Electrochemistry Test

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Electrochemistry Test - 9

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Question 1
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
Question 12
4 / -1

Comprehension Type
Direction : This section contains 3 paragraphs, each describing theory, experiments, data, etc. Six questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).
Passage I
The electrochemical cell shown below is a concentration cell :
M/M^z+ (saturated solution of a sparingly soluble salt(MX₂) || M²⁺(0.001 mol dm^-3)|M
The EMF of the cell depends on the difference in concentration of M²⁺ ions at the two electrodes .The EMF of the cell at 298 K is 0.059 V.
Q.
K_sp mol³ dm^-9 (the solubility product) of MX₂ at 298 K is
(2.303 RT/F = 0.059 V at 298 K)

A
1 x 10^-15

B
4 x 10^-15

C
1 x 10^-12

D
4 x 10^-12

Solution
Question 13
4 / -1

Passage I
The electrochemical cell shown below is a concentration cell :
M/M^z+ (saturated solution of a sparingly soluble salt(MX₂) || M²⁺(0.001 mol dm^-3)|M
The EMF of the cell depends on the difference in concentration of M²⁺ ions at the two electrodes .The EMF of the cell at 298 K is 0.059 V.
Q.
The value of ΔG (kJ mol^-1) is (1F = 96500 C mol^-1)

A
-5.7

B
5.7

C
11.4

D
-11.4

Solution
Question 14
4 / -1

Passage II
The standard half-cell reduction potential of
Fe³⁺(aq) | Fe is -0.036V and that of OH^- | Fe(OH₃)(s) | Feis -0.786V.
Q.
For the determination of solubility product of Fe(OH)₃ (K_sp) the appropriate cell representation and its EMF respectively are

A
Fe | Fe(OH)₃(s) || OH^-.(aq) Fe³⁺ (aq) | Fe -0.750 V

B
Fe | Fe³⁺ (aq), OH^-(aq) || Fe(OH)₃(s) | | Fe -0.750 V

C
Fe|Fe(OH₃)(s) || OH^-(aq), Fe³⁺(aq) | Fe +0.750V

D
Fe|Fe³⁺(aq), OH^-(aq)| Fe(OH₃)(s)|Fe -0.822V

Solution

Above reaction is the net reaction in a cell for determination of K_sp.
Question 15
4 / -1

Passage II
The standard half-cell reduction potential of
Fe³⁺(aq) | Fe is -0.036V and that of OH^- | Fe(OH₃)(s) | Feis -0.786V.
Q.
The value of log_e K_sp for Fe(OH)₃ at 298 K is

A
-38.1

B
+87.7

C
-96.0

D
-87.7

Solution

Above reaction is the net reaction in a cell for determination of K_sp.
Thus,
Anode
Fe → Fe³⁺ (aq) + 3e^-, E⁰_ox= 0.036V
Half - cell is Fe|Fe³⁺ (aq). OH^-(aq)
Cathode
Fe(OH)₃(s) + 3e^-→ Fe+ 3OH^-, E⁰_red= -0.786 V
Half-cell set up is Fe(OH)₃(s)|Fe
complete cell is Fe|Fe³⁺(aq), OH^- (aq)||Fe(OH)₃ (s)|Fe
and E⁰_cell= E⁰_ox+ E⁰_red= 0.036 + (-0.786) = -0.750V
By net reaction, K = K_sp= [Fe³⁺][OH^-]³
∴
Question 16
4 / -1

Passage III
Tollen's reagent is used for the detection of aldehydes.When a solution of AgNO₃is added to glucose with NH₄OH, then gluconic acid is formed.
Q.
For the reaction,

A
66.13

B
58.45

C
28.30

D
46.29

Solution

2Ag⁺ + 2e^- → 2Ag
C₆H₁₂O₆ +H₂O → C₆H₁₂O₇ + 2H⁺ + 2e^-
E⁰_r = 0.80V E⁰ = -0.05V
2Ag⁺ + C₆H₁₂O₆+ H₂O → 2Ag+ C₆H₁₂O₇+ 2H⁺
E⁰_cell= 0.75 V
At equilibrium, E_cell = 0

When NH₃ is added, pH increases to 11
Question 17
4 / -1

Passage III
Tollen's reagent is used for the detection of aldehydes.When a solution of AgNO₃is added to glucose with NH₄OH, then gluconic acid is formed.
Q.
In the presence of NH₃, pH is raised to 11.Thus,

A
E_ox will increase by a factor of 0.65 from E°_ox

B
E_ox will decrease by a factor of 0.65 from E°_ox

C
E_red will increase by a factor 0.65 from E°_red

D
E_red will decrease by a factor of 0.65 from E°_red

Solution

When NH₃ is added, pH increases to 11
Question 18
4 / -1

This section contains 2 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

Q.

For the following half-cell,

EMF at 298 K is 0.2364 V. What is pH?

A
4

B
5

C
6

D
9

Solution

It is oxidation half-cell, and anode reaction is
Question 19
4 / -1

For Cu(OH)₂|Cu hall-cell at pH = 12, electrode potential is 0.0455 at 298 K.

For Cu²⁺ (aq) + 2e^- → Cu, E° = 0.34 V. K_sp value of Cu(OH)₂ is x x 10^-4.

What is the value of x?

A
0

B
1

C
2

D
4

Solution