A1.0 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?

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/molFaraday'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/tamperage

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

⇒ I = 346466.267 C / 63000 s

⇒ I = 5.499 A

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na