[Pb^2+]=1.5×10^(-3) M, [Zn^2+]=0.55 M emf

Consider a voltaic cell whose overall reaction is

Pb^2+(aq) + Zn(s) → Pb(s) + Zn^2+(aq).

What is the emf generated by this voltaic cell

when the ion concentrations are

[Pb^2+] = 1.5×10^(-3) M and [Zn^2+] = 0.55 M?

(a) 0.71 V (b) 0.54 V (c) 0.49 V (d) 0.79 V (e) 0.64 V

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E°_cell = E°_red(환원전극) – E°_red(산화전극)

= (환원된 물질의 표준환원전위) – (산화된 물질의 표준환원전위)

= (–0.13) – (–0.76)

= +0.63 V

E = E° – (0.0592 V / n) × logQ

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

( 참고 https://ywpop.tistory.com/2900 )

= 0.63 – (0.0592 / 2) × log([0.55] / [1.5×10^(-3)])

= 0.554 V

답: (b) 0.54 V

[키워드] 네른스트 기준, nernst dic