2Ag(s) + 2H^+(aq) + 2Cl^-(aq) → 2AgCl(s) + H2(g)

2Ag(s) + 2H^+(aq) + 2Cl^-(aq) → 2AgCl(s) + H2(g)

 

 

다음 보기를 참고하여 전지 전압이 –0.420 V가 됨을 증명하시오.

Ag | AgCl(sat’d), HCl(0.020 M) | H2(0.8 atm) | Pt

AgCl(s) + e^- → Ag(s) + Cl^-(aq) ... E° = 0.222 V

H^+(aq) + e^- → 1/2 H2(g) ... E° = 0.000 V

 

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▶ 참고: Pt(s) | H2(g) | H^+(aq), Cl^-(aq) | AgCl(s) | Ag(s)

[ https://ywpop.tistory.com/16864 ]

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Ag | AgCl(sat’d), HCl(0.020 M) | H2(0.8 atm) | Pt

왼쪽이 산화전극이므로,

( 참고 https://ywpop.tistory.com/3072 )

 

> 산화: 2Ag(s) + 2Cl^-(aq) → 2AgCl(s) + 2e^-

> 환원: 2H^+(aq) + 2e^- → H2(g)

> 전체: 2Ag(s) + 2H^+(aq) + 2Cl^-(aq) → 2AgCl(s) + H2(g)

 

 

 

Q = [H2] / ([H^+]^2 [Cl^-]^2)

= (0.8) / ( (0.020)^2 (0.020)^2 )

 

 

 

E°cell = E°red(환원전극) – E°red(산화전극)

= 0.000 V – (0.222 V)

= –0.222 V

 

 

 

E = E° – (0.0592 V / n) × logQ

( 참고 https://ywpop.tistory.com/2900 )

 

= (–0.222) – (0.0592 / 2) × log((0.8) / ( (0.020)^2 (0.020)^2 ))

= –0.4203 V

= –0.420 V