE: Electrochemistry | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. For the reaction Ag → 5B(g). It initially 2 moles of A are present. What is ratio of total no of moles after 25% reaction is completed to initial no. of moles.

Ans. 2
A(g) → 5B(g)
2 moles 0
2 - 0.5 2.5 total no. of moles 1.5 + 2.5 = 4
Total of moles/Initial of moles = 4/2 = 2

Q.2. 5.9 gm of a sample of bleaching powder is treated with excess acetic acid and Kl solution. The liberated l₂ required 50 ml of M/10 hypo. What is the percent of available chlorine in the sample.

Ans. 3
Meq. of bleaching powder = Meq. of Cl₂ = Meq. of hypo
w/35.5 × 1000 = 50 × 1/10
^wCl₂ =0.1775g
∴ % chlorine = 0.1775/5.9 × 100 = 3

Q.3. 45 g of acid of mol. wt. 90 neutralized by 200 ml. of 5 N caustic potash. What is the basicity of the acid?

Ans. 2
meq. of acid = meq. of caustic potash
∴
∴ n = 2

Q.4. To make 0.01 N solution of a salt from its 0.1 N, 1 litre solution, the amount of H₂0 required is

Ans. 9
0.1 N × 1 = 0.01 × V
V = 0.1/0.01 = 10 litre
∴ volume of H20 required = 10 -1 = 9

Q.5. What is the n factor of IO^-₃ in the given equation I^- + IO^-₃ → I₂ + H₂O

Ans. 5

N factor = 5

Q.6. The charge on cobalt in [Co (CN )₆]^3- is

Ans. 3
In [Co (CN )₆]^3- complex Co shows + 3 oxidation state.

Q.7. To find the standard potential of X⁺³ | X electrode, the following cell is constituted
X(s) | X³⁺ (0.0018 M) || Ag⁺¹(0.01 M) | Ag(s)
The emf of this cell is found to be 0.42 volt. Calculate the standard potential of the half-cell reaction
X⁺³ + 3e  → X.
Express your answer in millivolts. 
(Given log 2 = 0.3010,  log 3 = 0.4771 and  = 0.8V)

Ans. 316
X  +  3Ag⁺ → 3 Ag + X⁺³

0.42 = E⁰_cell – 0.064
E⁰_cell = 0.484 volt
E⁰_anode = E⁰_cathode - E⁰_cell
E⁰_anode = [0.80 – .484] = 0.316 V
= 316 millivolt.

Q.8. A Daniel cell originally has 1.0 M Zn²⁺ in the anode half-cell and 1.0 M Cu²⁺ in the cathode half-cell. Estimate the cell potential of the Daniel cell in milli volt after sufficient ammonia has been added to the cathode compartment to make the NH₃ concentration 2.0 M. Given that = 0.76 V and  = -0.34 V. Also equilibrium constant for the formation of [Cu(NH₃)₄]²⁺ is 1 × 10¹².

Ans. 710 mV
Cu²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄]²⁺

∴ x = 6.25 × 10^-14 M
Note that due to high value of K_f almost all of the Cu²⁺ ions are converted to Cu (NH₃)²⁺₄ ion.
Now,


E_cell= 0.71 V = 710 mV.

Q.9. What is standard reduction potential in millivolt for the half cell Pt | MnO^-₄, MnO₂?
Given:

Ans. 1700
MnO^-₄  +  8H⁺ +  5e^- → Mn²⁺ +  4H₂O      E^° = 1.51 V
MnO₂ +  4H⁺ +  2e^– → Mn²⁺ +  2H₂O E^° = 1.225 V.
Subtracting MnO^-₄ +  4H⁺ +  3e^- → MnO₂ +  2H₂O
E^° = ?

= 1700 millivolt.

Q.10. In the electrolysis of KI, I₂ is formed at the anode by the reaction;
2I^- → I₂ + 2e^-
After the passage of current of 0.5 ampere for 9650 seconds, I₂ formed required 40 ml of 0.1 M Na₂S₂O₃.5H₂O solution in the reaction;
I₂ + 2S₂O^2-₃ → S₄O^2-₆ + 2I^-
What is the current efficiency?

Ans. 8%
Number of moles of hypo = MV/1000

∴ Number of moles of I₂ = 2 × 10^-3
Mass of I₂ = 2 × 10^-3 × 254 g

2 × 10^-3 × 254
i = 0.04 ampere
Current efficiency