Q: If the molar conductivity ( _ m ) of a 0.050 ~mol ~L ^ -1 solution of a monobasic weak acid is 90 ~S ~cm ^2 ~mol ^ -1 , its extent (degree) of dissociation will be [Assume _ + ^ =349.6 ~S ~cm ^2 ~mol

aligned & Degree of dissociation ( )= _m _m^ \\ & aligned _ m ^ & = _ + ^ + _ - ^ \\ & =349.6+50.4=400 Scm ^2 ~mol ^ -1 \\ & = _ m _ m ^ = 90 400 =0.225 aligned aligned

Question 1

The standard electrode potential (E^ ) for the half-cell reaction Fe ^ 3+ +e^- Fe ^ 2+ at 298 K is (Given: E^ ( Fe ^ 3+ / Fe )=-0.04 V and E^ ( Fe ^ 2+ / Fe )=-0.44 V at 298 K)

Accepted Answer

The given half-cell reactions are: Fe ^ 3+ + 3e^- Fe , E^ _1 = -0.04 V G^ _1 = -n_1FE^ _1 = -3 F (-0.04) = 0.12F Fe ^ 2+ + 2e^- Fe , E^ _2 = -0.44 V G^ _2 = -n_2FE^ _2 = -2 F (-0.44) = 0.88F The required half-cell reaction is: Fe ^ 3+ + e^- Fe ^ 2+ This reaction can be obtained by subtracting the second reaction from the first reaction: G^ _3 = G^ _1 - G^ _2 -1 F E^ _3 = 0.12F - 0.88F -FE^ _3 = -0.76F E^ _3 = +0.76 V Answer: +0.76 V

Question 2

For a salt XY , which is a strong electrolyte, the plot of _m versus c has a slope of -90.0 S cm^2 mol^ -3/2 L^ 1/2 at 298 K. At 0.01 M concentration of XY , the value of _m is 145.0 S cm^2 mol^ -1 .

Accepted Answer

According to the Debye-Huckel-Onsager equation for strong electrolytes: _m = _m^0 - A c The slope of the plot of _m versus c is -A. Given slope = -90.0, we have A = 90.0. At c = 0.01 M, c = 0.1 M^ 1/2 . Substituting the given values into the equation: 145.0 = _m^0 - 90.0 0.1 145.0 = _m^0 - 9.0 _m^0 = 154.0 S cm^2 mol^ -1 According to Kohlrausch's law of independent migration of ions: _m^0( XY ) = ^0_ X ^+ + ^0_ Y ^- 154.0 = 74.0 + ^0_ Y ^- ^0_ Y ^- = 154.0 - 74.0 = 80.0 S cm^2 mol^ -1 Answer: 80

Question 3

A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is : (Given : Molar mass of Cu = 63 g mol^ -1 ; 1 F = 96487 C mol^ -

Accepted Answer

The reaction at the cathode is Cu ^ 2+ + 2 e ^ - Cu . The quantity of charge Q passed through the solution is given by Q = I t. Substituting the given values: Q = 1.5 A (10 60) s = 900 C From Faraday's first law of electrolysis, the mass of copper deposited is: m = M Q n F Here, M = 63 g mol ^ -1 , n = 2, and F = 96487 C mol ^ -1 . m = 63 900 2 96487 m = 56700 192974 0.2938 g Answer: 0.2938 g

Question 4

Calculate emf of the half cell given below : Pt (s) | H _2 (g, 2 atm) | HCl (aq, 0.02 M) E^ _ H _2/ H ^ + = 0 V ( Given : 2.303\ RT F = 0.059, 2 = 0.3010 )

Accepted Answer

The given half-cell is represented as an oxidation electrode: Pt (s) | H _2 (g) | H ^ + (aq) . The corresponding oxidation half-cell reaction is: H _2 (g) 2 H ^ + (aq) + 2 e ^ - Using the Nernst equation for the oxidation potential: E = E^ _ H _2/ H ^ + - 0.059 n [ H ^ + ]^2 P_ H _2 Given values: E^ _ H _2/ H ^ + = 0 V n = 2 [ H ^ + ] = 0.02 M (since HCl is a strong monoprotic acid) P_ H _2 = 2 atm Substituting the values into the Nernst equation: E = 0 - 0.059 2 (0.02)^2 2 E = -0.0295 4 10^ -4

Question 5

If the molar conductivity ( _ m ) of a 0.050 ~mol ~L ^ -1 solution of a monobasic weak acid is 90 ~S ~cm ^2 ~mol ^ -1 , its extent (degree) of dissociation will be [Assume _ + ^ =349.6 ~S ~cm ^2 ~mol

Accepted Answer

aligned & Degree of dissociation ( )= _m _m^ \ & aligned _ m ^ & = _ + ^ + _ - ^ \ & =349.6+50.4=400 Scm ^2 ~mol ^ -1 \ & = _ m _ m ^ = 90 400 =0.225 aligned aligned

Question 6

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 ~g ~mol ^ -1 , 1 ~F =96487 C )

Accepted Answer

Cu ^ 2+ ( aq )+2 e ^ - Cu ( s ) aligned Mass of Cu deposited (w) & = M i t n F \ & = 63 9.6487 100 2 96487 \ & =0.315 ~g aligned

Question 7

From the following select the one which is not an example of corrosion.

Accepted Answer

Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal. The rusting of iron, tarnishing of silver, development of green coating on copper and bronze are some of the examples of corrosion. - Production of H _2 by electrolysis of water is an example of electrolytic cell.

Question 8

The standard cell potential of the following cell Zn | Zn ^ 2+ ( aq ) | Fe ^ 2+ ( aq ) Fe is 0.32 V . Calculate the standard Gibbs energy change for the reaction : Zn ( s )+ Fe ^ 2+ ( aq ) Zn ^ 2+ ( a

Accepted Answer

_r G^ =-n F E_ cell ^ For the given reaction, n=2 _ r G ^ =-2 96487 0.32 =-61751.68 ~J ~mol ^ -1 =-61.751 ~kJ ~mol ^ -1

Electrochemistry — Previous Year Questions & Solutions