Question

A galvanic is set up from a zinc bar weighing 50 g and 1.0 litre, 1.0 M `CuSO_(4)` solution. How long would the cell run assuing it delivers a steady current of 1.0 ampere?

Answer 1

We have to calculate the time in which all Cu or Zn will be consumed on producing a steady current of 1 ampere.
`ZntoZn^(2+)+2e^(-)`
`underline(Cu^(2+)+2e^(-)toCu" ")`
`Zn+Cu^(2+)toZn^(2+)+Cu`
1L, 1M `CuSO_(4)` contain `CuSO_(4)=1` mole=1 mole of `Cu^(2+)`
`Zn=50g=(50)/(65.4)`mole `lt1`
Thus, Zn will be consumed first completely
`ZntoZn^(2+)+2e^(-)`
1 mole Zn (65.4g) requires charge=`2xx96500C`
`therefore50g` Zn will require charge`=(2xx96500)/(65.4)xx50C=147554C`
Q=Ixxtthereforet=(Q)/(I)=(147554)/(1)sec=41` hrs