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A current of 10 ampere is passed for 80 minutes 27 seconds through a cell containing dilute H_{2}SO_{4} : - How many moles of oxygen gas is liberated at anode?
- Calculate the amount of zinc deposited at the cathode when another cell containing ZnSO
_{4} solution is connected in series ?

_{4}solution is connected in series ?Time, t = 80 min and 27 second

= (80 $\times $60) + 27

= 4827 seconds

Charge, Q = I t

Q = 10$\times $ 4827

= 48270 F

Now in electrolysis of H

_{2}SO

_{4}, H

_{2}and O

_{2}liberated as follows:

At cathode:

2H

^{+}+ 2e

^{-}→ H

_{2}

At anode:

2H

_{2}O → O

_{2}+ 4H

^{+}+ 4e

^{-}

4F liberate 1 mole of Oxygen gas or 32 g of Oxygen.

4F or 4$\times $96500 C will liberate 32 g of Oxygen gas

48270 C will liberate $48270\times \frac{32}{4\times 96500}$

= 4 g of Oxygen gas

(ii) Reaction of Zic is as foloows:

Zn

^{2}

^{+}+ 2e

^{- }→ Zn

2F of electricity willl liberate 1 mole of Zinc or 65 g of Zinc.

2$\times $96500 C liberate 65 g of Zinc

48270 C will liberate $48270\times \frac{65}{2\times 96500}$

= 16.25 g

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