A galvanic cell is set up from a zinc bar weighing 50 gram and 1.0 litre of 1.0 M, CuSO4 solution. How long would the cell run, assuming it delivers a steady current of 1.0 ampere (a) 48hrs (b) 41hrs (ć) 21 hrs (d) 1 hrs

Question

Arun · Answer

As u know According to Faraday's 1st law if electrolysis….

W=Z×q or W=Z×I×t

We know Z ( electrochemical equivalent):—

Z= Equivalent weight / Faraday const. (F)

So, Equivalent wt of CuSO4 = molecular mass of CuSO4 / valency.

Z= 159/2 = 79.5

Now we r given with,

W= 50g

I=1A

t=???

So put values in Formula and time will be founded

arun kumar · Answer

According to faraday 1st law of electrolysis W=zit W=50g Molecular mass of cuso4 solution is 179 N factor 2 Z=159÷2=79.5 W=ZIt 50=79.5×1×t T~0.63 ×3600(in sec) T~21 hrs

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A galvanic cell is set up from a zinc bar weighing 50 gram and 1.0 litre of 1.0 M, CuSO4 solution. How long would the cell run, assuming it delivers a steady current of 1.0 ampere (a) 48hrs (b) 41hrs (ć) 21 hrs (d) 1 hrs

A galvanic cell is set up from a zinc bar weighing 50 gram and 1.0 litre of 1.0 M, CuSO4 solution. How long would the cell run, assuming it delivers a steady current of 1.0 ampere
(a) 48hrs (b) 41hrs (ć) 21 hrs (d) 1 hrs

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A galvanic cell is set up from a zinc bar weighing 50 gram and 1.0 litre of 1.0 M, CuSO4 solution. How long would the cell run, assuming it delivers a steady current of 1.0 ampere (a) 48hrs (b) 41hrs (ć) 21 hrs (d) 1 hrs

A galvanic cell is set up from a zinc bar weighing 50 gram and 1.0 litre of 1.0 M, CuSO4 solution. How long would the cell run, assuming it delivers a steady current of 1.0 ampere(a) 48hrs (b) 41hrs (ć) 21 hrs (d) 1 hrs

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A galvanic cell is set up from a zinc bar weighing 50 gram and 1.0 litre of 1.0 M, CuSO4 solution. How long would the cell run, assuming it delivers a steady current of 1.0 ampere
(a) 48hrs (b) 41hrs (ć) 21 hrs (d) 1 hrs