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The overall cell reaction occurring in an alkaline battery isZn(s) + MnO₂(s) + H₂O(l) → ZnO(s) + Mn(OH)₂(s) (d) How many coulombs are produced in part (b)?(b) If 4.50 g of zinc is oxidized, how many grams of manganese dioxide and of water are consumed?

Respuesta :

shaikhmohd0119

d) Columbs produced = 132763.36 C

b) Mass of MnO₂ = 5.981 g

    Mass of H₂O = 1.2384 g

b) Given Reaction

           Zn(s) + MnO₂(s) + H₂O(l) → ZnO(s) + Mn(OH)₂(s)

 Mass of Zn = 4.50 g

 Moles of Zn = 0.0688 moles

  Now,

   Moles of Zn = moles of MnO₂ = moles of H₂O = moles of ZnO = moles of                   Mn(OH)₂

Hence ,

Moles of MnO₂ = 0.0688 moles

Mass of MnO₂ = 0.0688 × 86.9368 g

                         = 5.981 g

Similarly,

   Moles of H₂O = 0.0688 moles

    Mass of H₂O = 0.0688 × 18 g

                          = 1.2384 g

d)  Now ,

        1 Mole of Zn in given reaction = 2 Faraday

        0.0688 Moles of Zn in given reaction = 0.1376 Faraday

         1 Faraday = 96485 C

       Hence ,

           0.0688 Moles of Zn in given reaction = 0.1376 Faraday

                                                                           = 0.13760× 96485 C

                                                                            = 132763.36 C

Thus from above Conclusion we can say that no. of coulombs produced in the given reaction is 132763.36 C

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