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# Test: Reference Half Cells & Nernst Equations

## 29 Questions MCQ Test Chemistry for JEE | Test: Reference Half Cells & Nernst Equations

Description
This mock test of Test: Reference Half Cells & Nernst Equations for Class 12 helps you for every Class 12 entrance exam. This contains 29 Multiple Choice Questions for Class 12 Test: Reference Half Cells & Nernst Equations (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Reference Half Cells & Nernst Equations quiz give you a good mix of easy questions and tough questions. Class 12 students definitely take this Test: Reference Half Cells & Nernst Equations exercise for a better result in the exam. You can find other Test: Reference Half Cells & Nernst Equations extra questions, long questions & short questions for Class 12 on EduRev as well by searching above.
QUESTION: 1

### Only One Option Correct Type This section contains 18 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?

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

### For the following cell with hydrogen electrodes at two different  pressure p1 and p2   emf is given by

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

### For the cell, Thus (x/y) is

Solution:

This is a type of concentration cell using hydrogen electrode as anode and cathode.

QUESTION: 4

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

Solution:

When Fe2+ is 50% titrated

=

where Fe2+ is 90% titrated

*Multiple options can be correct
QUESTION: 5

Two half-cells are given

Ag | AgCl | KCl (0.2 M), Ag | AgBr | KBr (0.001 M),

Ksp(AgCl) = 2.8 x 10-10, Ksp (AgBr) = 3.3 x 10 -13

For a spontaneous cell reaction, cell set up is

Solution:

Ksp (AgCI) = 2.8 x 10-10
[Ag+] [Cl-] = 2 .8 x 10-10

∴ [Ag+]left =
Ksp (AgBr) = 3.3 x 10-12
[Ag+][Br-] = 3.3 x 10-13

∴ [Ag+]left =

Net

= -0.037 V
Thus, cell reaction is non-spontaneous
(b) Cell is reversed of (a), thus spontaneous.

QUESTION: 6

In the following cell at 298 K,two weak acids (HA) and (HB) with pKa (HA) = 3 and pKa (HB) = 5 of equal molarity have been used as shown.

Thus, emf of the cell is

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: 7

A concentration cell reversible to anion (Cl-) is set up

cell reaction is spontaneous ,if

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, Ecell > 0, hence C, > C2

QUESTION: 8

Which has the maximum potential at 298 K (numerical value) for the half-cell reaction?

2H+ + 2e-  → H2 (1 bar)

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: 9

For the cell,  and for the cell Pt(H2) | H+ (1M)| Ag,

Thus Ecell for the

Ag|Ag+ (0.1M) || Zn2+ (0.1M) | Zn is  ....................and cell reaction is...............

Solution:

Ecell < 0, hence reaction is non-spontaneous.

QUESTION: 10

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

Solution:

pH = 3, [H+] = 10-3M

Electrons involved n = 4

QUESTION: 11

For the following cell with gas electrodes at p1 and p2 as shown:

Cell reaction is spontaneous , if

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
cell = 0.00V

QUESTION: 12

For the half-cell, Cl-|Hg2Cl2, Hg(l), E = 0.280V at 298K electrode potential has maximum value when KCl used is

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 Ecaiomel.

QUESTION: 13

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

Solution:

QUESTION: 14

E°red (standard reduction electrode potentials) of different half-cell are given

In which cell , is ΔG° most negative?

Solution:

In (b) E°cell is most positive, then ΔG° is most negative.

QUESTION: 15

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)?

Solution:

(a)

Oxidation

C3- → C- + 2e-        E0 = 1.25 V

Reduction

QUESTION: 16

For

Then E° for the reaction

Solution:

QUESTION: 17

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

Solution:

Pt(H2)| H2O
This is oxidation half-cell. Instead of
[H+] = 1 M and pH2 = 1 bar
H2O has been taken as a source of [H+]

QUESTION: 18

Electrode potential of the following half-cell is dependent on
Hg, HgO |OH-(aq)

Solution:

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

*Multiple options can be correct
QUESTION: 19

One or More than One Options Correct Type

This section contains 4 multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q.

Select the correct statement(s).

Solution:

a) If salt-bridge is removed, anodic and cathodic solution intermix and potential difference falls to zero — correct.

Quinhydrone is a mixture of A and S in (1:1) molar ratio joined by H-bonding. Above equilibrium is set up in aqueous solution.

If [H+] changes, EQH also changes. Thus, quinhydrone electrode is reversible to H+ ion correct
(c) Liquid junction potential is minimised by use of concentration cell— correct.
(d) Calomel electrode is Hg(/), Hg2Cl2 (s)| Cl-.
Thus, (d) is incorrect.

*Multiple options can be correct
QUESTION: 20

In the following electrochemical cell.
Zn cell Zn2+ ||H| Pt (H2)

Ecell = E°cell if

Solution:

*Multiple options can be correct
QUESTION: 21

For Daniell cell E°cell = 1.10 V. If state of equilibrium is attained, then

Solution:

Reaction taking place in a Daniell cell is

Zn(s) + Cu2+ (1M)  Cu(s) + Zn2+ (1M),

*Multiple options can be correct
QUESTION: 22

Select the half-cell for the half-cell reaction(s) to be spontaneous

Solution:

QUESTION: 23

Comprehension Type

This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

Passage I

1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecell of Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 V

Q.

Pure [Ag+] in the ore is

Solution:

Cell Reaction

∴ x = 8.4 x 10-6 M

[Ag+] = 8.4 x 10-6 mol L-1

QUESTION: 24

Passage I

1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecell of Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 V

Q.

Percentage of silver in the sample is

Solution:

Cell Reaction

∴ x = 8.4 x 10-6 M

[Ag+] = 8.4 x 10-6 mol L-1

= 8.4 x 10-6 x 0.350 mol in 350 mL

= 8.4 x 10-6 x 0.350 x 108 g in 350 mL

= 3.1752 x 10-4 g in 1.05 g sample

QUESTION: 25

Passage II

For the following,

Q.

pH of the solution in the half-cell containing 0.02 HA is(HA is a weak monobasic acid)

Solution:

Anodic

Cathodic

Reaction quotient (Q)

QUESTION: 26

Passage II

For the following,

Q.

pKa of the weak monobasic acid is

Solution:

Anodic

Cathodic

Reaction quotient (Q)

*Answer can only contain numeric values
QUESTION: 27

One Integer Value Correct Type

This section contains 3 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

Q.

For the following cell with metal X electrodes,

Ecell = -0.028 V at 298 K, if there is no liquid juncton potential,valency of X is.......

Solution:

*Answer can only contain numeric values
QUESTION: 28

Osmatic pressure of a 0.001 M weak monobasic acid (HA) at 300 K is 2.5 x 10-2 atm.Thus,emf of the follwing in decivolt is| 0.001 M HA........

Solution:

This question relates colligative properties to emf of the cell

*Answer can only contain numeric values
QUESTION: 29

pH of the solution in the anode compartment of the follwoing cell at 298 K is x

when Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.

Derive the value of x.

Solution: