Q: An aqueous solution of hydrazine ( N _2 H _4 ) is electrochemically oxidized by O _2, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electro

$ Oxidation \\ (Anode) -2 N_2H_4 + Reduction \\ (Cathode) 0 O_2 0 N_2 + H _2 O ^ -2 $ At anode: $ N _2 H _4+4 OH ^ - N _2+4 H _2 O +4 e^ - $ At cathode: $ O _2+2 H _2 O +4 e^ - 4 OH ^ - $ Complete reaction: $ N _2 H _4+ O _2 N _2+2 H _2 O $ Statements (A) and (C) are correct.

Q: Consider the strong electrolytes Z m X n , U m Y p and V m X n . Limiting molar conductivity of U m Y p and V m X n are 250 and 440 S cm 2 mol - 1 , respectively

λ m = λ m ° - A C For electrolyte Z m X n and from given curve λ m Z m X n = λ m 0 Z m X n - A C - A = 336 - 339 0 . 04 - 0 . 01 = - 3 0 . 03 ⇒ A = 100 ∴ For λ m = 336 S cm 2 mol - 1 ⇒ 336 = λ m 0 Z m X n - 100 × 0 . 04 λ m 0 = 336 + 4 = 340 S cm 2 mol - 1 Z m X n ⟶ mZ n + + nX m - ∴ 50 m + 80 n = 340 ⇒ 5 m + 8 n = 34 &#

Q: The reduction potential E 0 , in V of MnO 4 - aq / Mn s is [Given: E MnO 4 - aq / MnO 2 s o = 1 . 68 V ; E MnO 2 s / Mn 2 + aq o = 1 . 21 V ; E Mn 2 + aq

Given (1) MnO 4 - aq + 4 H + + 3 e ⟶ MnO 2 s + 2 H 2 O ; E ° = 1 . 68 V ΔG 1 ° = - 3 F 1 . 68 = - 5 . 04 F (2) MnO 2 s + 4 H + + 2 e ⟶ Mn 2 + ( aq ) + 2 H 2 O ; E ° = 1 . 21 V ΔG 2 ° = - 2 F 1 . 21 = - 2 . 42 F (3) Mn 2 + aq + 2 e ⟶ Mn s ; E ° = - 1 . 03 V ΔG 3 ° = - 2 F - 1 . 03 = + 2 . 06 F Adding 1 , 2 and 3 , MnO 4 - aq + 8 H + + 7 e

Q: Some standard electrode potentials at 298 K are given below: Pb 2 + / Pb - 0 . 13 V Ni 2 + / Ni - 0 . 24 V Cd 2 + / Cd - 0 . 40 V Fe 2 + / Fe - 0 . 44 V To a solutio

Since X is getting dissolved, so reaction is X → X 2 + + 2 e - Anode Y 2 + + 2 e - → Y cathode Y 2 + + X → X 2 + + Y E cell = E cell ° - 0 . 06 2 log 10 - 3 10 - 1 E cell = E cell ° + 0 . 06 (A) Cd & Ni $=0.16+0.05912 2$ $=.16+0.0591$ $=0.21(+ ve )$ E cell < 0 (Non-spontaneous) (B) Cd & Fe E cell = − 0 .044 − − 0 .40 + 0 .06 = - 0 . 04 + 0 . 06 = 0

Q: At 298 ~K , the limiting molar conductivity of a weak monobasic acid is 4 10^ 2 ~S ~cm ^ 2 ~mol ^ -1 . At 298 ~K , for an aqueous solution of the acid the degree of dissociation is and the molar condu

α = λ m λ m ∘ α 1 = y × 10 + 2 4 × 10 2 = y 4 . . . 1 k a = C 1 α 1 2 1 − α 1 . . . 2 After dilution α 2 = 3 y × 10 2 4 × 10 2 = 3 y 4 . . . 3 k a = C 2 α 2 2 1 − α 2 . . . 4 eq. 1 / 3 ⇒ α 1 α 2 = 1 3 eq 2 / 4 ⇒ &#160

Question 1

At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below: Salt Concentration (M) Molar conductivity (S cm^2 mol^ -1 ) NaNO _3 0.01 111

Accepted Answer

Using the Debye-Hückel-Onsager equation, _m = ^ _m - b C , we can find the limiting molar conductivity ( ^ _m) for each salt. For NaNO _3: ^ _m - b 0.01 = 111 ^ _m - 0.1b = 111 ^ _m - b 0.04 = 101 ^ _m - 0.2b = 101 Subtracting the two equations gives 0.1b = 10 b = 100. ^ _m( NaNO _3) = 111 + 100(0.1) = 121 S cm ^2 mol ^ -1 . For NaCl : ^ _m - 0.1b = 117 ^ _m - 0.2b = 107 Subtracting gives 0.1b = 10 b = 100. ^ _m( NaCl ) = 117 + 100(0.1) = 127 S cm ^2 mol ^ -1 . For AgNO _3: ^ _m - 0.1b = 125 ^ _

Question 2

In an electrochemicalcell, dichromate ions in aqueous acidic medium are reduced to Cr ^ 3+ . The current (in amperes) that flows through the cell for 48.25 minutes to produce 1 mole of Cr ^ 3+ is ____

Accepted Answer

For reduction of dichromate, balanced reaction is : Cr _2 O _7^ -2 ( aq )+6 e ^ - +14 H ^ + ( aq ) 2 Cr ^ 3+ ( aq )+7 H _2 O ( l ) 3 Mole 1 Mole Number of Farads required =3 ~mol Let current = I amperes aligned & I 48.25 60 96500 =3 \ & I=100 ~A aligned

Question 3

An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is X F 10^3 volts, where F is the Faraday constant. The value of X is . Use : Standard Gibbs energ

Accepted Answer

aligned & C _4 H _ 10 ( ~g )+ 13 2 O _2( ~g ) 4 CO _2( ~g )+5 H _2 O (l) \ & _ r G ^ =4 _ f G _ CO _2 ^ +5 _ f G _ H _2 O ^ - _ f G _ C _4 H _ 10 ^ \ & =4 (-394)+5(-237)+18 \ & =-2743 ~kJ / mol \ & _ r G ^ =- nFE ^ \ & -2743 1000=-26 FE ^ \ & E ^ = 105.5 ~F 10^3=105.50 aligned

Question 4

An aqueous solution of hydrazine ( N _2 H _4 ) is electrochemically oxidized by O _2, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electro

Accepted Answer

$ Oxidation \ (Anode) -2 N_2H_4 + Reduction \ (Cathode) 0 O_2 0 N_2 + H _2 O ^ -2 $ At anode: $ N _2 H _4+4 OH ^ - N _2+4 H _2 O +4 e^ - $ At cathode: $ O _2+2 H _2 O +4 e^ - 4 OH ^ - $ Complete reaction: $ N _2 H _4+ O _2 N _2+2 H _2 O $ Statements (A) and (C) are correct.

Question 5

In a conductometric titration, small volume of titrant of higher concentration is added stepwise to a larger volume of titrate of much lower concentration, and the conductance is measured after each a

Accepted Answer

Option (P) : On adding AgNO _3 solution to KCl solution precipitation of AgCl will occur due to which Cl ^ - already present will be replaced by NO _3^ - ions. So conductance of solution will decrease till equivalence point. After complete precipitation of AgCl , further added AgNO _3 will increase the number of ions in resulting solution so conductance will increase. Option ( Q ) : On adding KCl solution to AgNO _3 solution precipitation of AgCl will occur due to which already present Ag ^ + io

Question 6

Plotting 1 / &#923; m against c&#923; m for aqueous solutions of a monobasic weak acid ( HX ) resulted in a straight line with y - axis intercept of P and slope of S . The ratio P / S  is &#923;

Accepted Answer

The degree of dissociation is the ratio of molar conductivity to the limiting molar conductivity. &#945; = &#923; m &#923; m o The equilibrium reaction is given below, &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; HX &#8652; H + + X - initial &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; c

Question 7

Consider the strong electrolytes Z m X n , &#160; U m Y p  and V m X n . Limiting molar conductivity of U m Y p and V m X n are 250 and 440 &#160; S &#160; cm 2 &#160; mol - 1 , respectively

Accepted Answer

&#955; m = &#955; m &#176; - A C For electrolyte Z m X n  and from given curve &#955; m Z m X n = &#955; m 0 Z m X n - A C - A = 336 - 339 0 . 04 - 0 . 01 = - 3 0 . 03 &#8658; A = 100 &#8756; For &#955; m = 336 &#160; S &#160; cm 2 &#160; mol - 1 &#8658; 336 = &#955; m 0 Z m X n - 100 &#215; 0 . 04 &#955; m 0 = 336 + 4 = 340 &#160; S &#160; cm 2 &#160; mol - 1 Z m X n &#10230; mZ n + + nX m - &#8756; &#160;&#160; 50 m + 80 n = 340 &#8658; 5 m + 8 n = 34 &#160; &#160; &#160; &#160; &#160; &#

Question 8

The reduction potential E 0 , in  V of MnO 4 - aq / Mn s is [Given:  E MnO 4 - aq / MnO 2 s o = 1 . 68 &#160; V ; &#160; E MnO 2 s / Mn 2 + aq o = 1 . 21 &#160; V ; &#160; &#160; E Mn 2 + aq

Accepted Answer

Given (1) MnO 4 - aq + 4 H + + 3 e &#10230; MnO 2 s + 2 H 2 O ;  E &#176; = 1 . 68 &#160; V &#916;G 1 &#176; = - 3 &#160; F 1 . 68 = - 5 . 04 &#160; F (2) MnO 2 s + 4 H + + 2 e &#10230; Mn 2 + ( aq ) &#160; + 2 H 2 O ;  E &#176; = 1 . 21 &#160; V &#916;G 2 &#176; = - 2 &#160; F 1 . 21 = - 2 . 42 &#160; F (3) Mn 2 + aq + 2 e &#10230; Mn s ;  E &#176; = - 1 . 03 &#160; V &#916;G 3 &#176; = - 2 &#160; F - 1 . 03 = + 2 . 06 &#160; F Adding  1 , 2 and 3 , MnO 4 - aq + 8 H + + 7 e

Question 9

Some standard electrode potentials at 298 &#160; K are given below: Pb 2 + / Pb - 0 . 13 &#160; V Ni 2 + / Ni - 0 . 24 &#160; V Cd 2 + / Cd - 0 . 40 &#160; V Fe 2 + / Fe - 0 . 44 &#160; V To a solutio

Accepted Answer

Since  X  is getting dissolved, so reaction is X &#8594; X 2 + + 2 e - &#160; &#160; Anode Y 2 + + 2 e - &#8594; Y &#160; &#160; cathode Y 2 + + X &#8594; X 2 + + Y E cell = E cell &#176; - 0 . 06 2 log 10 - 3 10 - 1 E cell = E cell &#176; + 0 . 06 (A)  Cd &#160; &#38; &#160; Ni $=0.16+0.05912 2$ $=.16+0.0591$ $=0.21(+ ve )$ E cell &#60; 0  (Non-spontaneous) (B)  Cd &#160; &#38; &#160; Fe E cell = &#8722; 0 .044 &#8722; &#8722; 0 .40 + 0 .06 = - 0 . 04 + 0 . 06 = 0

Question 10

At 298 ~K , the limiting molar conductivity of a weak monobasic acid is 4 10^ 2 ~S ~cm ^ 2 ~mol ^ -1 . At 298 ~K , for an aqueous solution of the acid the degree of dissociation is and the molar condu

Accepted Answer

&#945; = &#955; m &#955; m &#8728; &#945; 1 = y &#215; 10 + 2 4 &#215; 10 2 = y 4 &#160; &#160; &#160; &#160; &#160; . . . 1 k a = C 1 &#945; 1 2 1 &#8722; &#945; 1 &#160; &#160; &#160; &#160; &#160; &#160; . . . 2 After dilution &#945; 2 = 3 y &#215; 10 2 4 &#215; 10 2 = 3 y 4 &#160; &#160; &#160; &#160; &#160; . . . 3 k a = C 2 &#945; 2 2 1 &#8722; &#945; 2 &#160; &#160; &#160; &#160; &#160; &#160; . . . 4 eq.  1 / 3 &#8658; &#160; &#945; 1 &#945; 2 = 1 3 eq  2 / 4 &#8658; &#160

Question 11

At 298 ~K , the limiting molar conductivity of a weak monobasic acid is 4 10^ 2 ~S ~cm ^ 2 ~mol ^ -1 . At 298 ~K , for an aqueous solution of the acid the degree of dissociation is and the molar condu

Accepted Answer

&#945; = &#955; m &#955; m &#8728; &#945; 1 = y &#215; 10 + 2 4 &#215; 10 2 = y 4 &#160; &#160; &#160; &#160; &#160; . . . 1 k a = C 1 &#945; 1 2 1 &#8722; &#945; 1 &#160; &#160; &#160; &#160; &#160; &#160; . . . 2 After dilution &#945; 2 = 3 y &#215; 10 2 4 &#215; 10 2 = 3 y 4 &#160; &#160; &#160; &#160; &#160; . . . 3 k a = C 2 &#945; 2 2 1 &#8722; &#945; 2 &#160; &#160; &#160; &#160; &#160; &#160; . . . 4 eq.  1 / 3 &#8658; &#160; &#945; 1 &#945; 2 = 1 3 eq  2 / 4 &#8658; &#160

Question 12

Consider a 70 % efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 &#160; K . Its cell reaction is H 2 ( g ) + 1 2 O 2 ( g ) &#8594; H 2 O ( l ) The work derived fr

Accepted Answer

For given reaction  H 2 g + 1 2 O 2 g &#10230; 2 e - H 2 O l    E &#176; = 1 . 23 &#160; V &#916;G &#176; = - nFE cell&#160; &#176; = [ - 2 &#215; 96500 &#215; 1 . 23 ] &#160; 1 &#215; 10 - 3 &#215; 0 . 7 = - 166 . 173 &#160; J W = 166 . 173 &#160; J W = nR&#916;T &#947; - 1 166 . 173 = 1 &#215; 8 . 314 &#215; &#916;T 5 3 - 1 = 8 . 314 &#215; 3 2 &#916;T &#916;T = 166 . 173 &#215; 2 8 . 314 &#215; 3 = 13 . 32

Electrochemistry — Previous Year Questions & Solutions