Electrochemistry

NCERT Solution
Part 2

Question 11: Conductivity of 0.00241 M acetic acid is 7.896 `xx` 10-5 S cm-1. Calculate its molar conductivity. If Λm° for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?

Answer: `Λ_m=(κxx1000)/text(Molarity)`

`=(7.896xx10^(-5) S cm^(-1)xx1000 cm^3L^(-1))/(0.00241 mol L^(-1))`

`=32.76` S cm2 mol-1

`α=(Λ_m)/(Λ_m°)`

`=(32.76)/(390.5)=8.4xx10^(-2)`

Now, dissociation constant can be calculated as follows:

`K_a=(Cα^2)/(1-α)`

`=(0.0024xx(8.4xx10^(-2))^2)/(1-0.084)=1.86xx10^(-5)`


Question 12: How much charge is required for the following reductions:

1 mol of Al3+ to Al?

Answer: The electrode reaction is `Al^(3+)+3e^(-)→Al`

Hence, charge required for 1 mol of Al3+ `=3F=3xx96500 C=289500 C`

1 mol of Cu2+ to Cu?

Answer: The electrode reaction is `Cu^(2+)+2e^(-)→Cu`

Hence, charge required for reduction of 1 mol of Cu2+ `=2F=2xx96500 C=193000 C`

1 mol of MnO4- to Mn2+?

Answer: The electrode reaction is `Mn\O_4^(-)→Mn^(2+)`

This means, `Mn^(7+)+5e^(-)→Mn^(2+)`

Hence, charge required = 5F

`=5xx96500 C=482500 C`

Question 13: How much electricity in terms of Faraday is required to produce

20.0 g of Ca from molten CaCl2?

Answer: Reaction is `Ca^(2+)+2e^(-)→Ca`

So, charge required for 1 mol of Ca, i.e. 40 g Ca = 2F

Hence, charge required for 20 g Ca = 1F

40.0 g of Al from molten Al2O3?

Answer: Reaction is `Al^(3+)+3e^(-)→Al`

So, charge required for 1 mol of Al, i.e. 27 g Al = 3F

Hence, charge required for 40 g Al

`=3/(27)xx40=4.44` F

Question 14: How much electricity is required in coulomb for the oxidation of

1 mol of H2O to O2?

Answer: Reaction is `H_2O→H_2+1/2O_2`

This means `O^(2-)→1/2O_2+2e^(-)`

So, charge required = 2F

`=2xx96500=193000` C

1 mol of FeO to Fe2O3

Answer: Reaction is `Fe\O+1/2O_2→1/2Fe_2O_3`

This means `Fe^(2+)→Fe^(3+)+e^(-)`

So, charge required = 1F = 96500 C

Question 15: A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 ampere for 20 minute. What mass of Ni is deposited at the cathode?

Answer: Charge `= It`

`=5A\xx20xx60 s=6000 C`

Reaction is `Ni^(2+)+2e^(-)→Ni`

Hence, charge required for 1 mol of Ni = 2F = 2 `xx` 96500 C

We know that 1 mol of Ni = 58.7 g

Mass of Ni deposited by 6000 C charge

`=(58.7xx6000)/(2xx96500)=1.825` g

Question 16: Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4 respectively are connected in series. A steady current of 1.5 ampere was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Answer: Current I = 1.5A, mass of Ag = 1.45 g, t = ?, E = 108, n = 1

Using Faraday’s 1st law of electrolysis

`W=ZI\t`

Or, `W=E/(nf)It`

Or, `t=(1.45xx96500)/(1.5xx108)=863.73` second

Now, for Cu we have, W1 = 1.45 g of Ag, E1 = 108, W2 = ?, E2 = 31.75

From Faraday’s second law of electrolysis

`(W_1)/(W_2)=(E_1)/(E_2)`

`=(1.45)/(W_2)=(108)/(31.75)`

Or, `W_2=(1.45xx31.75)/(108)=0.426` g of Cu

Now, for Zn, W1 = 1.45 g of Ag, E1 = 108, W2 = ?, E2 = 32.65

Using the formula `(W_1)/(W_2)=(E_1)/(E_2)`

`(1.45)/(W_2)=(108)/(32.65)`

Or, `W_2=(1.45xx32.65)/(108)=0.438` g of Zn


Question 17: Using the standard electrode potentials given in Table 3.1 predict if the reaction between the following is feasible:

Fe3+ (aq) and I-(aq)

Answer: We know that the reaction is feasible if EMF of the cell reaction is positive

`Fe^(3+)(aq)+I^(-)(aq)→Fe^(2+)(aq)+1/2I_2(g)`

Standard electrode potential for `Fe^(3+)+e^(-)→Fe^(2+) = 0.77`

Standard electrode potential for `I_2+2e^(-)→2I^(-)=0.54`

So, cell potential `=0.77-0.54=0.23` V

Hence, this reaction is possible

Ag+(aq) and Cu(s)

Answer: `2Ag^(+)(aq)+Cu(s)→2Ag(s)+Cu^(2+)`

Standard electode potential for `Ag^(+)+e^(-1)→Ag=0.80`

Standard electrode potential for `Cu^(2+)+2e^(-)→Cu=0.34`

So, cell potential `=0.80-0.34=0.46` V

Hence, this reaction is possible

Fe3+ (aq) and Br-(aq)

Answer: `Fe^(3+)(aq)+Br^(-)(aq)→Fe^(2+)(aq)+1/2Br_2(g)`

Standard electrode potential for `Fe^(3+)+e^(-)→Fe^(2+) = 0.77`

Standard electrode potential for `Br_2+2e^(-)=2Br^(-1)=1.09`

So, cell potential `=0.77-1.09=-0.32`

Hence, this reaction is not feasible.

Ag(s) and Fe3+(aq)

Answer: `Ag(s)+Fe^(3+)(aq)→Ag^(+)(aq)+Fe^(2+)(aq)`

Cell potential `= 0.77-0.80=-0.03`

Hence, this reaction is not feasible.

Br2 (aq) and Fe2+ (aq)

Answer: `1/2Br_2(g)+Fe^(2+)(aq)→Br^(-)(aq)+Fe^(3+)`

Cell potential `=1.09-0.77=0.32`

Hence, this reaction is feasible


Question 18: Predict the products of electrolysis in each of the following

An aqueous solution of AgNO3 with silver electrodes.

Answer: `AgNO_3(s)+aq→Ag^(+)(aq)+NO_2^(-)(aq)`

`H_2O⇄H^(+)+OH^(-)`

At cathode: Ag+ ions have lower discharge potential than H+ ions. So, Ag+ ions will be deposited as Ag in preference to H+ ions.

At anode: Since Ag anode is attacked by nitrate ions, hence Ag of the anode will dissolve to form Ag+ ions in the solution.

An aqueous solution of AgNO3 with platinum electrodes.

Answer: At cathode: Ag+ will be deposited as Ag (as explained in previous question)

At anode: As anode is not being attacked by nitrate or OH- ions, so OH- ions will be discharged in preference over nitrate ions. After that, OH- ions decompose to give out O2.

A dilute solution of H2SO4 with platinum electrodes

Answer: `H_2SO_4(aq)→2H^(+)(aq)+SO_4^(2-)(aq)`

Hydrogen gas is liberated at cathode and oxygen gas is liberated at anode.

An aqueous solution of CuCl2 with platinum electrodes.

Answer: `Cu\Cl_2(s)+aq→Cu^(2+)(aq)+2Cl^(-)(aq)`

At cathode: Cu2+ ions are reduced and copper is deposited.

At anode: Chloride ions will be discharged in preference to OH- ions.



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