Numerical Problems

Numerical problems on electrode potential and EMF of a cell

Ques: 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

(c) Zn + H2SO4 = ZnSO4 + H2

(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


(c) In this case Zn is oxidized to Zn2+ and H+ is reduced to H2. The cell can be represented as

Zn|ZnSO4||H2SO4|Cu

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


(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


Ques: 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

+/Cd

2 = +0.40 volt

Eo = EoCd

+/Cd

2 + Eo

Ag+

/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.

Ques: 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 = Eo

Zn/Zn2+ + Eo

Cu2+

/Cu

= Oxi. Potential of zinc + Red Potential of copper

EoZn/Zn

2+ = -0.763 (Reduction potential)

EoZn

2+/Zn= +0.763 (Oxidation potential)

and EoCu

2+/Cu = 0.350 (Reduction potential)

So Ecello= 0.763+0.350 = 1.113 volt

Oxidation potential is EoM/M

a+ while reduction potential is represented as EoM

a+/M. The value of Eo

Zn/Zn2+ (oxidation potential of Zn) is +0.76 volt and the value of Eo

Cu2+

/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.

Ques: 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

(c) Pt |H2, H+||Cu2+|Cu

(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

(c) Oxidation half reaction, H2 --> 2H+ + 2e-

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

Ques: 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/Fe

2+ = 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.

Ques: 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 ?

(c) Whether the reaction between FeCl3 and SnCl2 occurs or not ?

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.

(c) Chemical reaction

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.

Ques: Calculate the electrode potential at a copper electrode dipped in a 0.1 M solution of copper sulphate at 25o C. The standard electrode potential of Cu2+/Cu system is 0.34 volt at 298 K.

Solution: We know that Ered = Eored + 0.0591/n log10[ion]

Putting the values of Eored =0.34 V, n = 2 and [Cu2+]= 0.1 M

Eored = 0.34+0.0591/2 log10[0.1]

= 0.34 + 0.02955 × (-1)

= 0.34 - 0.02955 = 0.31045 volt

Ques: What is the single electrode potential of a half-cell foe zinc electrode dipping in 0.01 M ZnSO4 solution at 25o C? The standard electrode potential of Zn/Zn2+ system is 0.763 volt at 25o C.

Solution: We know that Eox = Eored - 0.0591/n log10[ion]

Putting the value of Eoox=0.763 V,n=2 and

[Zn2+]=0.01 M

Eoox = 0.763-0.0591/2 log_10 [0.01]

= 0.763 - 0.02955 × (-2)

= (0.763 + 0.0591) volt = 0.8221 volt

Ques: The standard oxidation potential of zinc is 0.76 volt and of silver is -0.80 volt. Calculate the emf of the cell:

Zn|Zn(NO3)2||AgNO3|Ag

0.25 M 0.1 M

at 250C.

Solution: The cell reaction is

Zn + 2 Ag+ --> 2Ag + Zn2+

Eoox of Zn = 0.76 volt

Eoox of Ag = 0.80 volt

Eocell = Eoox of Zn + of Ag = 0.76 + 0.80

= 1.56 volt

= 1.56 - 0.0591/2×1.3979

= (1.56-0.0413) volt

= 1.5187 volt

Alternative method:

First of all, the single electrode potentials of both the electrodes are determined on the basis of given concentrations.

Eox(Zinc) = Eoox -0.0591/2 log 0.25

= 0.76 + 0.0177 = 0.7777 volt

Ered(Silver) = Eored -0.0591/2 log 0.1

= 0.80 + 0.0591 = 0.7409 volt

Ecell = Eox(Zinc) + Ered(Silver)

Ques: The emf (E°) of the following cells are:

Ag|Ag|(1 M)||Cu2+(1 M)|Cu; E° = -0.46 volt

Zn|Zn2(1 M)||Cu2|(1 M)|Cu; E° = +1.10 volt

Calculate the emf of the cell:

Zn|Zn2+(1 M)||Ag+(1 M)|Ag

Solution : Zn|Zn2+(1 M)||Ag+(1 M)|Ag

Ecell = Eox(Zn/Zn2+) + Ered (Ag+/Ag)

With the help of the following two cells, the above equation can be obtained.

Ag|Ag+(1 M)||Cu2+(1 M)|Cu, E° = -0.46 volt

or Cu|Cu2+(1 M)||Ag+(1 M)|Ag, E° will be +0.46 volt

or +0.46 = Eox(cwcu2+

) + Ered (Ag+/Ag) .... (i)

Zn|Zn2+(1 M)||Cu2+1|Cu, E° = +1.10 volt

+ 1.10 = Eox(Zn/Zn+

) + Ered(Cu2+/Cu) ....... (ii)

Adding Eqs. (i) and (ii),

+ 1.56 = Eox(Cu/Cu2+

) + Ered(Ag+

/Ag) + Eox(Zn/Zn2+

) + Ered(Cu2+/Cu)

Since Eox(Cu/Cu2+) - Ered(Cu2+/Cu)

So +1.56 = Em(Zn/Zn+

) + Ered(Ag+

/Ag)

Thus, the emf of the following cell is

Zn|Zn2+(1 M)||Ag+(1 M)|Ag is +1.56 volt.

Ques: Calculate the e.m.f of the cell.

Mg(s)|Mg2+(0.2M)||Ag+(1×10-3)|Ag

EoAg+/Ag = +0.8 volt, Eo

Mg2+/Mg = -2.37 volt

What will be the effect on e.m.f. if concentration of Mg2+ ion is decreased to 0.1 M?

Solution: Eocell = Eo

Cathode - Eoanode

= 0.80-(-2.37) = 3.17 volt

Cell reaction

Mg + 2Ag+ --> 2Ag + Mg2+

Ecell = Ecello - 0.0591/n log(Mg2+)/[Ag+]2

= 3.17 -0.0591/2 log 0.2/[1× 10-3 ]2

= 3.17 - 0.1566 = 3.0134 volt

when Mg2+ = 0.1 M

Ecell = Eocell - 0.0591/n log(0.1)/[1 x 10-3]2

= (3.17 - 0.1477) volt

= 3.0223 volt

Ques: To find the standard potential of M3+/M electrode, the following cell is constituted:

Pt|M|M3+(0.0018 mol-1L)||Ag+(0.01 mol-1L)|Ag

The emf of this cell is found to be 0.42 volt. Calculate the standard potential of the half reaction M3+ + 3e- M3+. = 0.80 volt.

Solution: The cell reaction is

M + 3Ag+ ---> 3Ag + M3+

Applying Nernst equation,

Ecell = Ecello - 0.0591/n log(Mg2+)/[Ag+]3

0.42 = Ecello - 0.0591/n log (0.0018)/(0.01)3 = Ecell

o - 0.064

Ecello =(0.042+0.064)= 0.484 volt

Eocell = Eo

cathode - Eoanode

or Eoanode = Eo

cathode - Eocell

= (0.80-0.484) = 0.32 volt

Numerical Problems

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