A current of 5.21 A is passed through a Pb(NO3)2 solution for 1.40 hours. How much lead is plated out?

Question

So I can't seem to get the correct answer. What I've done is:

5.21*1.4hr*60mins*60secs=26258
Then divide by 1 faraday
26258/96485= 0.27 moles

Using the half reaction of Pb2+ only 1/2 moles (this might be where I'm going wrong)
Thus 0.27moles * 0.5 * 331.21 (MW of Pb(NO3)2) = 45.07g but this is coming out wrong. So I can't seem to get the correct answer. What I've done is:

5.21*1.4hr*60mins*60secs=26258
Then divide by 1 faraday
26258/96485= 0.27 moles

Using the half reaction of Pb2+ only 1/2 moles (this might be where I'm going wrong)
Thus 0.27moles * 0.5 * 331.21 (MW of Pb(NO3)2) = 45.07g but this is coming out wrong.

Explanation / Answer

oulombs = 5.21 amp x 1.4 hr x (60 min/hr)(60 sec/min) = estimated 26258 Coulombs
96,485 C will plate out 207.2/2 grams Pb(about 104 g) so
(207.2/2) x (26258/96,485) = 275.18 g Pb plated out.

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