Solved Examples on Electrochemistry

Example 1.

Find the charge in coulomb on 1 g-ion of N^3-.

Solution:

Charge on one ion of N^3-

= 3 × 1.6 × 10^-19 coulomb 

Thus, charge on one g-ion of N^3-
= 3 × 1.6 10^-19 × 6.02 × 10²³
= 2.89 × 10⁵  coulomb

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Example 2.

How much charge is required to reduce (a) 1 mole of Al³⁺ to Al and (b)1 mole  of to Mn²⁺ ? 

Solution:

(a) The reduction reaction is 
Al³⁺ + 3e-  →  Al 

Thus, 3 mole of electrons are needed to reduce 1 mole of Al3+
Q = 3 × F   = 3 × 96500 = 289500 coulomb 

(b) The reduction is 

Mn^4-+ 8H+ 5e-  →  Mn²+ + 4H₂O 
1 mole 5 mole 
Q = 5 × F   = 5 × 96500 = 48500 coulomb 

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Example 3.

How much electric charge is required to oxidise (a) 1 mole of H₂O to O₂ and  (b)1 mole of FeO to Fe₂O₃? 

Solution:

(a) The oxidation reaction is 

H₂O →  1/2 O₂ + 2H⁺  + 2e-
Q = 2 × F   = 2 × 96500 =193000 coulomb 

(b) The oxidation reaction is 

FeO + 1/2 H₂O → 1/2 Fe₂O₃ + H⁺  + e^-
Q = F = 96500 coulomb 

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Example 4.

Exactly 0.4 faraday electric charge is passed through three electrolytic cells in series, first containing AgNO3, second CuSO4 and third FeCl3 solution. How many gram of  rach metal will be deposited assuming only cathodic reaction in each cell? 

Solution:

The cathodic reactions in the cells are respectively. 
 Ag⁺ + e-  →  Ag 

Cu²⁺ + 2e-  → >Cu 

and Fe³⁺ + 3e- →  Fe 

Hence, Ag deposited = 108 × 0.4 = 43.2 g 

Cu deposited = 63.5/2×0.4=12.7 g 

and Fe deposited = 56/3 ×0.4=7.47 g 

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Example 5.

An electric current of 100 ampere is passed through a molten liquid of sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the electrode at NTP. 

Solution:

The reaction taking place at anode is 
2Cl^-  →  Cl₂ + 2e-
Q = I × t = 100 × 5 × 600 coulomb 
The amount of chlorine liberated by passing 100 × 5 × 60 × 60 coulomb of electric charge.   =1/(2×96500)×100×5×60×60=9.3264 mole 
Volume of Cl₂ liberated at NTP = 9.3264 × 22.4 = 208.91 L

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Example 6. 

A 100 watt, 100 volt incandescent lamp is connected in series with an  electrolytic cell containing cadmium sulphate solution. What mass of cadmium will be deposited by the current flowing for 10 hours? 

Solution:

We know that 

Watt = ampere × volt 

100 = ampere × 110 

Ampere = 100/110 
Quantity of charge = ampere × second = 100/110×10×60×60 coulomb 

The cathodic reaction is 
Cd²⁺ + 2e- →  Cd 
Mass of cadmium deposited by passing 100/110×10×60×60 
Coulomb charge = 112.4/(2×96500)×100/110×10×60×60=19.0598 g 

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Example 7.

In an electrolysis experiment, a current was passed for 5 hours through two cells  connected in series. The first cell contains a solution gold salt and the second cell contains  copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of  gold is +3, find the amount of copper deposited on the cathode in the second cell. Also calculate 
the magnitude of the current in ampere. 

Solution:

We know that 
(Mass of Au deposited)/(Mass f Cu deposited)=(Eq.mass of Au)/(Eq.Mass of Cu) 
Eq. mass of Au = 197/3;

Eq. mass of Cu 63.5/2 
Mass of copper deposited   = 9.85 × 63.5/2 x 3/197 g = 4.7625 g 
Let Z be the electrochemical equivalent of Cu. 
E = Z × 96500 
or Z =E/96500=63.5/(2×96500) 
Applying W = Z × I × t 
T = 5 hour = 5 × 3600 second 
4.7625 = 63.5/(2×96500) × I × 5 × 3600 
or I = (4.7625 × 2 × 96500)/(63.5 × 5 × 3600)=0.0804 ampere 

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Example 8.

How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a metal surface of 80 cm² with 0.005 cm thick layer? Density of silver is 10.5 
g/cm³.

Solution:

Mass of silver to be deposited  = Volume × density  = Area ×thickness × density 
Given: Area = 80 cm²
thickness = 0.0005 cm and density = 10.5 g/cm³
 
Mass of silver to be deposited = 80 × 0.0005 × 10.5 = 0.42 g 
Applying to silver E = Z × 96500 
Z = 108/96500 g 
Let the current be passed for r seconds. 
We know that 
W = Z × I × t 
So, 0.42 = 108/96500 x 3 x t 
or t = (0.42 × 96500)/(108×3)=125.09 second 

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Example 9.

What current strength in ampere will be required to liberate 10 g of chlorine  from sodium chloride solution in one hour? 

Solution:

Applying E = Z × 96500 (E for chlorine = 35.5) 
35.5 = Z × 96500 
or Z = 35.5/96500 g 
Now, applying the formula

W = Z × I × t

Where W = 10 g, Z= 35.5/96500 t = 60×60 =3600 second 

I = 10x96500/35.5x96500 = 7.55 ampere 

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Example 10.

0.2964 g of copper was deposited on passage of a current of 0.5 ampere for 30 minutes through a solution of copper sulphate. Calculate the atomic mass of copper. (1 faraday = 96500 coulomb) 

Solution:

Quantity of charge passed 

0.5 × 30 × 60 = 900 coulomb 
900 coulomb deposit copper = 0.2964 g 
96500 coulomb deposit copper = 0.2964/900×96500=31.78 g 
Thus, 31.78 is the equivalent mass of copper. 
At. mass = Eq. mass × Valency = 31.78 × 2 = 63.56

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Example 11.

19 g of molten SnCI2 is electrolysed for some time using inert electrodes until 0.119 g of Sn is deposited at the cathode. No substance is lost during electrolysis. Find the ratio of the masses of SnCI2 : SnCI4 after electrolysis. 

Solution:

The chemical reaction occurring during electrolysis is 
 2SnCl₂ →  SnCl4 + Sn 
2×190 g 261 g 119 g 

119 g of Sn is deposited by the decomposition of 380 g of SnCl₂

So, 0.119 g of SnCl₂ of Sn is deposited by the decomposition of 
380/119×0.119=0.380 g of SnCl₂
Remaining amount of SnCl₂ = (19-0.380) = 18.62 g 

380 g of SnCl₂ produce = 261 g of SnCl₄
So 0.380 g of SnCl2 produce = 261/380×0.380=0.261 g of SnCl 
Thus, the ratio SnCl2 : SnCl4 =18.2/0.261 , i.e., 71.34 : 1 

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Example 12.

A current of 2.68 ampere is passed for one hour through an aqueous solution of copper sulphate using copper electrodes. Calculate the change in mass of cathode and that of the anode. (At. mass of copper = 63.5). 

Solution:

The electrode reactions are: 
Cu2+ + 2e- →  Cu (Cathode) 
1 mole 2 × 96500 C 
Cu →  Cu2+ + 2e-
(Anode) 
Thus, cathode increases in mass as copper is deposited on it and the anode decreases in mass as copper from it dissolves. 
Charge passed through cell = 2.68 × 60 × 60 coulomb 
Copper deposited or dissolved = 63,5/(2×96500)×2.68×60×60 =3.174 g 
Increase in mass of cathode = Decrease in mass of anode = 3.174 g 

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Example 13.

An ammeter and a copper voltameter are connected in series through which a  constant current flows. The ammeter shows 0.52 ampere. If 0.635 g of copper is deposited in one hour, what is the percentage error of the ammeter? (At. mass of copper = 63.5) 

Solution :

The electrode reaction is: 
Cu²⁺ + 2e →  Cu 

1 mole 2 × 96500 C 
63.5 g of copper deposited by passing charge = 2 × 96500 Coulomb 
0.635 g of copper deposited by passing charge =(2×96500)/63.5×0.653 coulomb  = 2 × 965 coulomb  = 1930 coulomb 

We know that 
Q = l × t 
1930 = I × 60 × 60 
I= 1930/3600=0.536 ampere 
Percentage error = ((0.536-0.52))/0.536×100=2.985 

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Example 14.

A current of 3.7 ampere is passed for 6 hours between platinum electrodes in  0.5 litre of a 2 M solution of Ni(NO₃)₂. What will be the molarity of the solution at the end of  electrolysis?

What will be the molarity of solution if nickel electrodes are used? (1 F = 96500  coulomb; Ni = 58.7)

Solution: 

The electrode reaction is 

Ni²⁺ + 2e- →  Ni 

1 mole 2 × 96500 C 

Quantity of electric charge passed  = 3.7 × 6 × 60 × 60 coulomb = 79920 coulomb 
Number of moles of Ni(NO3)2 decomposed or nickel deposited = (1.0 - 0.4140) = 0.586 
Since 0.586 moles are present in 0.5 litre, 
Molarity of the solution = 2 × 0.586 = 1.72 M 
When nickel electrodes are used, anodic nickel will dissolve and get deposited at the cathode. 
The molarity of the solution will, thus, remain unaffected

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Example 15:

An acidic solution of Cu²⁺ salt containing 0.4 g of Cu²⁺ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with volume of  solution kept at 100 mL and the current at 1.2 amp. Calculate the gases evolved at NTP during the entire electrolysis. 

Solution:

0.4 g of Cu2+ = 0.4/31.75 = 0.0126 g equivalent 
At the same time, the oxygen deposited at anode = 0.0126 g equivalent  = 8/32 × 0.0126 = 0.00315 g mol 
After the complete deposited of copper, the electrolysis will discharge hydrogen at cathode 
and oxygen at anode. The amount of charge passed = 1.2 × 7 × 60 = 504 coulomb 
So, Oxygen liberated = 1/96500 × 504 = 0.00523 g equivalent 
= 8/32 × 0.00523 = 0.001307 g mole 

Hydrogen liberated = 0.00523 g equivalent 
= 1/2 × 0.00523 = 0.00261 g mole 
Total gases evolved = (0.00315 + 0.001307 + 0.00261) g mole 
= 0.007067 g mole 
Volume of gases evolved at NTP = 22400 × 0.007067 mL = 158.3 mL

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Example 16:     

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 E^o 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)

E^o_Ag⁺/Ag =0.80  volt          (Reduction potential)

Cd → Cd²⁺  + 2e^-,

Oxidation

(Anode)

E^o_Cd⁺/Cd = -0.40 volt               (Reduction potential)

or     E^o_Cd⁺/Cd² = +0.40 volt

E^o = E^o_Cd⁺/Cd² + E^o_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.

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Example 17:

Calculate the electricity that would be required to reduce 12.3 g of nitrobenzene  to aniline, if the current efficiency for the process is 50 per cent. If the potential drop across the cell is 3.0 volt, how much energy will be consumed?

Solution:

The reduction reaction is 
C₆H₅NO₂ + 3H₂ C₆H₅NH₂ + 2H₂O  
Hydrogen required for reduction of 12.3/123 or 0.1 mole of nitrobenzene = 0.1 × 3 = 0.3 mole 
Amount of charge required for liberation of 0.3 mole of hydrogen = 2 × 96500 × 0.3 = 57900  coulomb 

Actual amount of charge required as efficiency is 50%  = 2 × 57900 = 115800 coulomb 

Energy consumed = 115800 × 3.0 = 347400 J  = 347.4 kJ

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Example 18:

After electrolysis of a sodium chloride solution with inert electrodes for a certain period of time, 600 mL of the solution was left which was found to be 1 N in NaOH. During the same period 31.75 g of copper was deposited in the copper voltameter in series with the electrolytic cell. Calculate the percentage theoretical yield of NaOH obtained. 

Solution:

Equivalent mass of NaOH = 40/1000 × 600 = 24 g 
Amount of NaOH formed = 40/1000 × 600 = 24 g 
31.75 g of Cu = 1 g equivalent of Cu. 
During the same period, 1 g equivalent of NaOH should have been formed. 
1 g equivalent of NaOH = 40 g 
% yield = 24/40 × 100 = 60

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Example 19: 

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,

E_cell = E_cell^o - 0.0591/n log(Mg²⁺)/[Ag⁺]³

0.42 =  E_cell^o - 0.0591/n log (0.0018)/(0.01)³ =  E_cell^o - 0.064

E_cell^o =(0.042+0.064)= 0.484 volt

E^o_cell = E^o_cathode - E^o_anode

 or E^o_anode  = E^o_cathode  - E^o_cell = (0.80-0.484) = 0.32 volt

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Example 20:

Cadmium amalgam is prepared by electrolysis of a solution of CdCl₂ using a mercury c0thode. Find how long a current of 5 ampere should be passed in order to prepare 12%  Cd-Hg amalgam on a cathode of 2 g mercury. At mass of Cd = 112.40. 

Solution:

2 g Hg require Cd to prepare 12% amalgam = 12/88 × 2 = 0.273 g 
Cd2+ + 2e- →  Cd 
1 mole 2 × 96500C 
112.40g 
Charge required to deposit 0.273 g of Cd  = 2*96500/112.40 × 0.273 coulomb 
Charge = ampere × second 
Second = 2*96500*0.273/112.40*5 = 93.75

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