Electrochemistry Test

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Electrochemistry Test - 19

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Question 1
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Only One Option Correct Type
This section contains 17 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.
Which cell will measure the standard electrode potential of zinc electrode?

A
P t(s)|H₂(g, 0.1 bar) | H⁺ (aq, 0.1 M) || Zn²⁺(aq, 0.1 M)|

B
P t(s)|H₂(g, 0.1 bar) | H⁺ (aq, 0.1 M) || Zn²⁺(aq, 0.2 M)| Zn

C
P t(s)|H₂(g, 1 bar) | H⁺ (aq, 0.1 M) || Zn²⁺(aq, 0.1 M)| Zn

D
P t(s)|H₂(g, 0.1 bar) | H⁺ (aq, 1.0 M) || Zn²⁺(aq, 0.1 M)| Zn

Solution

Standard electrode potential is the potential difference under standard state, i.e.
When [Mⁿ⁺] = 1M
It can be measured by coupling it with SHE electrode
Pt (s) | H₂(gr. 1 bar) | H⁺ (1.0 M)
Question 2
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For the following cell with hydrogen electrodes at two different pressure p₁and p₂
emf is given by

A

B

C

D

Solution

For SHE E°_SHE = 0.00 V
Oxidation at anode (left)

Reduction at cathode (right)
Net
This is the type of the cell in which electrodes at different pressures are dipped in same electrolyte and connectivity is made by a salt-bridge.
Reaction Quotient (Q)
∵
Question 3
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For the cell,
Thus (x/y) is

A
100

B
10

C
0.01

D
0.1

Solution

This is a type of concentration cell using hydrogen electrode as anode and cathode.
Question 4
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A solution of Fe²⁺ is titrated potentiometrically using Ce⁴⁺solution.
Fe²⁺ → Fe³⁺ + e^- , E⁰ = -0.77 V
emf of the Pt | Fe²⁺ , Fe³⁺ pair at 50% and 90% titration of Fe²⁺ are

A
0.77 V , 0.826 V

B
-0.826 V , -0.77 V

C
-0.77 V ,-0.826 V

D
0.00 V , 0.00 V

Solution

When Fe²⁺ is 50% titrated
=
where Fe²⁺ is 90% titrated
Question 5
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In the following cell at 298 K,two weak acids (HA) and (HB) with pK_a (HA) = 3 and pK_a (HB) = 5 of equal molarity have been used as shown.
Thus, emf of the cell is

A
0.059 V

B
-0.059 V

C

D

Solution

For weak acid by Ostwald's dilution law

= 0 - 0.0591 [log [H+]_L - log [H⁺]_R]
= 0.0591 [-log (H+)_L - (-log (H⁺)_R]
= 0.0591 [(pH)_HA - (pH)_HB]
= + 0.0591 [1.5 - 2.5]= - 0.0591
Question 6
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A concentration cell reversible to anion (Cl^-) is set up

cell reaction is spontaneous ,if

A
C₁ > C₂

B
C₁ < C₂

C
C₁ = C₂

D
C₁ = 0

Solution

This is a type of concentration cell with gas electrodes at the same pressure (1 bar) but dipped in aqueous solution of different concentration. Hence, a potential difference is set up.
At anode
At cathode

=

To make cell reaction spontaneous, E_cell > 0, hence C, > C₂
Question 7
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Which has the maximum potential at 298 K (numerical value) for the half-cell reaction?
2H⁺ + 2e^- → H₂ (1 bar)

A
1.0 M HCl

B
A solution having pH 4

C
Pure water

D
1.0 M NaOH

Solution

2H⁺ + 2e^- → H₂
This represents reduction half-cell reaction

(a) [H⁺] = 1 M, E = 0
(b) pH = 4, [H⁺] = 10^-4M
∴ E = 0.0591 log 10^-4
= - 4 x 0.0591 = - 0.2364 V
(c) Pure water, [H⁺] = 10^-7M
∴ E = 0.0591 log 10^-7
= - 7 x 0,0591 = - 0.4137 V
(d) 1.0 M NaOH
[OH^-] = 1 M

∴ E = 0.0591 log 1 x 10^-14
= -14 x 0.0591
= - 0.8274 V (maximum)
Question 8
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For the cell, and for the cell Pt(H₂) | H⁺(1M)| Ag,
Thus E_cell for the
Ag|Ag⁺ (0.1M) || Zn²⁺(0.1M) | Zn is ....................and cell reaction is...............

A
1.44 V ,spontaneous

B
0.4 V ,spontaneous

C
-1.44 V ,non-spontaneous

D
-1.53 V ,non-spontaneous

Solution

E_cell < 0, hence reaction is non-spontaneous.
Question 9
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Consider the following cell reaction,
2Fe(s) + O₂(g) + 4H⁺(aq) →2Fe²⁺(aq) + 2H₂O(l) , E^°= 1.67 V
At [Fe²⁺] = 1 x 10^-3 M. and pH = 3,the cell potential at 298 K is

A
1.47 V

B
1.77 V

C
1.87 V

D
1.57 V

Solution

pH = 3, [H⁺] = 10^-3M

Electrons involved n = 4
Question 10
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For the following cell with gas electrodes at p₁ and p₂ as shown:
Cell reaction is spontaneous , if

A
p₁ = p₂

B
p₁ > p₂

C
p₁ < p₂

D
p₂= 0

Solution

This is a type of concentration cell in which two electrodes are at different pressure but dipped in same electrolyte. Salt-bridge is used to make connectivity and liquid junction potential is minimised and
E°_cell = 0.00V
Question 11
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For the half-cell, Cl^-|Hg₂Cl₂, Hg(l), E = 0.280V at 298K electrode potential has maximum value when KCl used is

A
1 M

B
0.1 M

C
0.01 M

D
10 M

Solution

Cl^- / Hg₂CI₂, Hg (/)
This is reduction half-cell
Hg₂CI₂(s) + 2e^- → 2Hg (/) + 2Cl^-
Reaction quotient (Q) = [Cl^-]²

Thus, larger the value of [Cl^-], smaller the value of E_caiomel.
Question 12
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Given at 298 K standard oxidation potential of quinhydrone electrode = -0.699 V Standard oxidation potential of calomel electrode = -0.268 V
Thus, emf of the cell at 298 K is

A
0.82 V

B
0.16 V

C
-0.17 V

D
-0.82 V

Solution
Question 13
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E^°_red (standard reduction electrode potentials) of different half-cell are given

In which cell , is ΔG^° most negative?

A
Zn|zn²⁺(1M)||Mg²⁺(1M)|Mg

B
Zn|zn²⁺ (1M) ||Ag⁺ (1M)|Ag

C
Cu|Cu²⁺ (1M)||Ag⁺(1M) | Ag

D
Ag|Ag⁺ (1M) || Mg²⁺ (1M)|Mg

Solution

In (b) E°_cell is most positive, then ΔG° is most negative.
Question 14
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Given the following half-cell reactions and corresponding reduction potentials:
Which combination of two half-cell would result in a cell with largest (E^°_cell > 0)?

A
II and III

B
I and II

C
I and III

D
II and Iv

Solution

(a)
Oxidation
C^3-→ C^-+ 2e^-E⁰ = 1.25 V
Reduction
Question 15
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For

Then E^° for the reaction

A
+ 0.255 V

B
- 0.255 V

C
+ 1.055 V

D
- 1.055 V

Solution
Question 16
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Following half-cell, Pt (H₂)|H₂O behaves as SHE at a pressure of

A
1 bar

B
10^-14bar

C
10⁷bar

D
10¹⁴bar

Solution

Pt(H₂)| H₂O
This is oxidation half-cell. Instead of
[H⁺] = 1 M and p_H2 = 1 bar
H₂O has been taken as a source of [H⁺]
Question 17
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Electrode potential of the following half-cell is dependent on
Hg, HgO |OH^-(aq)

A
temperature

B
pH

C
Both (a) and (b)

D
None of these

Solution

Thus, electrode potential is dependent on (i) pH, (ii) temperature.