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Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

Consider electrolysis of aqueous silver ion.Consider electrolysis of aqueous silver ion.

AgAg++ (aq) + e- ---> Ag(s) (aq) + e- ---> Ag(s)

1 mol e-1 mol e- ---> 1 mol Ag---> 1 mol Ag

If we could measure the moles of e-, we If we could measure the moles of e-, we could know the quantity of Ag formed.could know the quantity of Ag formed.

But how to measure moles of e-?But how to measure moles of e-?

Current = charge passing

timeCurrent =

charge passingtime

I (amps) = coulombsseconds

I (amps) = coulombsseconds

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But how is charge related to moles of But how is charge related to moles of electrons?electrons?

Current = charge passing

timeCurrent =

charge passingtime

I (amps) = coulombsseconds

I (amps) = coulombsseconds

Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

= = 96,500 C/mol e- 96,500 C/mol e- = = 1 Faraday1 Faraday

Charge on 1 mol e -

= 1.60 x 10-19 Ce -

6.02 x 1023

e -mol

33Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 (aq) solution for 15.0 min. What mass of Ag metal is deposited?min. What mass of Ag metal is deposited?

SolutionSolution

(a)(a) Calc. chargeCalc. charge

Charge (C) = current (A) x time (t)Charge (C) = current (A) x time (t)

= (1.5 amps)(15.0 min)(60 s/min) = 1350 C= (1.5 amps)(15.0 min)(60 s/min) = 1350 C

I (amps) = coulombsseconds

I (amps) = coulombsseconds

44Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

SolutionSolution

(a)(a) Charge = 1350 CCharge = 1350 C

(b)(b) Calculate moles of e- usedCalculate moles of e- used

I (amps) = coulombsseconds

I (amps) = coulombsseconds

1350 C • 1 mol e -96, 500 C

0.0140 mol e -1350 C • 1 mol e -96, 500 C

0.0140 mol e -

0.0140 mol e - • 1 mol Ag1 mol e -

0.0140 mol Ag or 1.51 g Ag0.0140 mol e - • 1 mol Ag1 mol e -

0.0140 mol Ag or 1.51 g Ag

1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 min. What (aq) solution for 15.0 min. What mass of Ag metal is deposited?mass of Ag metal is deposited?

(c)(c) Calc. quantity of AgCalc. quantity of Ag

55Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is

Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-

If a battery delivers 1.50 amp, and you have 454 g of Pb, how If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?long will the battery last?

SolutionSolution

a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb

b)b) Calculate moles of e-Calculate moles of e-

2.19 mol Pb • 2 mol e -1 mol Pb

= 4.38 mol e -2.19 mol Pb • 2 mol e -1 mol Pb

= 4.38 mol e -

c)c) Calculate chargeCalculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C4.38 mol e- • 96,500 C/mol e- = 423,000 C

66Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is

Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-

If a battery delivers 1.50 amp, and you have 454 g of Pb, how If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?long will the battery last?

SolutionSolution

a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb

b)b) Mol of e- = 4.38 molMol of e- = 4.38 mol

c)c) Charge = 423,000 CCharge = 423,000 C

Time (s) = Charge (C)

I (amps)Time (s) =

Charge (C)I (amps)

Time (s) = 423, 000 C1.50 amp

= 282,000 sTime (s) = 423, 000 C1.50 amp

= 282,000 s About 78 hoursAbout 78 hours

d)d) Calculate timeCalculate time