What quantity of charge is required to cause reduction of 0.20 mole of Cr3 to Cr?

Solved Examples on Electrochemistry

Example 1.

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

Solution:

Charge on ane ion of Northward^3-

= 3 × 1.6 × ten^-one9 coulomb

Thus, accuse on one yard-ion of North^iii-
= 3 × ane.half-dozen 10^-one9 × 6.02 × 10²³
= 2.89 × 10⁵  coulomb

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Example 2.

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

Solution:

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

Thus, iii 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 v mole
Q = 5 × F   = v × 96500 = 48500 coulomb

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Instance iii.

How much electric accuse is required to oxidise (a) 1 mole of H₂O to O₂ and  (b)1 mole of FeO to Atomic number 26₂O₃?

Solution:

(a) The oxidation reaction is

H_twoO →  one/2 O_ii + 2H⁺  + 2e-
Q = 2 × F   = two × 96500 =193000 coulomb

(b) The oxidation reaction is

FeO + i/ii H₂O → ane/2 Fe₂O_three + H⁺  + e^-
Q = F = 96500 coulomb

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Example iv.

Exactly 0.four faraday electrical accuse is passed through three electrolytic cells in series, commencement containing AgNO3, second CuSO4 and third FeCl3 solution. How many gram of  rach metallic will be deposited assuming only cathodic reaction in each jail cell?

Solution:

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

Cuⁱⁱ⁺ + 2e-  → >Cu

and Fe³⁺ + 3e- →  Fe

Hence, Ag deposited = 108 × 0.four = 43.ii g

Cu deposited = 63.5/2×0.4=12.7 thousand

and Iron deposited = 56/3 ×0.4=seven.47 g

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Case 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 × five × 600 coulomb
The corporeality of chlorine liberated past passing 100 × v × 60 × threescore coulomb of electric charge.   =1/(two×96500)×100×5×60×60=nine.3264 mole
Volume of Cl₂ liberated at NTP = ix.3264 × 22.iv = 208.91 L

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Instance 6.

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

Solution:

Nosotros know that

Watt = ampere × volt

100 = ampere × 110

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

The cathodic reaction is
Cdⁱⁱ⁺ + 2e- →  Cd
Mass of cadmium deposited by passing 100/110×10×60×60
Coulomb charge = 112.iv/(2×96500)×100/110×ten×60×lx=19.0598 g

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Case seven.

In an electrolysis experiment, a current was passed for 5 hours through two cells  connected in series. The first jail cell contains a solution gold salt and the 2nd cell contains  copper sulphate solution. 9.85 g of golden was deposited in the first cell. If the oxidation number of  gold is +3, discover the amount of copper deposited on the cathode in the second cell. As well summate
the magnitude of the electric 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/three;

Eq. mass of Cu 63.5/2
Mass of copper deposited   = 9.85 × 63.5/2 x three/197 g = 4.7625 one thousand
Let Z be the electrochemical equivalent of Cu.
Due east = Z × 96500
or Z =E/96500=63.five/(2×96500)
Applying West = Z × I × t
T = 5 hour = v × 3600 second
4.7625 = 63.v/(two×96500) × I × v × 3600
or I = (4.7625 × 2 × 96500)/(63.5 × 5 × 3600)=0.0804 ampere

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Example 8.

How long has a electric current of iii ampere to exist applied through a solution of argent nitrate to glaze a metal surface of 80 cm^two with 0.005 cm thick layer? Density of silver is 10.5
g/cm ³ .

Solution:

Mass of silver to exist deposited  = Volume × density  = Expanse ×thickness × density
Given: Area = 80 cm²
thickness = 0.0005 cm and density = ten.v m/cm³

Mass of silver to be deposited = eighty × 0.0005 × ten.five = 0.42 thousand
Applying to silver E = Z × 96500
Z = 108/96500 chiliad
Let the current be passed for r seconds.
We know that
W = Z × I × t
So, 0.42 = 108/96500 ten 3 ten t
or t = (0.42 × 96500)/(108×3)=125.09 2d

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Example nine.

What current forcefulness in ampere will be required to liberate 10 g of chlorine  from sodium chloride solution in 1 hour?

Solution:

Applying E = Z × 96500 (E for chlorine = 35.five)
35.5 = Z × 96500
or Z = 35.v/96500 g
Now, applying the formula

Westward = Z × I × t

Where W = 10 one thousand, Z= 35.5/96500 t = lx×lx =3600 second

I = 10x96500/35.5x96500 = 7.55 ampere

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Case 10.

0.2964 g of copper was deposited on passage of a current of 0.v 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 × sixty = 900 coulomb
900 coulomb deposit copper = 0.2964 thou
96500 coulomb deposit copper = 0.2964/900×96500=31.78 thou
Thus, 31.78 is the equivalent mass of copper.
At. mass = Eq. mass × Valency = 31.78 × 2 = 63.56

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Instance xi.

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 afterwards electrolysis.

Solution:

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

119 yard of Sn is deposited by the decomposition of 380 1000 of SnCl_two

So, 0.119 g of SnCl_two of Sn is deposited by the decomposition of
380/119×0.119=0.380 k of SnCl_two
Remaining amount of SnCl_ii = (xix-0.380) = eighteen.62 g

380 one thousand of SnCl_ii produce = 261 thou of SnCl_iv
So 0.380 g of SnCl2 produce = 261/380×0.380=0.261 1000 of SnCl
Thus, the ratio SnCl2 : SnCl4 =xviii.2/0.261 , i.east., 71.34 : ane

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Case 12.

A current of 2.68 ampere is passed for one hour through an aqueous solution of copper sulphate using copper electrodes. Summate 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 ii × 96500 C
Cu →  Cu2+ + 2e-
(Anode)
Thus, cathode increases in mass equally copper is deposited on it and the anode decreases in mass as copper from it dissolves.
Charge passed through jail cell = 2.68 × 60 × 60 coulomb
Copper deposited or dissolved = 63,5/(two×96500)×two.68×60×sixty =3.174 g
Increase in mass of cathode = Decrease in mass of anode = three.174 g

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Example xiii.

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 ii × 96500 C
63.5 g of copper deposited past passing accuse = ii × 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 = fifty × t
1930 = I × lx × 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 electric current of 3.7 ampere is passed for 6 hours between platinum electrodes in  0.5 litre of a two Grand solution of Ni(NO_three)₂. What will exist the molarity of the solution at the end of  electrolysis?

What volition exist the molarity of solution if nickel electrodes are used? (one F = 96500  coulomb; Ni = 58.7)

Solution:

The electrode reaction is

Ni²⁺ + 2e- →  Ni

1 mole two × 96500 C

Quantity of electrical charge passed  = 3.7 × half dozen × sixty × lx 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 volition dissolve and get deposited at the cathode.
The molarity of the solution will, thus, remain unaffected

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Example xv:

An acidic solution of Cu²⁺ salt containing 0.4 g of Cu²⁺ is electrolysed until all the copper is deposited. The electrolysis is connected for seven more than minutes with book of  solution kept at 100 mL and the current at ane.2 amp. Calculate the gases evolved at NTP during the entire electrolysis.

Solution:

0.four 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 yard mol
Later on the consummate deposited of copper, the electrolysis volition discharge hydrogen at cathode
and oxygen at anode. The corporeality of charge passed = 1.ii × 7 × 60 = 504 coulomb
Then, Oxygen liberated = 1/96500 × 504 = 0.00523 g equivalent
= 8/32 × 0.00523 = 0.001307 1000 mole

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

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Example sixteen:

Consider the reaction,

2Ag⁺ + Cd →   2Ag + Cd²⁺

The standard electrode potentials for Ag⁺ --> Ag and Cd²⁺ --> Cd couples are 0.eighty 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 identify which electrode is negative electrode?

Solution:

(i) The half reactions are:

2Ag⁺  + 2e^- →   2Ag.

 Reduction

Cathode)

Eastward^o _Ag⁺/Ag =0.80  volt          (Reduction potential)

Cd → Cd²⁺  + 2e^-,

Oxidation

(Anode)

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

or     Eastward^o _Cd⁺/Cd²  = +0.40 volt

E^o = E^o _Cd⁺/Cd²+ E^o _Ag⁺/Ag= 0.xl+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 thousand of nitrobenzene  to aniline, if the current efficiency for the procedure is 50 per cent. If the potential drop across the prison cell is 3.0 volt, how much free energy will be consumed?

Solution:

The reduction reaction is
C_{half dozen}H_vNO₂ + 3H₂ C₆H₅NH₂ + 2H_iiO
Hydrogen required for reduction of 12.iii/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 = two × 96500 × 0.3 = 57900  coulomb

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

Energy consumed = 115800 × iii.0 = 347400 J  = 347.4 kJ

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Example 18:

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

Solution:

Equivalent mass of NaOH = 40/thou × 600 = 24 one thousand
Amount of NaOH formed = 40/thousand × 600 = 24 one thousand
31.75 g of Cu = 1 g equivalent of Cu.
During the same flow, 1 g equivalent of NaOH should have been formed.
i g equivalent of NaOH = 40 g
% yield = 24/twoscore × 100 = 60

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Instance nineteen:

To find the standard potential of One thousand³⁺/M electrode, the following prison cell is constituted:

Pt|One thousand|M³⁺(0.0018 mol^-1L)||Ag⁺(0.01 mol^-aneFifty)|Ag

The emf of this cell is found to be 0.42 volt. Calculate the standard potential of the half reaction G³⁺ + 3e^-  M³⁺. = 0.80 volt.

Solution:

The cell reaction is

M + 3Ag⁺ →  3Ag + M³⁺

Applying Nernst equation,

E_cell= E_{jail cell} ^o - 0.0591/n log(Mg²⁺)/[Ag⁺]³

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

Due east_cell ^o =(0.042+0.064)= 0.484 volt

E^o _{prison cell} = E^o _cathode- E^o _anode

 or E^o _anode= E^o _cathode- Due east^o _cell = (0.80-0.484) = 0.32 volt

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Example twenty:

Cadmium amalgam is prepared past electrolysis of a solution of CdCl_two using a mercury c0thode. Find how long a electric current of 5 ampere should exist passed in club to fix 12%  Cd-Hg amalgam on a cathode of 2 chiliad mercury. At mass of Cd = 112.forty.

Solution:

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

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