[Cd^2+] = 0.20 M, [H^+] = 0.16 M, P_H2 = 0.80 atm

[Cd^2+] = 0.20 M, [H^+] = 0.16 M, P_H2 = 0.80 atm

 

 

[Cd^2+] = 0.20 M, [H^+] = 0.16 M, P_H2 = 0.80 atm 이라면,

Cd^2+/Cd 반쪽 전지와 Pt/H^+/H2 반쪽 전지로 구성된 전지의 기전력은 얼마인가?

 

 

What is the emf of a galvanic cell

consisting of a Cd^2+/Cd half-cell and a Pt/H^+/H2 half-cell

at the following concentrations and pressures?

[Cd^2+] = 0.20 M, [H^+] = 0.16 M, P_H2 = 0.80 atm

 

 

 

표준 환원 전위

> 2H^+ + 2e^- → H2 ... E° = 0.00 V

> Cd^2+ + 2e^- → Cd ... E° = –0.40 V

( 참고 https://ywpop.tistory.com/7027 )

 

 

 

전지 반응

Cd(s) + 2H^+(aq) → Cd^2+(aq) + H2(g)

 

 

 

표준 기전력

E°cell = E°red(환원전극) – E°red(산화전극)

= (환원된 물질의 표준환원전위) – (산화된 물질의 표준환원전위)

( 참고 https://ywpop.tistory.com/4558 )

 

= (0.00) – (–0.40) = +0.40 V

 

 

 

반응비 (reaction quotient)

Q = [Cd^2+] P_H2 / [H^+]^2

= (0.20) (0.80) / (0.16)^2 = 6.25

 

 

 

네른스트 식. Nernst equation

E = E° – (0.0592 V / n) × logQ

( 참고 https://ywpop.tistory.com/2900 )

 

= (+0.40 V) – (0.0592 V / 2) × log(6.25)

= +0.38 V

 

 

 

답: +0.38 V

 

 

 

 

[ 관련 글 https://ywpop.blogspot.com/2024/07/pb2-010-m-h-0050-m-ph2-10-atm.html ]

[Pb^2+] = 0.10 M, [H^+] = 0.050 M, P_H2 = 1.0 atm

만일 [Pb^2+] = 0.10 M, [H^+] = 0.050 M이고, P_H2 = 1.0 atm이라면 Pb^2+/Pb 반쪽 전지와 Pt/H^+/H2 반쪽 전지로 구성된 전지의 기전력은 얼마인가?

What is the emf of a cell formed by the half cells of Pb^2+/Pb and Pt/H^+/H2 if [Pb^2+] = 0.10 M, [H^+] = 0.050 M and P_H2 = 1.0 atm?

 

 

 

[키워드] Pt/H^+/H2 반쪽 전지 기준