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Chemistry I

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Class 10 Class 12
Three electrolytic cells A, B and C containing solutions of ZnSO4(zinc sulphate), AgNO(silver nitrate) and CuSO4  (copper sulphate),  respectively are connected in series. A steady current of 1.5 ampere was passed through them until 1.45 g of silver is deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? (Atomic masses: Ag = 108, Zn = 65.4, Cu = 63.5, all in amu). 

Ag++e-  Ag  1 mol            1 mol ( = 108 g)              (Cathode reaction in cell B)
108 g of silver is deposited at cathode when 1 mol of electrons are passed or 108 g of silver deposit needs = 1 Faraday = 96,500 coulombs. Therefore, 1.45 g of silver needs
                     = 96,500×1.45108  = 1295.6 columbs
       But quantity of electricity passed
                    = current x time
                    = 1295.6 C  = 1.5 A x time (in sec.)
       or time for which current is passed
               = 1295.61.5  seconds = 863.7 s = 14.40 min.
          The cathode reaction in copper sulphate cell is
                         Cu2++2e-    Cu     2 mol                   1 mol (63.5 g)    (2×96500 c)
              2 x 96500 coulombs gives a deposit of 63.5 g of Cu.
          Therefore, 1295.6 coulombs will deposit of 
                       = 63.5 × 1295.62×96, 500 = 0.4263 g
Similarly,
                        Zn2++2e-  Zn     2 mol              1 mol (65.4 g)   (2×96500)
               (cathode reaction in zinc sulphate cell)
Mass of zinc deposited
                         =65.4 x 1295.62×96, 500 g = 0.44 g.
341 Views

Consult the table of standard electrode potentials and suggest three substance that can oxidize ferrous ions under suitable conditions.

Answer:


oxidation of ferrous ion means :

Fe2+--> Fe3+ +e-EFe3+Fe2+0 = 0.77V
Any substance which standard electrode potential is more than that of Fe+3 /F+2 can oxidise ferrous ions. 
(refer to the table given in book)

The EMF of the substance whose reduction potentials greater than 0.77V will oxidised ferrous ion.
for example Br2, Cl2,and F2 .

879 Views

Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M)  Ni2+ (0.160 M) + 2Ag(s)
Given that EoCell = 1.05 V

Answer:

Ni(s)+2Ag+(aq)Ni2+(aq)+2As(s)
Ecell = Eocell + RT2FlnAg+(aq)2Ni2+ (aq)
or      Ecell = Eocell+0.0592 logAg+(aq)2Ni2+ (aq)
 
The equation is also written as

or    Ecell = Eocell-0.0592logNi2+(aq)Ag+ (aq)2         = 1.05 V - 0.0295 log 0.1600.002  

= 1.05 V – 0.0295 x log 80

= 1.05 V – 0.0295 x 1.9031

= 1.05 V – 0.056 = 0.99 V.

1585 Views

How would you determine the standard electrode potential of the system Mg2+/Mg?

Answer:

Use standard hydrogen electrode as anode and Mg2+ | Mg as a cathode we can measure the standard electrodepotential of systemMg2+ | Mg. Standard hydrogen electrode, represented by Pt(s), H2(g) (1 atm) | H+ (aq) and dip the electrode of Magnesium wire in a 1M MgSO4 solution .The standard hydrogen electrode is always zero.
Use formula
Eocell = Eo right  – Eoleft
The standard hydrogen electrode is always zero.
So that the value of
Eoleft =0
Hence
Eocell = Eo Mg|Mg2+
Or
Eo Mg|Mg2+= Eocell
2203 Views

Can you store copper sulphate solutions in a Zinc pot?

Answer:

No. Because zinc is more reactive than copper and thus holes will be developed in zinc pot.

Cu2+(aq) + Zn(s) → Zn2+ (aq) + Cu(s)
1384 Views

Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Answer:

For hydrogen electrode, ,
given that      pH = 10
use formula [H+] = 10–pH
so that  [H+] = 10−10 M
Electrode reaction will
H+  + e –  →1/2 H2
Use the formula


EH+/H2  = EoH+/H2 - RTnFlnH2[H+]2              = 0-8.314×2982×96500ln IH+2              = 0.05915 log [H+]              =-0.05915 pH             =-0.05915 × 10 =- 0.59 V
1279 Views