Electrochemistry Quiz 5 – Punjab Meritorious School Preparation Portal

Welcome to your 12th Oct Electrochemistry Quiz 5 CBT TEST

Fill the Details mentioned below and Attempt the Test.

Name

School Name with City

Phone Number

A button cell used in watches function as following :

Zn_(s) + Ag₂O_(s) + H2O(l) 2Ag(s) + Zn²⁺(aq) + 2OH^–(aq)

If half cell potentials are

Zn²⁺(aq) + 2e^– → Zn_(s); E° = – 0.76 V

Ag₂O(s) + H₂O(l) + 2e^– → 2Ag(s) + 2OH^–(aq) ; E° = 0.34 V

The cell potential will be

0.84 V

1.34 V

1.10 V

0.42 V

A hypothetical electrochemical cell is shown below :

A/ A⁺ (x M)||B⁺(y M)|B

The emf measured is + 0.20 V. The cell reaction is

On the basis of the following E° values, the strongest oxidizing agent is

[Fe(CN)₆]^4– → [Fe(CN)₆]^3– + e^– ; E° = –0.35 V

Fe²⁺ → Fe³⁺ + e^– ; E° = –0.77 V

The standard electrode potential (E°) values of Al³⁺/Al, Ag⁺/Ag, K⁺/K and Cr³+/Cr are –1.66 V, 0.80V, –2.93 V and –0.74 V, respectively. The correct decreasing order of reducing power of the metal is

Ag > Cr > Al > K

K > Al > Cr > Ag

K > Al > Ag > Cr

Al > K > Ag > Cr

Standard electrode potential for Sn⁴⁺/Sn²⁺ couple is +0.15 V and that for the Cr³⁺/Cr couple is –0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be

+ 1.19 V

+ 0.89 V

+ 0.18 V

+ 1.83 V

Standard reduction potentials of the half reactions are given below :

F_2(g) + 2e^– → 2F^–_(aq) ; E° = + 2.85 V

Cl_2(g) + 2e^– → 2Cl^–_(aq); E° = + 1.36 V

Br_2(l) + 2e^– → 2Br^–_(aq) ; E° = + 1.06 V

I_2(s) + 2e^– → 2I^–_(aq) ; E° = + 0.53 VThe strongest oxidising and reducing agents respectively are

A solution contains Fe²⁺, Fe³⁺ and I^– ions. This solution was treated with iodine at 35°C. E° for

Fe³⁺/Fe²⁺ is + 0.77 V and E° for I₂/2I^– = 0.536 V. The favourable redox reaction is

E°_Fe2+/Fe = – 0.441 V and E°_Fe3+/Fe2+ = 0.771 V, the standard EMF of the reaction Fe + 2Fe³⁺ → 3Fe²⁺ will be

0.111 V

0.330 V

1.653 V

1.212 V

Standard electrode potentials of three metals X, Y and Z are –1.2 V, + 0.5 V and – 3.0 V respectively.

The reducing power of these metals will be

Y > Z > X

Y > X > Z

Z > X > Y

X > Y > Z

Consider the following relations for emf of an electrochemical cell

(i) EMF of cell = (Oxidation potential of anode) – (Reduction potential of cathode)

(ii) EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode)

(iii) EMF of cell = (Reductional potential of anode) + (Reduction potential of cathode)

(iv) EMF of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)

Which of the above relations are correct?

(iii) and (i)

(i) and (ii)

(iii) and (iv)

(ii) and (iv)

Please fill in the comment box below.