Test: Reference Half Cells & Nernst Equations - JEE MCQ

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   

Two half-cells are given

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

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

For a spontaneous cell reaction, cell set up is    

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

cell reaction is spontaneous ,if 

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

2H⁺ + 2e^-  → H₂ (1 bar)   

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...............

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 

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

Cell reaction is spontaneous , if  

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

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 

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

In which cell , is ΔG^° most negative?

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)?   

For

Then E^° for the reaction    

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

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

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).

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

E_cell = E°_cell if

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

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

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 E_cell 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    

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 E_cell 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      

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)   

Passage II

For the following,

Q.

pK_a of the weak monobasic acid is  

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, 

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

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........                                                                                                    

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

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

Derive the value of x.