pH = 3.00 이고 P(AsH3) = 1.00 mbar

pH = 3.00 이고 P(AsH3) = 1.00 mbar

 

 

pH = 3.00 이고 P(AsH3) = 1.00 mbar 일 때

아래의 반쪽전지전위(V)는?

As(s) + 3H^+ + 3e^- ⇌ AsH3(g)

E° = –0.238 V

 

① –0.592

② –0.415

③ –0.356

④ –0.120

 

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pH = –log[H^+] = 3.00 이므로,

[H^+] = 10^(-3.00) M

 

 

 

P(AsH3) = 1.00 mbar

= 0.00100 bar ≒ 0.00100 atm

 

 

 

또는

1 atm = 1.01325 bar 이므로,

0.00100 bar × (1 atm / 1.01325 bar) = 0.000987 atm

 

 

 

Nernst 식

E = E° – (0.0592 V / n) × logQ

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

( 참고 https://ywpop.tistory.com/2900 )

 

= (–0.238) – (0.0592 / 3) × log(P_AsH3 / [H^+]^3)

= (–0.238) – (0.0592 / 3) × log(0.00100 / [10^(-3.00)]^3)

= –0.3564 V

 

 

 

또는

= (–0.238) – (0.0592 / 3) × log(0.000987 / [10^(-3.00)]^3)

= –0.3563 V

 

 

 

답: ③ –0.356