Electrochemistry Test

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Electrochemistry Test - 23

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Question 1
1 / -0

Only One Option Correct Type
This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct
Q.
For, Pt(H₂)/H₂O, electrode potential at 298 K and 1 bar is

A
-0.2364 V

B
0.4137 V

C
-0.4137 V

D
0.00 V

Solution

It is an anode and thus oxidation takes place:
Question 2
1 / -0

For the cell reaction at 298K, Cu²⁺(aq) + 2e^- → Cu(s)
Variation of log[Cu²⁺ ] with is a straight line of intercept 0.34 V on axis .Then electrode potential of the half-cell. Cu/Cu²⁺(0.1 M) will be

A

B

C
0.34 V

D

Solution

Cu²⁺(aq) + 2e^- → Cu

Given half-cell is oxidation half-cell (anode)
Question 3
1 / -0

Given, Cu²⁺ + 2e^- → Cu, E° = + 0.34V, K_sp of Cu(OH)₂ = 1.0 x 10^-19
What i s E°_red of Cu²⁺/Cu couple at pH = 12?

A
0.78 V

B
0.44 V

C
-0.10 V

D
0.10 V

Solution

pH = 12
pOH = 2, [OH-] = 10^-2M
Cu(OH)_2,(s) cu²⁺ (aq) + 2OH^-(aq)
∴ K_sp = [Cu²⁺][OH^-]²
1 x 10^-19 = [Cu²⁺][10^2-]²
∴ Cu²⁺ + 2e^-→ cu
Reaction Quotient (Q) =
Question 4
1 / -0

EMF of the following cell is 0.67 V at 298K.
Thus, pH of the solution = 0.28 V

A
6.6

B
4.2

C
7.0

D
1.0

Solution

Hydrogen electrode has been coupled to calomel electrode.
Anode
H₂(g) → 2H⁺(aq) + 2e^- E⁰_SHE = 0.00
Cathode
Hg₂Cl₂(s) + 2e^{- .}→ 2 Hg(e)+2Cl^-(aq) E⁰_cal,= 0.28 V
Hg₂Cl₂(s) + H₂(g) → 2Hg(l) + 2H⁺(aq) + 2Cl^- (aq) E⁰_cell,= 0.28 V
Reaction quotient (Q) =
Question 5
1 / -0

Given,
A³⁺ + e^- A²⁺ , E^{0. = 1.42 V}
B⁴⁺+ 2e^- B²⁺,E^{0. = 0.40 V}
In the potentiometric titration of B²⁺ with A³⁺ the potential at the equivalence point is

A
0.61 V

B
0.74 V

C
0.91 V

D
1.08 V

Solution
Question 6
1 / -0

For the cell of 298 K,
Zn(s) + Cu²⁺ (aq)  Cu(s) + Zn²⁺(aq)

Variation of E_cell with logQ (where Q is reaction quotient ) is of the type


At what value of ratio of molar concentration of ions E_cell would be 1.1591?

A

B

C

D

Solution

This equation represents a straight line
Question 7
1 / -0

Given, Ag⁺ + e^- → Ag,

Thus, E° of the following half-cell reaction at 298 K.

A
0.80 V

B
-0.37 V

C
0.30 V

D
0.37 V

Solution
Question 8
1 / -0

Ag⁺/Ag electrode is immersed in 1.00 M KCl at 298 K. K_sp (AgCl) = 1.0 x 10^-10

Thus, EMF of the cell set up is

A
0.209 V

B
-0.209 V

C
0.505 V

D
-0.505 V

Solution

In this case Ag⁺ ion collects as solid AgCI on the electrode itself. Some Ag⁺ exists in equilibrium with AgCI(s) in solution.
Question 9
1 / -0

For the following cell, Pt(H₂) | HCI(aq) || AgCI | Ag
E_cell = 0.2650 V at 298 K and E_cell = 0.2595 V at 308 K
Thus, heat of reaction at 298 K is

A
0.00 kJ

B
-82.8 kJ

C
+82.8 kJ

D
-41.4 kJ

Solution

By Gibbs Helmholtz equation
Question 10
1 / -0

Quinhydrone electrode is an indicator electrode in contact with platinum metal and can be used in acid-base potentiometric titration
At the equivalence point, [A] = [B] and thus, EMF of the cell is dependent on pH. When NaOH is added to HCl in potentiometric titration, potentiogram is

A

B

C

D

Solution

Initially in acid (HCI), pH < 7 at neutral point pH = 7 and after HCI is neutralised, pH increases due to further addition of NaOH. Hence, E_cell also increases.
Question 11
1 / -0

Potential at the equivalence point in the potentiometric titration of Fe²⁺ with MnO₄^- in acidic medium is

A

B

C

D

Solution

When equivalence point is reached