Zn | Zn^2+(1.00 M) || Ag^+(1.00 M) | Ag [Zn^2+] 0.20 mol/L

Zn | Zn^2+(1.00 M) || Ag^+(1.00 M) | Ag [Zn^2+] 0.20 mol/L

Consider the cell described below:

Zn | Zn^2+(1.00 M) || Ag^+(1.00 M) | Ag

Calculate the cell potential after the reaction has operated long

enough for the [Zn^2+] to have changed by 0.20 mol/L.

(Assume temperature = 25℃)

---------------------------------------------------

▶ 표준환원전위

> Ag^+ + e^- → Ag ... E° = +0.80 V

> Zn^2+ + 2e^- → Zn ... E° = –0.76 V

---> Ag가 환원된다.

( 참고 https://ywpop.tistory.com/7027 )

Zn(s) + 2Ag^+(s) → Zn^2+(aq) + 2Ag(s)

K = [Zn^2+] / [Ag^+]^2

E°_cell = E°_red(환원전극) – E°_red(산화전극)

= (환원된 물질의 표준환원전위) – (산화된 물질의 표준환원전위)

( 참고 https://ywpop.tistory.com/4558 )

= (+0.80) – (–0.76)

= +1.56 V

[Zn^2+] to have changed by 0.20 mol/L

---> 전지 반응이 일어날수록,

Zn(s) + 2Ag^+(s) → Zn^2+(aq) + 2Ag(s)

Zn^2+의 농도는 증가하므로,

( Ag^+의 농도는 감소. )

[Zn^2+]가 0.20 mol/L만큼 변했다는 것은

[Zn^2+]가 증가했다는 의미이므로,

> 1.00 M + 0.20 M = 1.20 M Zn^2+

> 1.00 M – 2(0.20 M) = 0.60 M Ag^+

( 반응물인 Ag^+의 계수가 2임에 주의할 것. )

Q = [Zn^2+] / [Ag^+]^2

= 1.20 / 0.60^2

= 3.3333

E = E° – (0.0592 V / n) × logQ

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

( 참고 https://ywpop.tistory.com/2900 )

= (+1.56) – (0.0592 / 2) × log(1.20 / 0.60^2)

= +1.5445 V

또는

E = E° – (RT / nF) lnQ

= (+1.56) – [(8.314 × (273.15+25)) / (2 × 96485)] ln(1.20 / 0.60^2)

= +1.5445 V

답: +1.54 V

[키워드] Zn-Ag 전지 기준