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A 1.0 L solution of B e F 2 was electrolyzed for 17.5 h to give 16.178 g of beryllium. Assuming the minimum voltage needed was available, what amperage would be needed to complete the electrolysis in the given time?

Respuesta :

Answer:

I = 5.499 A

Explanation:

  • BeF2 → Be2+  + 2F-

∴ m Be = 16.178 g

∴ t = 17.5 h

  • m = (Q/F)(M/z)........mass of the substance produced at the electrode

⇒ Q = (m×F×z)/M.......... total electrical charge

∴ F ≅ 96500 C/mol....Faraday's constant

∴ M Be = 9.012 g/mol

∴ z Be = 2

⇒ Q = (16.178 g Be)(96500 C/mol)(2)/(9.012 g/mol)

⇒ Q = 346466.267 C

  • Q = I×t

⇒ I = Q/t............amperage

∴ t = (17.5 h)(60 min/h)(60 s/min) = 63000 s

⇒ I = 346466.267 C / 63000 s

⇒ I = 5.499 A

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