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How much charge is required for the following reductions:
1 mol of Cu2+ to Cu?


Cu2+ +  2e-   Cu1 mol      2 mol        1 mol           (reduction of Cu2+ cathode)

Formula required charge n × F
n = difference of charge on ions   
F is constant and equal to 96487 Coulombs
Here n = 2
 
plug the value in formula we get
Required charge= 2 × 96487 Coulombs
                        = 192974 Coulombs
                        = 1.93 × 105 Coulombs
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How much electricity is required in coulomb for the oxidation of
1 mol of H2O to O2 ?

2H2O   4H++O2+4e  2 mol                       4 mol or 4F

1 mol of H2O  will give one atom of O 
Charge on O = – 2
Electricity required         =n F
= 2 × 96487 Coulombs
= 192974 Coulombs
= 1.93 × 105 Coulombs


 
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How much electricity in terms of Faraday is required to produce
40.0 g of Al from molten Al2O3?

Al2O3  2Al3++3O2-
or       Al3++3e-      Al      3 mol               1 mol ( = 27 g)                                     (reduction of Al3+)
Charge on Al in Al2O3
2Al  + 3O = 0
Oxygen has –2
2Al +3(–2)   = 0
Al         = 3
Change transfer  n = 3
Charge required for 1 mol of Al    = 3F
Number of moles of Al  = 40 /27  = 1.48
 
The electricity required to produce 1.48 mol of Al = 1.48  × 3 F
                                                                                = 4.44 F
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How much charge is required for the following reductions:
1 mol of MnO4 to Mn2+?


MnO4-(aq) + 8H++5e-             Mn2+(aq) + 6H2O. 1 mol                            5 moles             1 mol

Formula required charge n × F
n = difference of charge on ions   
F is constant and equal to 96487 Coulombs

Charge on Mn in MnO4
Charge on Oxygen is – 2
Mn + 4O     = – 1
Mn +4(–2)   = – 1
Mn              = +7
So our reaction is
MN 7+ à Mn 2+
n = 7– 2  = 5  
Here n= 5
Required charge will  = 5 × 96487 Coulombs
                                = 482435 Coulombs
                                = 4.82 × 105 Coulombs
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How much electricity in terms of Faraday is required to produce
20.0 g of Ca from molten CaCl2?


            CaCl2   Ca2+    +   2Cl-     Ca2++2e-              Ca              2 mol                        1 mol ( = 40)                                                     (reduction of Ca2+) 
        40g of calcium needs = 2 mol of electrons = 2 Faraday.                                                                ( 1 mole of e = 1 F)
The balance reaction will
The charge on Ca in CaCl2
Ca + 2Cl      = 0
Cl has –1 charge so that
Ca  + 2(–1) = 0
Ca               = 2
We have to get 2o g Ca from Ca 2+
Number of required moles  = mass / molar mass
Molar mass of Ca is 40 g/mol and required mass of Ca is 20 g
Hence number of moles           = 20/40      = 0.5 mol
Electricity required to produce 1 mol of calcium = 2 F
The electricity required to produce 0.5 mol of calcium
= 0.5 × 2 F
   = 1 F
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