The following electrochemical cell has been set up
Pt(1) | Fe3+, Fe2+ (a = 1) || Ce4+, Ce3+ (a = 1) | Pt (2)
(Fe2+) = 0.77 V, and E°(Ce4+,Ce3+) = 1.61 V
If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time?


For the electrochemical cell
Pt(1) | Fe3+, Fe2+ (a = 1) || Ce4+, Ce3+ (a = 1) | Pt (2)
the cell regions are

  Right-half cell: reduction
                     Ce4++e-   Ce3+
 
Left-half cell: oxidation
                      Fe2+   Fe3++e-
______________________________
Add Ce4++Fe2+ Ca3++Fe3+

The net cell potential is
E° Cell = E° R – E° L = 1.61 V – 0.77 V = 0.84 V.
Since E°cell is positive, the cell reaction will be spontaneous.
The current in the external circuit will flow from Pt (1) (which serves as anode to Pt(2) which serves as cathode.

With the passage of time, Ecell will decrease and so is the current in the external circuit.

220 Views

In a fuel cell, H2 and O2 react to produce electricity. In the process H2 gas is oxidised at the anode and O2 at cathode. If 67.2 litre of H2 at STP reacts in 15 minutes, what is the average current produced? If the entire current is used for electro-deposition of Cu from Cu2+, how many grams of copper are deposited?

The redox changes in fuel cell are
               2H2(g) + O2(g)   2H2O (l)
At anode:     H2+2OH-    2H2O + 2e
At cathode:  O2+2H2O+4e   4OH-
Therefore, moles of H2 reacting = 67.222.4 = 3

Therefore, equivalent of H2 used  = 67.222.4=3
Now,            wE= i×t96500
Therefore,        
                   6 = i×15×6096500i = 643.33 ampere
or
   Also Eq. of H2 = Eq. of Cu formed  = 6.
Hence,           WCu = 6×63.52=190.5 g.



195 Views

Calculate the maximum possible electric work that can be obtained from the following cell under the standard conditions at 25°C:
     Fe|Fe2+(aq) || Cu2+(aq)|Cu
E°Fe2+(aq)/Fe = -0.44 VE°Cu2+(aq)/Cu = + 0.34 V and F = 96, 500 C.


Wmax = nFE°cell

The cell is
                Fe|Fe2+(aq)|| Cu2+(aq)|Cu
                         Fe  Fe2+ + 2e
         Cu2++2e  Cu
__________________________
      
      Fe+Cu2+   Fe2++Cu
___________________________

Here,              n = 2
Therefore,   
                 E°cell = E°cathode - E°anode           = E°Cu2+/Cu - E°Fe2+/Fe           = + 0.34 V - (-0.44 V)            = 0.78 V

                Wmax =2×96500 C × 0.78 V            = 150540 J = 150.5 kJ.
163 Views

Advertisement

When a certain conductivity cell was filled with 0.1 M KCl, it has a resistance of 85 Q at 25°C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 Ω. Calculate the molar conductivity of the unknown electrolyte at this concentration. (Specific conductivity of 0.1 M KCl = 1.29 x 10–2 ohm–1cm–1).


Resistance of KCl solution,
         R = 85 Ω
Cell constant  = K x R
                       =1.29 ×10-2Ω-1cm-1×85Ω= 1.1 cm-1
Resistance of unknown electrolyte solution,
                  R = 96 Ω
Specific conductance
                  K = Cell constantR     = 1.1 cm-196 Ω = 11960 Ω-1 cm-1
Concentration,   
                   C = 0.052 molL    = 0.052 mol1000 cm3     = 5.2 × 10-5 mol cm-3
Molar conductance,
                Λm = KC=11 Ω-1 cm-1960×5.2×10-5 mol cm-3       = 220.2 Ω-1 cm2 mol-1.




931 Views

Advertisement
Zn | Zn2+ (α = 0.1 M) || Fe2+ (α = 0.01 M) | Fe. The emf of the above cell is 0.2905 V. What is the equilibrium constant for the cell reaction?

For cell
Zn vertical line Zn to the power of 2 plus end exponent left parenthesis straight alpha space equals space 0.1 space straight M right parenthesis space vertical line vertical line space Fe to the power of 2 plus end exponent space left parenthesis straight alpha space equals space 0.01 space straight M right parenthesis space vertical line space Fe
The cell reaction
 left parenthesis straight i right parenthesis space Zn left parenthesis straight s right parenthesis space rightwards arrow space Zn to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space 2 straight e
left parenthesis ii right parenthesis space Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space 2 straight e space rightwards arrow space Fe left parenthesis straight s right parenthesis

For cellThe cell reaction On applying Nernst equation         
On applying Nernst equation
                    straight E subscript cell space equals space straight E degree subscript cell space minus space fraction numerator 0.0591 over denominator straight n end fraction log space 10 fraction numerator open square brackets Zn to the power of 2 plus end exponent close square brackets over denominator open square brackets Fe to the power of 2 plus end exponent close square brackets end fraction
0.2905 space equals space straight E degree subscript cell minus fraction numerator 0.0591 over denominator 2 end fraction log subscript 10 fraction numerator 0.1 over denominator 0.01 end fraction

or              0.2905 space equals space straight E degree subscript cell minus 0.0295 cross times log subscript 10 10
or              0.2905 space space equals space straight E degree subscript cell space minus space 0.0295 space cross times space 1
or                straight E degree subscript cell space equals space 0.2905 space plus space 0.0295 space equals space 0.32 space straight V
    At equilibrium left parenthesis straight E subscript cell space equals space 0 right parenthesis
                    straight E subscript cell space equals space straight E degree subscript cell space minus space fraction numerator 0.0591 over denominator straight n end fraction log subscript 10 straight K subscript straight c
∴         0 space equals space straight E degree subscript cell space minus space fraction numerator 0.0591 over denominator straight n end fraction log subscript 10 space straight K subscript straight c
or      straight E degree subscript cell space equals space fraction numerator 0.0591 over denominator straight n end fraction log subscript 10 space straight K subscript straight c
or      0.32 space equals space fraction numerator 0.0591 over denominator straight n end fraction space log subscript 10 space straight K subscript straight c
or               straight K subscript straight c space equals space 10 to the power of 0.32 divided by 0.0295 end exponent.

785 Views

Advertisement