Question

¿que cantidad de cobre de una solucion de sulfato cuprico CuSO4 se depositara cuando pasa una corriente de 5 amperios en 30 minutos ?

Answer 1

             Cu⁺² + 2e⁻ → Cu

Mm Cu = 63.5 g/mol

1. aplicar la fórmula m = Mm x  I  x  t
                                       `````````````````
                                       carga x 96500

t = 1 min ------- 60 s
   30 min -------  x
         x = 1800 s

m = 63.5 g/mol x 5 c/s x 1800 s
       `````````````````````````````````````
                  2 x 96500 c/eq

m = 2.96 g de cobre

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otra forma de calcular gramos

Mm Cu = 63.5 g/mol

calcular eq-g del Cu = 63.5 g/mol / 2

eq-g = 31.75 g/eg

calcular Q = I x T

Q = 5 c/s  x 1800 s

Q = 9000 coulombios

calcular faraday

1 faraday ------- 96500 c
       x       --------  9000 c

x = 0.0932 faraday

3. calcular gramos

           1 f ------- 31.75 g
  0.0932 f -------     x
            x = 2.96 g de Cu