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Chapter 17: Q10P (page 428)

A $0.326 8 - g$ unknown containing $Pb {({CH}_{3} {CHOHCO}_{2})}_{2}$ (leadLactate, FM 385.3) plus inert material was electrolyzed to produce $0.1111 g of {PbO}_{2} (FM 239.2)$ . 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 $w (lead lactate) = 54.76 %$ .

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

$Pb {({CH}_{3} CHOHCO)}_{2} + 2 H_{2} O 鈫� {PbO}_{2} + 2 lactate + 4 H^{+} + 2 e^{-}$

${Pb}^{2 +}$ Is oxidized to ${Pb}^{4 +}$ at the anode

First, we need to calculate n of ${PbO}_{2}$

role="math" localid="1663650526992" $n ({PbO}_{2}) = \frac{m ({PbO}_{2})}{M ({PbO}_{2})} = \frac{0.1111 g}{239.2 g / mol} n ({PbO}_{2}) = 4.64 脳 10^{- 4} mol n ({PbO}_{2}) = n (lead lactate), the mass of the lead lactate is m (lead lactate) = n (lead lactate) 路 M (lead lactate) m (lead lactate) = 4.64 脳 10^{- 4} mol 脳 385.3 g / mol m (lead lactate) = 0.17896 g$

So, the w lead lactate is:

$w (lead lactate) = \frac{m (lead lactate)}{m} w (lead lactate) = \frac{0.17896 g}{0.3268 g} w (lead lactate) = 0.5476 = 54.76 %$

The lead lactate is $54.76 %$ .

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Most popular questions from this chapter

The cyclic voltammogram of the antibiotic chloramphenicol (abbreviated) is shown here. The first cathodic scan goes from 0 to -1.0 V. The first cathodic wave, , is from the reaction ${RNO}_{2} + 4 e^{-} + 4 H^{+} 鈫� RNHOH + H_{2} O$ . Peak B in the reverse anodic scan could be assigned to $RNHOH 鈫� RNO + 2 H^{+} + 2 e^{-}$ . In the second cathodic scan from +0.9 to -0.4 V, the new peak C appears. Write a reaction for peak C and explain why peak C was not seen in the initial scan.

Cyclic voltammogram of 3.7 脳10^-4 chloramphenicol in 0.1 M acetate buffer, pH 4.62. The voltage of the carbon paste working electrode was scanned at a rate of 350 mV/s. [Data from P. T. Kissinger and W. R. Heineman, 鈥淐yclic Voltammetry,鈥� J. Chem. Ed. 1983, 60, 702.]

Ions that react with $A g^{+}$ can be determined electrogravimetrically by deposition on a silver working anode:

$A g (s) + X^{-} 鈫� A g X (s) + e^{-}$

(a) What will be the final mass of a silver anode used to electrolyze 75.00 mL of 0.02380 M KSCN if the initial mass of the anode is 12.4638 g?

(b) At what electrolysis voltage (versus S.C.E.) will AgBr(s) be deposited from 0.10M Br? (Consider negligible current flow, so that there is no ohmic potential, concentration polarization, or overpotential.)

(c) Is it theoretically possible to separate 99.99% of0.10M Klfrom0.10MKBr by controlled-potential electrolysis?

Consider the following electrolysis reactions.

Cathode: $H_{2} O (l) + e^{-} 鈬� \frac{1}{2} H_{2} (g, 1.0 b a r) + O H^{-} (a q, 0.10 M)$

Anode: $B r^{-} (a q, 0.10 M) 鈬� \frac{1}{2} B r_{2} (l) + e^{-}$

Calculate the voltage needed to drive the net reaction if current is negligible.
Suppose that the cell has a resistance of $2.0 惟$ and a current of 100 mA. How much voltage is needed to overcome the cell resistance? This is the ohmic potential.
Suppose that the anode reaction has an overpotential of 0.20 V and that the cathode overpotential is 0.40 V. What voltage is needed to overcome these effects combined with those of parts (a) and (b)?
Suppose that concentration polarization occurs $[O H^{-}]_{s}$ . at the cathode surface increases to 1.0 M and $[B r^{-}]_{s}$ at the anode surface decreases to 0.010 M. What voltage is needed to overcome these effects combined with those of (b) and (c)?

Suppose that the diffusion current in a polarogram for reduction of ${Cd}^{2 +}$ at a mercury electrode is $14 渭础$ If the solution containsof 25mLof $0.50 {mMCd}^{2 +}$ what percentage of ${Cd}^{2 +}$ is reduced in the 3.4 min required to scan from $- 0.6 to - 1.2 V$ ?

Write the chemical reactions that show that 1 mol of I₂ is required for 1 mol of H₂O in a Karl Fischer titration.

See all solutions

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