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  • Stuck here, help me understand: - Redox Reaction and Electrochemistry - JEE Main-4

Cathodic raection in the dry  cell is represented by 

2 MnO_2+ Zn^{2+} +2e^-\rightarrow ZnMn_2O_4(s)

Let there be 8 g of  2 MnO_2 in the cathodic compartment . How many days will the dry cell continue to give a current of 4\times 10^{-3} ampere ? 

  • Option 1)

    23.2

  • Option 2)

    25.7

  • Option 3)

    29.79

  • Option 4)

    25.05

 

Answers (1)

best_answer

As we have learned

Dry Cell -

Anode - Zinc container 

Cathode - Graphite rod surrounded by MnO_{2} powder.

Electrolyte - Paste of NH_{4}Cl\:+\:ZnCl_{2}

- wherein

Cathode reaction:

2MnO_{2} + 2NH_{4}^{+}(aq)+ 2e^{-}\rightarrow Mn_{2}O_{3}(s)+2NH_{3}(g)+H_{2}O(l)

Anode reaction:

Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}

Cell potential 1.25 v to 1.5 v

 

 

  Equuivalent mass of Mn O_2= \frac{molecular \; \; \; mass}{change\; \; in \; \; oxidation \; \;state }= 87/1=87

cathodic reaction : 2MnO_4+ Zn^{2+}+2e^-\rightarrow Zn ^{+3}Mn_2O_4

from law of electrolysis 

W= \frac{ItE}{96500}

8= \frac{4\times 10^{-3}t\times 87}{96500}

t = 25.675 days \approx 25.7 days 

 

 

 

 

 

 


Option 1)

23.2

Option 2)

25.7

Option 3)

29.79

Option 4)

25.05

Posted by

prateek

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