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Chapter 17: Q10P (page 428)

A0.3268-gunknown containing Pb(CH3CHOHCO2)2(leadLactate, FM 385.3) plus inert material was electrolyzed to produce 0.1111gofPbO2(FM239.2). Was the PbO2deposited at the anode or at the cathode? Find the weight percent of lead lactate in theUnknown.

Short Answer

Expert verified

PbO2was deposited at the anode. The weight percent of lead lactate is wleadlactate=54.76%.

Step by step solution

01

Electrogravimetric Analysis

  • Separation along with the quantification of metals are mainly observed in electrogravimetricanalysis.
  • Electrolysis of the analyte solution is done during this procedure.
02

Determine the lead lactate

The reaction is

PbCH3CHOHCO2+2H2O→PbO2+2lactate+4H++2e-

Pb2+Is oxidized to Pb4+at the anode

First, we need to calculate n of PbO2

role="math" localid="1663650526992" nPbO2=mPbO2MPbO2=0.1111g239.2g/molnPbO2=4.64×10-4molnPbO2=nleadlactate,themassoftheleadlactateismleadlactate=nleadlactate·Mleadlactatemleadlactate=4.64×10-4mol×385.3g/molmleadlactate=0.17896g

So, the w lead lactate is:

wleadlactate=mleadlactatemwleadlactate=0.17896g0.3268gwleadlactate=0.5476=54.76%

The lead lactate is 54.76%.

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