ECS是否適用於conc. acid

1.
[1] ---- NO3(−) + 2 H(+) +  e− <======> NO2 +  H2O E = +0.80V
[2] ---- NO3(−) + 4 H(+) +  e− <======> NO + 2 H2O E = +0.96V

[3] ---- Hg(2+) + 2 e− <======> Hg E = +0.85V

[1]-[3] = -0.05V
[2]-[3] = +0.11V
因此, 可以預期standard conditions 下, 會出NO 多過會出NO2 .


2.
與其叫ECS, 不如寫做standard reduction potential, 有數得計.


3.
考慮此反應:
2NO3(−) + 4 H(+) +  Hg <======> 2NO2 + 2 H2O + Hg(2+)
根據Le Chatelier's principle 預測, H(+) & NO3(−) 的濃度提升, 會將平衡位置 equilibrium position 推向product side.


4.
請用Nernst equation, 估計濃度改變後的reduction potential.

n<ox> + ne- <======> m<red>

E = E(o) - (RT/nF) ln ( [red]^m / [ox]^n )
E = 半反應的還原電勢
E(o) = 半反應的標準還原電勢
R = universal constant = 8.314
T = 溫度, 當298K
n = 該半反應中涉及到的電子數
F = Faraday constant = 96485
[red] = 半反應中還原劑reductant 的濃度
m = 還原劑的系數 coefficient(s)
[ox] = 半反應中氧化劑oxidant 的濃度
n = 氧化劑的系數

考慮NO3(−) + 2 H(+) +  e− <======> NO2 +  H2O , 硝酸濃度=18M, 其他條件則維持於standard conditions
E = E(o) - (RT/nF) ln ( [NO2]^1 / [NO3(-)]^1 [H(+)]^2 )
= (+0.80) - (8.314 x 298)/(1 x 96485) x ln[ (1)/(18 x 18^2) ]
= +0.80 - 0.0256 x ln(1/5832)
= +0.80 - 0.0256 x (-8.6711)
= +0.80 + 0.2220 = +1.02 V ---- [4]

[4] - [3] = +1.02 - +0.85 = +0.17V
電勢差是正數, 故可預測反應可以進行. 有反應.


2012-10-30 23:17:27 補充：
的確有此顧慮.
其實上面寫的Nernst equation 是簡化過: "濃度" 部份嚴格來說應是activity
= concentration x activity coefficient
不過, 用以作大約的估算也是夠了.
Activity 要計要查也是太麻煩.

actually 咁高既concentration入面硝酸既activity coefficient會唔會唔等於佢個濃度?(因為愈conc.愈唔似ideal solution嘛)