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When zinc is refined by electrolysis, the desired halfreaction at the cathode isZn²⁺(aq) + 2e⁻ → Zn(s) A competing reaction, which lowers the yield, is the formation of hydrogen gas:2H⁺(aq) + 2e⁻ → H₂(g) If 91.50% of the current flowing results in zinc being deposited, while 8.50% produces hydrogen gas, how many liters of H₂,measured at STP, form per kilogram of zinc?

Respuesta :

The volume of hydrogen gas is 31.6L.

Desired Half reaction

Zn²⁺(aq) + 2e⁻ →Zn(s)

Competing Reaction

2H⁺(aq) + 2e⁻ → H₂(g)

As we know that, 2moles of electrons are transferred to deposit completely 1 mole of Zinc.

So, the total number of moles used to deposit 1Kg of Zn will be

(1000/65.41) × 2 = 30.6 mol.

As we know that,

1 mole = 96485C

So, 30.6 moles = 30.6 × 96485C

= 2.95 × 10^6 C.

Now, For 91.5% of current flowing, the total amount of current flowing can be

91.5 % × x = 2.95 × 10^6 C.

x = 2.95 × 10^6 / 0.915

x = 3.22 × 10^6 C

Now, calculate the quatity which is utilize in the production of hydrogen gas, we have

= 8.50% × 3.22 × 10^6 C

= 273700C

Calculation of number of moles

Now, calculating the moles transferred to charge,

273700/ 96485

= 2.83mole

Volume of hydrogen gas = 2.83× 22.4/2

= 31.6L

Thus, we calculated that the volume of hydrogen gas is 31.6L.

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