Q: Electrode potential data are given below: Fe ^ +3 ( aq )+ e ^ - Fe ^ +2 ( aq ) ; E ^ =+0.77 ~V Al ^ 3+ ( aq )+3 e ^ - Al _ ( s ) ; E ^ =-1.66 ~V Br _2( aq )+2 e ^ - 2 Br ^ - ( aq ) ; E ^ =+1.08 ~V Bas

array rrrr & Fe & A1 & B \\ E_ Red ^ & 0.77 & -1.66 & 1.08 \\ E_ Oxi ^ & -0.77 & 1.66 & -1.08 array Hence, reducing power Al > Fe ^ 2+ > Br

Question 1

The pH of 0.5 L of 1.0 M NaCl solution after electrolysis for 965 s using 5.0 A current is,

Accepted Answer

The reaction involved is, NaCl + H _2 O Electrolysis 1 2 H _2+ Cl _2+ NaOH Amount of NaCl present in 0.5 L of 1.0 M , NaCl =0.5 mol Quantity of electricity passed =965 5 coulombs. 4825 coulombs will decompose NaCl . aligned & = 4825 96500 ~mol \ & =0.05 ~mol aligned NaOH formed in the solution will also be 0.05 mol . Volume of solution =0.5 ~L Molarity of NaOH in the solution aligned & = 0.05 moles 0.5 ~L =0.1 M \ & =10^ -1 M aligned array rlrl & & pOH & =1 \ or & pH & =14-1=13 array

Question 2

Calculate the molarity of a solution containing 5 g of NaOH dissolved in the product of H _2- O _2 fuel cell operated at 1 A current for 595.1 hours. (Assume F =96500 C / mol of electron and molecular

Accepted Answer

The overall reaction of H _2- O _2 fuel is, 2 H _2+ O _2 2 H _2 O Thus 4 ~F =4 96500 C produce water =2 ~mol =36 ~g aligned Charge actually passed & =i t \ & =1 595.1 60 60 \ & =2142360 C aligned aligned & Water produced = 36 4 96500 2142360 \ & 200 ~g or 200 ~mL \ & Molarity of NaOH solution = 5 40 1 200 1000 \ & =0.625 M aligned

Question 3

When the same quantity of electricity is passed through the aqueous solutions of the given electrolytes for the same amount of time, which metal will be deposited in maximum amount on the cathode?

Accepted Answer

The equivalent weight of silver is highest among the given options.

Question 4

Which of the following solutions of KCl will have the highest value of specific conductance?

Accepted Answer

The specific conductance increases with concentration. The number of ions per cm ^3 increase with increase of concentration.

Question 5

Given that E_ O _2 / H _2 O ^ =+1.23 ~V ; aligned & E_ S _2 O _8^ 2- / SO _4^ 2- ^ =2.05 ~V ; E_ Br _2 / Br ^ - ^ =+1.09 ~V \ & E_ Au ^ 3+ / Au ^ =+1.4 ~V aligned The strongest oxidising agent is :

Accepted Answer

Reduction potential is maximum for S _2 O _8 ^ 2- therefore, it is the strongest oxidising agent amongst the given species.

Question 6

The conductivity of a weak acid HA of concentration 0.001 ~mol ~L ^ -1 is 2.0 10^ -5 ~S ~cm ^ -1 . If _ m ^ ( HA )=190 ~S ~cm ^2 ~mol ^ -1 , the ionization constant ( K _ a ) of HA is equal to _______

Accepted Answer

_ m =1000 K M =1000 2 10^ -5 0.001 =20 ~S ~cm ^2 ~mol ^ -1 = _ m _ m ^ = 20 190 = ( 2 19 ) 0.001(1- ) HA 0.001 H ^ + + 0.001 A ^ - K_ a =0.001 ( ^2 1- )= 0.001 ( 2 19 )^2 1- ( 2 19 ) =12.3 10^ -6

Question 7

Electrode potential data are given below: Fe ^ +3 ( aq )+ e ^ - Fe ^ +2 ( aq ) ; E ^ =+0.77 ~V Al ^ 3+ ( aq )+3 e ^ - Al _ ( s ) ; E ^ =-1.66 ~V Br _2( aq )+2 e ^ - 2 Br ^ - ( aq ) ; E ^ =+1.08 ~V Bas

Accepted Answer

array rrrr & Fe & A1 & B \ E_ Red ^ & 0.77 & -1.66 & 1.08 \ E_ Oxi ^ & -0.77 & 1.66 & -1.08 array Hence, reducing power Al > Fe ^ 2+ > Br

Question 8

Using the standard electrode potential, find out the pair between which redox reaction is not feasible. E^ values Fe ^ 3+ / Fe ^ 2+ =+0.77 ; I _2 / I ^ - =+054 Cu ^ 2+ / Cu =+0.34 ; Ag ^ + / Ag =0.80

Accepted Answer

(a) This reaction is feasible, since E_ cell ^ is positive. (b) This reaction is feasible, since E_ cell ^ is positive. (c) This reaction is feasible, since E_ cell ^ is positive. (d) This reaction is not feasible, since E_ cell ^ is negative.

Question 9

At 25^ C , the molar conductance at infinite dilution for the strong electrolytes NaOH , NaCl and BaCl _ 2 are 248 10^ -4 , 126 10^ -4 and 280 10^ -4 Sm ^ 2 ~mol ^ -1 respectively. _ m ^ 0 Ba ( OH )_

Accepted Answer

_ Na ^ + ^ + _ OH ^ - ^ =248 10^ -4 Sm ^ 2 ~mol ^ -1 _ Na ^ + ^ + _ Cl ^ - ^ =126 10^ -4 Sm ^ 2 ~mol ^ -1 _ Ba ^ ^ 2+ + _ 2 Cl ^ - ^ =280 10^ -4 Sm ^ 2 ~mol ^ -1 NoW , _ Ba ( OH )_ 2 ^ = _ BaCl _ 2 ^ +2 ^ _ NaOH ^ -2 _ NaC 1 ^ _ Ba ( OH )_ 2 ^ =280 10^ -4 +2 248 10^ -4 -2 126 10^ -4 _ Ba ( OH )_ 2 ^ =524 10^ -4 Sm ^ 2 ~mol ^ -1

Electrochemistry — Previous Year Questions & Solutions