# Electrochemistry

## NCERT SolutionPart 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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