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Find the charge in coulomb on 1 g-ion of N3-.
Charge on one ion of N3-
= 3 × 1.6 × 10-19 coulomb
Thus, charge on one g-ion of N3- = 3 × 1.6 10-19 × 6.02 × 1023 = 2.89 × 105 coulomb
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How much charge is required to reduce (a) 1 mole of Al3+ to Al and (b)1 mole of to Mn2+ ?
(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
Mn4-+ 8H+ 5e- → Mn2+ + 4H2O 1 mole 5 mole Q = 5 × F = 5 × 96500 = 48500 coulomb
How much electric charge is required to oxidise (a) 1 mole of H2O to O2 and (b)1 mole of FeO to Fe2O3?
(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
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?
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
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.
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
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?
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
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.
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
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.
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
What current strength in ampere will be required to liberate 10 g of chlorine from sodium chloride solution in one hour?
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
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)
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
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.
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
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).
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
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
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)
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
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.
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
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?
(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.
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?
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
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.
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
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.
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 = Ecello - 0.064
Ecello =(0.042+0.064)= 0.484 volt
Eocell = Eocathode - Eoanode
or Eoanode = Eocathode - Eocell = (0.80-0.484) = 0.32 volt
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.
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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