chemistry

EAMCET PROBLEMS WITH SOLUTIONS FROM ELECTROCHEMISTRY

PART-2

Example 11: The specific conductivity of 0.02 M KCl solution at 250C is 2.768 × 10-3 ohm-1 cm-1. The resistance of this solution at 250C when measured with a particular cell was 250.2 ohms. The resistance of 0.01 M CuSO4 solution at 250C measured with the same cell was 8331 ohms. Calculate the molar conductivity of the copper sulphate solution.

Solution: Cell constant = (Sp.cond.of KCl)/(Conductane of KCl)

= (2.768× 10-3)/(I/250.2)

= 2.768 × 10-3 × 250.2

For 0.01 M CuSO3 solution

Sp. conductivity = Cell constant × conductance

= 2.768 × 10-3 × 250.2 × 1/8331

Molar conductance = Sp. cond. × 1000/c

= (2.768×10-3 × 25.2)/8331 × 1000/(1/100)

Example 12: The values of Eo of some of the reactions are given below:

I2 + 2e- --> 2I-; Eo = +0.54 volt

Cl2 + 2e- --> 2Cl-; Eo = +1.36 volt

Fe3+ + e- --> Fe2+; Eo = +0.76 volt

Ce4+ + e- --> Ce3+; Eo = +1.60 volt

Sn4+ + 2e- --> Sn2+; Eo = +0.15 volt

On the basis of the above data, answer the following questions:

(a) Whether Fe3+ oxidizes Ce3+ or not ?

(b) Whether I2 displaces chlorine form KCl ?

Solution: (a) Chemical reaction,

Fe3+ + Ce3+ --> Ce4+ + Fe2+

Two half reactions,

Fe3+ + e --> Fe2+ Reduction Eo = 0.76 volt

Ce3+ --> Ce4+ + e- Oxidation Eoox = -1.60 volt

---------------------------

Adding = -0.84 volt

Since, emf is negative the reaction does not occur, i.e., Fe3+ does not oxidise Ce3+.

(b) Chemical reaction

I2 + 2KCl = 2Kl + Cl2

Half reactions

I2 + 2e- --> 2I- Reduction Eo = 0.54 volt

2Cl- --> Cl2 + 2e- Oxidation Eoox = -1.36 volt

---------------------------

Adding = -0.82 volt

Since, emf is negative, the reaction does not occur, i.e., I2 does not displace Cl2from KCl.

SnCl2 + 2FeCl3 --> SnCl4 + 2FeCl2

Half reactions

Fe3+ + e Fe2+ Reduction Eo = 0.76 volt

Ce2+ Sn4+ + 2e- Oxidation Eo = -0.15 volt

-------------------------

Adding = +0.61 volt

Since, emf is positive, the reaction will occur.

Example 13: The equivalent conductivity of N/10 solution of acetic acid at 250C is 14.3 ohm-1 cm2 equiv-1. Calculate the degree of dissociation of CH3COOH if is 390.71.

Solution:

/\∞CH3 COOH = 390.7 ohm-1 cm-2 equiv-1

/\∞CH3 COOH = 143.3 ohm-1 cm-2 equiv-1

Degree of dissociation, α = /\v//\∞ =14.3/390.71

= 0.0366 i.e., 3.66% dissociated

Example 14: A decinormal solution of NaCI has specific conductivity equal to 0.0092. If ionic conductances of Na+ and Cl- ions at the same temperature are 43.0 and 65.0 ohm-1 respectively, calculate the degree of dissociation of NaCl solution.

Solution: Equivalent conductance of N/10 NaCl solution

= Sp. conductivity × dilution

= 0.0092 × 10,000

= 92 ohm-1

/\∞ = λNa+ + λCl-

= 43.0 + 65.0

= 108 ohm-1

Degree of dissociation,

Example 15: At 180C, the conductivities at infinite dilution of NH4Cl, NaOH, NaCL are 129.8, 217.4, 108.9 mho respectively. If the equivalent conductivity of N/100 solution of NH4OH is 9.93 mho, calculate the degree of dissociation of NH4OH at this solution.

Solution: /\∞NH4 Cl = λNa4+ + λCl- =129.8 ..... (i)

/\∞NaOH = λNa+ + λOH- = 217.4 ..... (ii)

/\NaCl = λNa + λCl- = 108.9 ..... (iii)

Adding Eqs. (i) and (ii) and subtracting (iii),

λNa4+ + λCl- + λNa+ + λOH- - λNa- λCl-

λNa4+ + λOH- = 238.3 mho

Degree of dissociation, α=/\v//\∞ = 9.93/238.3 = 0.04167

or 4.17% dissociated

Example 16: Construct the cells in which the following reactions are taking place. Which of the electrodes shall act as anode (negative electrode) and which one as cathode (positive electrode)?

(a) Zn + CuSO4 = ZnSO4 + Cu

(b) Cu + 2AgNO3 = Cu(NO3)2 + 2 Ag

(d) Fe + SnCl2 = FeCl2 + Sn

Solution: It should always be kept in mind that the metal which goes into solution in the form of its ions undergoes oxidation and thus acts as negative electrode (anode) and the element which comes into the free state undergoes reduction and acts as positive electrode (cathode):

(a) In this case Zn is oxidized to Zn2+ and thus acts as anode (negative electrode) while Cl2+ is reduced to copper and thus acts as cathode (positive electrode). The cell can be represented

as Zn|ZnSO4||CuSO4|Cu

or Zn|Zn2+||Cu2+|Cu

Anode (-) Cathode (+)

(b) In this case Cu is oxidized to Cu2+ and Ag+ is reduced to Ag. The cell can be represented as

Cu|Cu(NO3)2||AgNO3|Ag

or Cu|Cu2+||Ag+|Ag

Anode (-) Cathode (+)

Zn|ZnSO4||H2SO4|Cu

or Zn|Zn2+||2H+|H2(Pt)

Anode (-) Cathode (+)

(d) Here Fe is oxidized to Fe2+ and Sn2+ is reduced to Sn. The cell can be represented as

Fe|FeCl2||SnCl2|Sn

or Fe|Fe2+||Sn2+|Sn

Anode (-) Cathode (+)

Example 17: Consider the reaction,

2Ag+ + Cd --> 2Ag + Cd2+

The standard electrode potentials for Ag+ --> Ag and Cd2+ --> Cd couples are 0.80 volt and -0.40 volt, respectively.

(i) What is the standard potential Eo for this reaction?

(ii) For the electrochemical cell in which this reaction takes place which electrode is negative electrode?

Solution: (i) The half reactions are:

2Ag+ + 2e- --> 2Ag.

Reduction

(Cathode)

EoAg+/Ag =0.80 volt (Reduction potential)

Cd --> Cd2+ + 2e-,

Oxidation

(Anode)

EoCd+/Cd = -0.40 volt (Reduction potential)

or EoCd+/Cd2 = +0.40 volt

Eo = EoCd+/Cd2 + EoAg+/Ag = 0.40+0.80 = 1.20 volt

(ii) The negative electrode is always the electrode whose reduction potential has smaller value or the electrode where oxidation occurs. Thus, Cd electrode is the negative electrode.

Example 18: Consider the cell,

Zn|Zn2+(aq)(1.0M)||Cu2+(aq)(1.0M)|Cu

The standard electrode potentials are

Cu2+ + 2e- --> Cu(aq) Eo = 0.350 volt

Zn2+ + 2e- --> Zn(aq) Eo = -0.763 volt

(i) Write down the cell reaction.

(ii) Calculate the emf of the cell

Solution: (i) Reduction potential of Zn is less than copper, hence Zn acts as anode and copper as cathode.

At anode Zn --> Zn2+ + 2e- (Oxidation)

At cathode Cu2+ + 2e- --> Cu (Reduction)

--------------------------------------------------------

Cell reaction Zn + Cu2+ --> Zn2+ + Cu

(ii) EoCell = EoZn/Zn2+ + EoCu2+/Cu

= Oxi. Potential of zinc + Red. Potential of copper

EoZn/Zn2+ = -0.763 (Reduction potential)

EoZn2+/Zn= +0.763 (Oxidation potential)

and EoCu2+/Cu = 0.350 (Reduction potential)

So Ecello= 0.763+0.350 = 1.113 volt

Oxidation potential is EoM/Ma+ while reduction potential is represented as EoMa+/M. The value of EoZn/Zn2+ (oxidation potential of Zn) is +0.76 volt and the value of EoCu2+/Cu (reduction potential of copper) is +0.34 volt. The electrode having lower value of reduction potential acts as an anode while that having higher value of reduction potential acts as cathode.

Example 19: Write the electrode reactions and the net cell reactions for the following cells. Which electrode would be the positive terminal in each cell?

(a) Zn|Zn2+||Br-, Br2|Pt

(b) Cr|Cr3+||I-, I2|Pt

(d) Cd|Cd2+||Cl-, AgCl|Ag

Solution: (a) Oxidation half reaction, Zn --> Zn2++2e-

Reduction half reaction, Br2 + 2e- --> 2Br-

-------------------------------------------------

Net cell reaction Zn + Br2 --> Zn2+ + 2Br-

Positive terminal-Cathode Pt

(b) Oxidation half reaction, [Cr --> Cr3+ + 3e-]× 2

Reduction half reaction, [I2 + 2e- --> 2Ir-] × 3

----------------------------------------------------

Net cell reaction 2Cr + 3I2 --> 2Cr3+ + 6I-

Positive terminal-Cathode Pt

Reduction half reaction, Cu2+ + 2e---> Cu

-------------------------------------------------

Net cell reaction H2 + Cu2+ --> Cu + 2H+

Positive terminal-Cathode Cu

(d) Oxidation half reaction, Cd --> Cd2+ + 2e-

Reduction half reaction, [AgCl+e- --> Ag+Cl- ]×2

------------------------------------------------------

Net cell reaction Cd+2AgCl --> Cd2++2Ag+2Cl-

Positive terminal-Cathode Ag

Example 20: Will Fe be oxidiesed to Fe2+ by reaction with 1.0 M HCl? Eo for Fe/Fe2+ = +0.44 volt.

Solution: The reaction will occur if Fr is oxidized to Fe2+.

Fe + 2HCl --> FeCl2 + H2

Writing two half reaction,

Fe --> Fe2+ + 2e- Oxidation EoFe/Fe2+ = 0.44 volt

2H++ 2e- --> H2 Reduction EoH+/H = 0.0 volt

--------------------------------------

Adding, emf = 0.44 volt

Since emf is positive, the reaction shall occur.

hari

Monday, April 21, 2014

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