concentrations NaF and KCl each 0.10 M in the cell

concentrations NaF and KCl each 0.10 M in the cell

 

 

Suppose that the concentrations of NaF and KCl

were each 0.10 M in the cell

Pb(s) | PbF2(s) | F^-(aq) || Cl^-(aq) | AgCl(s) | Ag(s)

 

a) Using the half-reactions

2AgCl(s) + 2e^- ⇌ 2Ag(s) + 2Cl^- and

PbF2(s) + 2e^- ⇌ Pb(s) + 2F^-,

calculate the cell voltage.

 

b) In which direction do electrons flow?

 

c) Now calculate the cell voltage by using the reactions

2Ag^+ + 2e^- ⇌ 2Ag(s) and

Pb^2+ + 2e^- ⇌ Pb(s). For this part,

you will need the solubility products for PbF2 and AgCl.

 

 

 

[참고] cell diagram. cell notation

산화 전극(anode) || 환원 전극(cathode)

( 참고 https://ywpop.tistory.com/3072 )

 

[참고] 표준 환원 전위

> AgCl + e^- → Ag + Cl^- ... E° = +0.222 V

> PbF2(s) + 2e^- ⇌ Pb(s) + 2F^- ... E° = –0.350 V

( 참고 https://ywpop.tistory.com/7027 )

 

 

 

 

a)

 

2AgCl(s) + 2e^- ⇌ 2Ag(s) + 2Cl^-

 

Nernst 식

E = E° – (0.0592 V / n) × logQ

= E° – (0.0592 V / n) × log([생성물]/[반응물])

( 참고 https://ywpop.tistory.com/2900 )

 

= (+0.222) – (0.0592 / 2) × log((0.10)^2)

= +0.281 V

= E_red (환원전극)

 

 

 

PbF2(s) + 2e^- ⇌ Pb(s) + 2F^-

 

E = E° – (0.0592 V / n) × log([생성물]/[반응물])

= (–0.350) – (0.0592 / 2) × log((0.10)^2)

= –0.291 V

= E_ox (산화전극)

 

 

 

E_cell = E_red(환원전극) – E_red(산화전극)

= (환원된 물질의 환원전위) – (산화된 물질의 환원전위)

( 참고 https://ywpop.tistory.com/4558 )

 

= E_red – E_ox

= (+0.281) – (–0.291)

= +0.572 V

 

 

 

[ 같은 글 https://ywpop.blogspot.com/2024/09/concentrations-naf-and-kcl-each-010-m.html ]

 

 

300x250

 

 

b)

 

The electrons are supplied by the oxidation reaction or

from the Pb/PbF2 electrode and go to the Ag/AgCl electrode.

 

 

 

 

c)

 

AgCl(s) ⇌ Ag^+(aq) + Cl^-(aq)

 

Ksp = [Ag^+] [Cl^-]

( 참고 https://ywpop.tistory.com/2966 )

 

[Ag^+] = Ksp / [Cl^-]

= (1.8×10^(-10)) / 0.10

= 1.8×10^(-9) M

 

 

 

PbF2(s) ⇌ Pb^2+(aq) + 2F^-(aq)

 

[Pb^2+] = (3.6×10^(-8)) / (0.1)^2

= 3.6×10^(-6) M

 

 

 

2Ag^+ + 2e^- ⇌ 2Ag(s)

 

E_red = 0.799 V – (0.0592 / 2) × log(1 / [Ag^+]^2)

= 0.799 – (0.0592 / 2) × log(1 / (1.8×10^(-9))^2)

= 0.281 V

 

 

 

Pb^2+ + 2e^- ⇌ Pb(s)

 

E_ox = (–0.126 V) – (0.0592 / 2) × log(1 / [Pb^2+])

= (–0.126) – (0.0592 / 2) × log(1 / (3.6×10^(-6)))

= –0.287 V

 

 

 

E_cell = E_red – E_ox

= 0.281 – (–0.287)

= 0.568 V

 

 

 

 

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