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.