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5. Calculation of One-Electron Reduction Potentials Using Radical Formation Constants
5. 使用自由基形成常数计算单电子还原电位

5.1 Introduction  5.1 介绍

Radicals, e.g. A − − A − − A^(--)\mathrm{A}^{--}may be present in equilibrium with oxidant, A and reductant, A 2 − A 2 − A^(2-)\mathrm{A}^{2-} or their protonated conjugates:
自由基，例如可能与 A − − A − − A^(--)\mathrm{A}^{--} 氧化剂 A 和还原剂或其 A 2 − A 2 − A^(2-)\mathrm{A}^{2-} 质子化共轭物平衡存在：
A + A 2 − ⇄ 2 A ⋅ − A + A 2 − ⇄ 2 A ⋅ − A+A^(2-)⇄2A^(*-)\mathrm{A}+\mathrm{A}^{2-} \rightleftarrows 2 \mathrm{~A}^{\cdot-}
and a radical formation constant can be defined:
并且可以定义一个自由基形成常数：
K f = ( A ⋅ ) 2 ( A ) ( A 2 − ) K f = ( A ⋅ ) 2 ( A ) A 2 − K_(f)=((A*)^(2))/(((A))(A^(2)-))K_{\mathrm{f}}=\frac{(\mathrm{A} \cdot)^{2}}{(\mathrm{~A})\left(\mathrm{A}^{2}-\right)}
The value of K f K f K_(f)K_{f} is obviously a measure of the steady-state concentrations of radicals, A − − A − − A^(--)\mathrm{A}^{--}obtained on mixing oxidant A A AA with reductant, A 2 − A 2 − A^(2-)\mathrm{A}^{2-}. When experimental conditions result in sufficiently high concentrations of radicals to be measured, estimates of K f K f K_(f)K_{\mathrm{f}} can be used in conjuction with the two-electron potentials, E ∘ ( A / A 2 ) E ∘ A / A 2 E^(@)(A//A^(2))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{2}\right) or E ∘ ( A E ∘ ( A E^(@)(AE^{\circ}(\mathrm{A}, 2 H + / AH 2 ) 2 H + / AH 2 {:2H^(+)//AH_(2))\left.2 \mathrm{H}^{+} / \mathrm{AH}_{2}\right) to obtain estimates of the one-electron couples, E ∘ ( A / A ∙ ) E ∘ A / A ∙ E^(@)(A//A^(∙))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right), etc.
的值 K f K f K_(f)K_{f} 显然是自由基稳态浓度的量度， A − − A − − A^(--)\mathrm{A}^{--} 通过将氧化剂 A A AA 与还原剂 混合而得到。 A 2 − A 2 − A^(2-)\mathrm{A}^{2-} 当实验条件导致需要测量足够高浓度的自由基时，可以将 的 K f K f K_(f)K_{\mathrm{f}} 估计值与双电子电位结合使用， E ∘ ( A / A 2 ) E ∘ A / A 2 E^(@)(A//A^(2))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{2}\right) 或者 E ∘ ( A E ∘ ( A E^(@)(AE^{\circ}(\mathrm{A} 2 H + / AH 2 ) 2 H + / AH 2 {:2H^(+)//AH_(2))\left.2 \mathrm{H}^{+} / \mathrm{AH}_{2}\right) 来获得单电子对的估计值， E ∘ ( A / A ∙ ) E ∘ A / A ∙ E^(@)(A//A^(∙))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right) 等等。

5.2. Derivation of Expressions
5.2. 表达式的推导

Reaction 34 (above) can be obtained by subtracting 33 a 33 a 33 a33 a from 32a:
反应 34（上图）可以通过 33 a 33 a 33 a33 a 从 32a 中减去得到：
32a  32 个
A + e − → A ⋅ − A ⋅ − + e − → A 2 − . A + e − → A ⋅ − A ⋅ − + e − → A 2 − . {:[A+e^(-)rarrA^(*-)],[A^(*-)+e^(-)rarrA^(2-).]:}\begin{aligned} & \mathrm{A}+e^{-} \rightarrow \mathrm{A}^{\cdot-} \\ & \mathrm{A}^{\cdot-}+e^{-} \rightarrow \mathrm{A}^{2-} . \end{aligned}
Eq. (59) is obtained by subtracting the corresponding free-energy changes:
方程（59）是通过减去相应的自由能变化得到的：
E ∘ ( A / A ⋅ ) − E ∘ ( A ⋅ − / A 2 − ) = R T F ln ⁡ K f . E ∘ A / A ⋅ − E ∘ A ⋅ − / A 2 − = R T F ln ⁡ K f . E^(@)(A//A^(*))-E^(@)(A*-//A^(2-))=(RT)/(F)ln K_(f).E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\cdot}\right)-E^{\circ}\left(\mathrm{A} \cdot-/ \mathrm{A}^{2-}\right)=\frac{R T}{F} \ln K_{\mathrm{f}} .
If we add reaction 32 a 32 a 32 a32 a to reaction 33 a 33 a 33 a33 a we obtain reaction 31 a 31 a 31 a31 a. Noting that n = 2 n = 2 n=2n=2 in the conversion of free energy to potential, Eq. (2), in the latter reaction:
如果我们在反应 33 a 33 a 33 a33 a 中加入反应 32 a 32 a 32 a32 a ，我们会得到反应 31 a 31 a 31 a31 a 。请注意， n = 2 n = 2 n=2n=2 在自由能到电势的转换中，方程（2），在后一个反应中：
E ∘ ( A / A ⋅ ) + E ∘ ( A − / A 2 − ) = 2 E ∘ ( A / A 2 − ) E ∘ A / A ⋅ + E ∘ A − / A 2 − = 2 E ∘ A / A 2 − E^(@)(A//A^(*))+E^(@)(A^(-)//A^(2-))=2E^(@)(A//A^(2)-)E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\cdot}\right)+E^{\circ}\left(\mathrm{A}^{-} / \mathrm{A}^{2-}\right)=2 E^{\circ}\left(\mathrm{A} / \mathrm{A}^{2}-\right)
(cf. Eq. (6)). Adding Eqs. (59) and (60) yields:
（参见方程 （6））。添加 eqs。（59） 和 （60） 得出：
E ∘ ( A / A ⋅ ) = E ∘ ( A / A 2 − ) + R T 2 F ln ⁡ K f E ∘ A / A ⋅ = E ∘ A / A 2 − + R T 2 F ln ⁡ K f E^(@)(A//A^(*))=E^(@)(A//A^(2-))+(RT)/(2F)ln K_(f)E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\cdot}\right)=E^{\circ}\left(\mathrm{A} / \mathrm{A}^{2-}\right)+\frac{R T}{2 F} \ln K_{\mathrm{f}}
while subtraction gives:  而 Subtraction 得到：
E ∘ ( A ⋅ − / A 2 − ) = E ∘ ( A / A 2 − ) − R T 2 F ln ⁡ K f . E ∘ A ⋅ − / A 2 − = E ∘ A / A 2 − − R T 2 F ln ⁡ K f . E^(@)(A*-//A^(2-))=E^(@)(A//A^(2-))-(RT)/(2F)ln K_(f).E^{\circ}\left(\mathrm{A} \cdot-/ \mathrm{A}^{2-}\right)=E^{\circ}\left(\mathrm{A} / \mathrm{A}^{2-}\right)-\frac{R T}{2 F} \ln K_{\mathrm{f}} .
Using Eq. (28) with potentials in mV and T ≈ 298 K T ≈ 298 K T~~298KT \approx 298 \mathrm{~K} :
使用方程 （28），电位单位为 mV 和 T ≈ 298 K T ≈ 298 K T~~298KT \approx 298 \mathrm{~K} ：
E ∘ ( A / A ⋅ − ) ≈ E ∘ ( A , 2 H + / AH 2 ) − 29.6 ( p K r 1 + p K r 2 + p K f ) E ∘ ( A ⋅ − / A 2 − ) ≈ E ∘ ( A , 2 H + / AH 2 ) − 29.6 ( p K r 1 + p K r 2 − p K f ) E ∘ ( A / A ⋅ − ) ≈ E ∘ A , 2 H + / AH 2 − 29.6 p K r 1 + p K r 2 + p K f E ∘ A ⋅ − / A 2 − ≈ E ∘ A , 2 H + / AH 2 − 29.6 p K r 1 + p K r 2 − p K f {:[E^(@)(A//A*-)~~E^(@)(A,2H^(+)//AH_(2))-29.6(pK_(r1)+pK_(r2)+pK_(f))],[E^(@)(A*-//A^(2-))~~E^(@)(A,2H^(+)//AH_(2))-29.6(pK_(r1)+pK_(r2)-pK_(f))]:}\begin{aligned} & E^{\circ}(\mathrm{A} / \mathrm{A} \cdot-) \approx E^{\circ}\left(\mathrm{A}, 2 \mathrm{H}^{+} / \mathrm{AH}_{2}\right)-29.6\left(\mathrm{p} K_{\mathrm{r} 1}+\mathrm{p} K_{\mathrm{r} 2}+\mathrm{p} K_{\mathrm{f}}\right) \\ & E^{\circ}\left(\mathrm{A} \cdot-/ \mathrm{A}^{2-}\right) \approx E^{\circ}\left(\mathrm{A}, 2 \mathrm{H}^{+} / \mathrm{AH}_{2}\right)-29.6\left(\mathrm{p} K_{\mathrm{r} 1}+\mathrm{p} K_{\mathrm{r} 2}-\mathrm{p} K_{\mathrm{f}}\right) \end{aligned}
where K r 1 , K r 2 K r 1 , K r 2 K_(r1),K_(r2)K_{\mathrm{r} 1}, K_{\mathrm{r} 2} are the dissociation constants for AH 2 AH 2 AH_(2)\mathrm{AH}_{2} and AH − AH − AH^(-)\mathrm{AH}^{-}respectively as defined in Eqs. (18) and (19).
其中 K r 1 , K r 2 K r 1 , K r 2 K_(r1),K_(r2)K_{\mathrm{r} 1}, K_{\mathrm{r} 2} 是 AH 2 AH 2 AH_(2)\mathrm{AH}_{2} 和 AH − AH − AH^(-)\mathrm{AH}^{-} 的解离常数，分别在 Eqs 中定义。（18） 和 （19）。
It may be difficult to measure K f K f K_(f)K_{\mathrm{f}} directly, e.g. because very high pH values may be required to ionize completely the reductant to A 2 − A 2 − A^(2-)\mathrm{A}^{2-}. It is much more convenient to define an apparent formation constant, K f i K f i K_(fi)K_{f i} at an experimentally accessible pH , i pH , i pH,i\mathrm{pH}, i :
它可能很难直接测量 K f K f K_(f)K_{\mathrm{f}} ，例如，因为可能需要非常高的 pH 值才能将还原剂完全电离到 A 2 − A 2 − A^(2-)\mathrm{A}^{2-} 。在实验可访问 K f i K f i K_(fi)K_{f i} pH , i pH , i pH,i\mathrm{pH}, i 的 ：
K fi = ( S s ) 2 ( S o ) ( S r ) . K fi = S s 2 S o S r . K_(fi)=((S_(s))^(2))/((S_(o))(S_(r))).K_{\mathrm{fi}}=\frac{\left(S_{\mathrm{s}}\right)^{2}}{\left(S_{\mathrm{o}}\right)\left(S_{\mathrm{r}}\right)} .
We follow previous symbolism and define S 0 S 0 S_(0)S_{0} and S r S r S_(r)S_{\mathrm{r}} as the sums of the oxidant (only A) and reductant ( AH 2 + AH − AH 2 + AH − (AH_(2)+AH^(-):}\left(\mathrm{AH}_{2}+\mathrm{AH}^{-}\right. + A 2 + A 2 +A^(2)+\mathrm{A}^{2} ] respectively, as before, and use S s S s S_(s)S_{\mathrm{s}} to represent the sum of the radical intermediate species. The subscript s is convenient because the radical will be a semiquinone in many examples. It is easily shown, using the approach already used in Sec. 4.3, that:
我们遵循前面的象征主义，像以前一样分别将 和 定义为 S 0 S 0 S_(0)S_{0} 氧化剂（只有 A）和还原剂 ( AH 2 + AH − AH 2 + AH − (AH_(2)+AH^(-):}\left(\mathrm{AH}_{2}+\mathrm{AH}^{-}\right. + A 2 + A 2 +A^(2)+\mathrm{A}^{2} ] 的总和，并用于 S s S s S_(s)S_{\mathrm{s}} 表示自由基中间物种的总 S r S r S_(r)S_{\mathrm{r}} 和。下标 s 很方便，因为在许多例子中，根式将是半醌。使用第 4.3 节中已经使用的方法，很容易证明：
K f = K f i ( K s K s + ( H + ) ) ( K r 1 K r 2 + K r 1 ( H + ) + ( H + ) 2 K r 1 K r 2 ) K f = K f i K s K s + H + K r 1 K r 2 + K r 1 H + + H + 2 K r 1 K r 2 K_(f)=K_(fi)((K_(s))/(K_(s)+(H^(+))))((K_(r1)K_(r2)+K_(r1)(H^(+))+(H^(+))^(2))/(K_(r1)K_(r2)))K_{\mathrm{f}}=K_{\mathrm{f} i}\left(\frac{K_{\mathrm{s}}}{K_{\mathrm{s}}+\left(\mathrm{H}^{+}\right)}\right)\left(\frac{K_{\mathrm{r} 1} K_{\mathrm{r} 2}+K_{\mathrm{r} 1}\left(\mathrm{H}^{+}\right)+\left(\mathrm{H}^{+}\right)^{2}}{K_{\mathrm{r} 1} K_{\mathrm{r} 2}}\right)
where K r 1 , K r 2 K r 1 , K r 2 K_(r1),K_(r2)K_{\mathrm{r} 1}, K_{\mathrm{r} 2} are defined in Eqs. (18) and (19) as before and K s = K 29 K s = K 29 K_(s)=K_(29)K_{\mathrm{s}}=K_{29}.
其中 K r 1 , K r 2 K r 1 , K r 2 K_(r1),K_(r2)K_{\mathrm{r} 1}, K_{\mathrm{r} 2} 定义在 Eqs 中。（18） 和 （19） 如前所述 和 K s = K 29 K s = K 29 K_(s)=K_(29)K_{\mathrm{s}}=K_{29} 。
As noted earlier, in practice, concentrations rather than activities are generally measured. We will usually obtain an estimate of K f K f K_(f)K_{\mathrm{f}} or K fi K fi K_(fi)K_{\mathrm{fi}} at some ionic strength, I I II. Using K f ′ , K f i ′ K f ′ , K f i ′ K_(f)^('),K_(fi)^(')K_{\mathrm{f}}^{\prime}, K_{\mathrm{f} i}^{\prime} as before to denote the apparent formation constants thus defined in concentration terms except for ( H + ) H + (H^(+))\left(\mathrm{H}^{+}\right), together with the mid-point potentials E mi E mi  E_("mi ")E_{\text {mi }} measured at the same ionic strength, it can be shown that:
如前所述，在实践中，通常测量的是浓度而不是活性。我们通常会获得离子强度或 K f K f K_(f)K_{\mathrm{f}} K fi K fi K_(fi)K_{\mathrm{fi}} 某个离子强度 的估计值。 I I II 如前所述，用 K f ′ , K f i ′ K f ′ , K f i ′ K_(f)^('),K_(fi)^(')K_{\mathrm{f}}^{\prime}, K_{\mathrm{f} i}^{\prime} 浓度项定义的表观形成常数，除了 ( H + ) H + (H^(+))\left(\mathrm{H}^{+}\right) 之外，再加上在相同离子强度 E mi E mi  E_("mi ")E_{\text {mi }} 下测得的中点电位，可以证明：
E m i ( A / A ⋅ − ) = E m i ( A / A 2 − ) + R T 2 F ln ⁡ K fi ′ E m i ( A − / A 2 − ) = E m i ( A / A 2 − ) − R T 2 F ln ⁡ K fi ′ E m i ( A / A ⋅ − ) = E m i A / A 2 − + R T 2 F ln ⁡ K fi ′ E m i A − / A 2 − = E m i A / A 2 − − R T 2 F ln ⁡ K fi ′ {:[E_(mi)(A//A*-)=E_(mi)((A)//A^(2-))+(RT)/(2F)ln K_(fi)^(')],[E_(mi)(A^(-)//A^(2-))=E_(mi)((A)//A^(2-))-(RT)/(2F)ln K_(fi)^(')]:}\begin{gathered} E_{\mathrm{m} i}(\mathrm{~A} / \mathrm{A} \cdot-)=E_{\mathrm{m} i}\left(\mathrm{~A} / \mathrm{A}^{2-}\right)+\frac{R T}{2 F} \ln K_{\mathrm{fi}}^{\prime} \\ E_{\mathrm{m} i}\left(\mathrm{~A}^{-} / \mathrm{A}^{2-}\right)=E_{\mathrm{m} i}\left(\mathrm{~A} / \mathrm{A}^{2-}\right)-\frac{R T}{2 F} \ln K_{\mathrm{fi}}^{\prime} \end{gathered}
The mid-point condition now refers to the sum of the concentrations of oxidant being equal to the sum of the concentrations of reductant. (The activity coefficient terms in Eqs. (36) and (51) cancel out the terms in Eq. (69)).
中间点条件现在是指氧化剂浓度之和等于还原剂浓度之和。（方程中的活性系数项。（36） 和 （51） 抵消了方程 （69） 中的项。
K f ′ = K f f A ⋅ 2 ⋅ f A f A 2 − K f ′ = K f f A ⋅ 2 ⋅ f A f A 2 − K_(f)^(')=K_(f)(f_(A*)^(2)*)/(f_(A)f_(A^(2-)))K_{\mathrm{f}}^{\prime}=K_{\mathrm{f}} \frac{f_{\mathrm{A} \cdot}^{2} \cdot}{f_{\mathrm{A}} f_{\mathrm{A}^{2-}}}

5.3. Examples of Calculations
5.3. 计算示例

The one-electron reduction potential of the oxidant, duroquinone (DQ) can be estimated using electrochemical data for the reduction potential of the two-electron couple: duroquinone/durohydroquinone, and spectrophotometric measurement of the semiquinone concentration present in mixtures of the quinone and hydroquinone at high pH . Interpolating Baxendale and Hardy’s data 44 , 45 44 , 45 ^(44,45){ }^{44,45} to yield values at 298 K give: p K r 1 ′ = 11.24 , p K r 2 ′ = 12.83 p K r 1 ′ = 11.24 , p K r 2 ′ = 12.83 pK_(r1)^(')=11.24,pK_(r2)^(')=12.83\mathrm{p} K_{\mathrm{r} 1}^{\prime}=11.24, \mathrm{p} K_{\mathrm{r} 2}^{\prime}=12.83 and p K f ′ = 0.11 p K f ′ = 0.11 pK_(f)^(')=0.11\mathrm{p} K_{\mathrm{f}}^{\prime}=0.11 at I = 0.65 I = 0.65 I=0.65I=0.65. Conant and Fieser 47 47 ^(47){ }^{47} indicate E ∘ ( DQ , 2 H + / DQH 2 ) = 480 mV E ∘ DQ , 2 H + / DQH 2 = 480 mV E^(@)(DQ,2H^(+)//DQH_(2))=480mVE^{\circ}\left(\mathrm{DQ}, 2 \mathrm{H}^{+} / \mathrm{DQH}_{2}\right)=480 \mathrm{mV} (but used 50 % 50 % 50%50 \% ethanol). Equation (63) then yields an estimate of E ∘ ( DQ ′ DQ ∙ ) = − 236 mV E ∘ DQ ′ DQ ∙ = − 236 mV E^(@)(DQ^(')DQ^(∙))=-236mVE^{\circ}\left(\mathrm{DQ}^{\prime} \mathrm{DQ}^{\bullet}\right)=-236 \mathrm{mV}, ignoring the use of practical rather then thermodynamic equilibrium constants. Alternatively, Michaelis et al. 48 48 ^(48){ }^{48} estimated E m i E m i E_(mi)E_{\mathrm{m} i} (duroquinone/durohydroquinone) using 20 % 20 % 20%20 \% pyridine in water at 303 K , for pH ( i ) = 7.4 pH ( i ) = 7.4 pH(i)=7.4\mathrm{pH}(i)=7.4 to 13.5 ; a value of E m 7 = 41 mV E m 7 = 41 mV E_(m7)=41mVE_{\mathrm{m} 7}=41 \mathrm{mV} is interpolated. Baxendale and Hardy’s data, 44 , 45 44 , 45 ^(44,45){ }^{44,45} and p K s ′ = 5.1 p K s ′ = 5.1 pK_(s)^(')=5.1\mathrm{p} K_{\mathrm{s}}^{\prime}=5.1 from pulse radiolysis, 3 3 ^(3){ }^{3} yields K f 7 ′ = 1.1 × 10 − 10 K f 7 ′ = 1.1 × 10 − 10 K_(f7)^(')=1.1 xx10^(-10)K_{\mathrm{f} 7}^{\prime}=1.1 \times 10^{-10}. Using Eq. (67), E 7 ( DQ / DQ ∙ ) = E 7 DQ / DQ ∙ = E_(7)(DQ//DQ^(∙))=E_{7}\left(\mathrm{DQ} / \mathrm{DQ}^{\bullet}\right)= -254 mV is estimated. These values are similar to those obtained quite independently by Wardman and Clarke 32 32 ^(32){ }^{32} using pulse radiolysis.
氧化剂杜醌 （DQ） 的单电子还原电位可以使用双电子对（杜醌/杜溴醌）的还原电位的电化学数据以及高 pH 值下醌和氢醌混合物中存在的半醌浓度的分光光度法测量来估计。将 Baxendale 和 Hardy 的数据 44 , 45 44 , 45 ^(44,45){ }^{44,45} 插值以生成 298 K 的值，得到： p K r 1 ′ = 11.24 , p K r 2 ′ = 12.83 p K r 1 ′ = 11.24 , p K r 2 ′ = 12.83 pK_(r1)^(')=11.24,pK_(r2)^(')=12.83\mathrm{p} K_{\mathrm{r} 1}^{\prime}=11.24, \mathrm{p} K_{\mathrm{r} 2}^{\prime}=12.83 和 p K f ′ = 0.11 p K f ′ = 0.11 pK_(f)^(')=0.11\mathrm{p} K_{\mathrm{f}}^{\prime}=0.11 。 I = 0.65 I = 0.65 I=0.65I=0.65 Conant 和 Fieser 47 47 ^(47){ }^{47} 表示 E ∘ ( DQ , 2 H + / DQH 2 ) = 480 mV E ∘ DQ , 2 H + / DQH 2 = 480 mV E^(@)(DQ,2H^(+)//DQH_(2))=480mVE^{\circ}\left(\mathrm{DQ}, 2 \mathrm{H}^{+} / \mathrm{DQH}_{2}\right)=480 \mathrm{mV} （但使用了 50 % 50 % 50%50 \% 乙醇）。然后，方程 （63） 得出 E ∘ ( DQ ′ DQ ∙ ) = − 236 mV E ∘ DQ ′ DQ ∙ = − 236 mV E^(@)(DQ^(')DQ^(∙))=-236mVE^{\circ}\left(\mathrm{DQ}^{\prime} \mathrm{DQ}^{\bullet}\right)=-236 \mathrm{mV} 的估计值，忽略了实际平衡常数而不是热力学平衡常数的使用。或者，Michaelis 等人 48 48 ^(48){ }^{48} 估计 E m i E m i E_(mi)E_{\mathrm{m} i} （杜罗醌/杜罗氢醌）在 303 K 的水中使用 20 % 20 % 20%20 \% 吡啶，为 pH ( i ) = 7.4 pH ( i ) = 7.4 pH(i)=7.4\mathrm{pH}(i)=7.4 13.5 ;值 of E m 7 = 41 mV E m 7 = 41 mV E_(m7)=41mVE_{\mathrm{m} 7}=41 \mathrm{mV} 是插值的。Baxendale 和 Hardy 的数据以及 44 , 45 44 , 45 ^(44,45){ }^{44,45} p K s ′ = 5.1 p K s ′ = 5.1 pK_(s)^(')=5.1\mathrm{p} K_{\mathrm{s}}^{\prime}=5.1 脉冲放射分解 3 3 ^(3){ }^{3} 得出 K f 7 ′ = 1.1 × 10 − 10 K f 7 ′ = 1.1 × 10 − 10 K_(f7)^(')=1.1 xx10^(-10)K_{\mathrm{f} 7}^{\prime}=1.1 \times 10^{-10} 。使用方程 （67）， E 7 ( DQ / DQ ∙ ) = E 7 DQ / DQ ∙ = E_(7)(DQ//DQ^(∙))=E_{7}\left(\mathrm{DQ} / \mathrm{DQ}^{\bullet}\right)= 估计为 -254 mV。这些值与 Wardman 和 Clarke 32 32 ^(32){ }^{32} 使用脉冲放射裂解获得的非常独立的值相似。
(A number of authors have used p K r 2 ′ = 13.2 p K r 2 ′ = 13.2 pK_(r2)^(')=13.2\mathrm{p} K_{\mathrm{r} 2}^{\prime}=13.2 for duroquinone, as tabulated from Bishop and Tong 46 46 ^(46){ }^{46} from Baxendale and Hardy’s measurements. The original data 44 44 ^(44){ }^{44} clearly show p K r 2 ′ p K r 2 ′ pK_(r2)^(')\mathrm{p} K_{\mathrm{r} 2}^{\prime} varying between 13.17 at 14.9 ∘ C 14.9 ∘ C 14.9^(@)C14.9^{\circ} \mathrm{C} to 12.70 at 29.8 ∘ C 29.8 ∘ C 29.8^(@)C29.8{ }^{\circ} \mathrm{C}, from which the present author interpolates a value of 12.83 at 298 K ).
（许多作者都使用 p K r 2 ′ = 13.2 p K r 2 ′ = 13.2 pK_(r2)^(')=13.2\mathrm{p} K_{\mathrm{r} 2}^{\prime}=13.2 了杜罗醌，如 Baxendale 和 Hardy 的测量中的 Bishop 和 Tong 46 46 ^(46){ }^{46} 所列。原始数据 44 44 ^(44){ }^{44} 清楚地显示在 p K r 2 ′ p K r 2 ′ pK_(r2)^(')\mathrm{p} K_{\mathrm{r} 2}^{\prime} 13.17 at 14.9 ∘ C 14.9 ∘ C 14.9^(@)C14.9^{\circ} \mathrm{C} 到 12.70 at 29.8 ∘ C 29.8 ∘ C 29.8^(@)C29.8{ }^{\circ} \mathrm{C} 之间变化，本文作者从 298 K 处插入值 12.83 ）。
Electron spin resonance measurements 49 49 ^(49){ }^{49} of the steady-state concentrations of ascorbyl radicals produced on mixing the reductant, ascorbic acid with the corresponding oxidant, dehydroascorbic acid gave estimates of K fi K fi  K_("fi ")K_{\text {fi }} between pH 4.0 and 6.4. An estimate of K f = 1.2 × 10 − 3 K f = 1.2 × 10 − 3 K_(f)=1.2 xx10^(-3)K_{\mathrm{f}}=1.2 \times 10^{-3} is obtained using Eq. (66) and p K r 1 ′ = 4.21 , p K r 2 ′ = 11.52 p K r 1 ′ = 4.21 , p K r 2 ′ = 11.52 pK_(r1)^(')=4.21,pK_(r2)^(')=11.52\mathrm{p} K_{\mathrm{r} 1}^{\prime}=4.21, \mathrm{p} K_{\mathrm{r} 2}^{\prime}=11.52 (representative literature values) and p K s = − 0.45 . 50 A p K s = − 0.45 . 50 A pK_(s)=-0.45.^(50)A\mathrm{p} K_{\mathrm{s}}=-0.45 .{ }^{50} \mathrm{~A} value of E m 0 = 400 mV E m 0 = 400 mV E_(m0)=400mVE_{\mathrm{m} 0}=400 \mathrm{mV} for the two-electron reduction (see Clark, 22 22 ^(22){ }^{22} p.470), will be close to E ∘ ( A , 2 H + / AH 2 ) E ∘ A , 2 H + / AH 2 E^(@)(A,2H^(+)//AH_(2))E^{\circ}\left(\mathrm{A}, 2 \mathrm{H}^{+} / \mathrm{AH}_{2}\right), from Eq. (24). Eq. (64) yields E ∘ ( A − / A 2 ) ≈ 19 mV E ∘ A − / A 2 ≈ 19 mV E^(@)(A^(-)//A^(2))~~19mVE^{\circ}\left(\mathrm{A}^{-} / \mathrm{A}^{2}\right) \approx 19 \mathrm{mV} for A 2 − = A 2 − = A^(2-)=\mathrm{A}^{2-}= ascorbic acid. Steenken and Neta, 8 8 ^(8){ }^{8} using the pulse radiolysis redox equilibrium method, estimated E 13.5 ( A ∙ / A 2 ) = E 13.5 A ∙ / A 2 = E_(13.5)(A^(∙)//A^(2))=E_{13.5}\left(\mathrm{~A}^{\bullet} / \mathrm{A}^{2}\right)= 15 mV . This is well within the uncertainty of the independent calculation. (Because p K r 2 ′ ≈ 11.5 , E 13.5 ( A − / A 2 ) ≈ p K r 2 ′ ≈ 11.5 , E 13.5 A − / A 2 ≈ pK_(r2)^(')~~11.5,E_(13.5)(A^(-)//A^(2))~~\mathrm{p} K_{\mathrm{r} 2}^{\prime} \approx 11.5, E_{13.5}\left(\mathrm{~A}^{-} / \mathrm{A}^{2}\right) \approx E ∘ ( A − / A 2 ) E ∘ A − / A 2 E^(@)(A^(-)//A^(2))E^{\circ}\left(\mathrm{A}^{-} / \mathrm{A}^{2}\right) ).
将还原剂抗坏血酸与相应的氧化剂脱氢抗坏血酸混合时产生的抗坏血酸自由基稳态浓度的电子自旋共振测量 49 49 ^(49){ }^{49} 得出的 pH 值 K fi K fi  K_("fi ")K_{\text {fi }} 在 4.0 和 6.4 之间。使用方程（66）和 p K r 1 ′ = 4.21 , p K r 2 ′ = 11.52 p K r 1 ′ = 4.21 , p K r 2 ′ = 11.52 pK_(r1)^(')=4.21,pK_(r2)^(')=11.52\mathrm{p} K_{\mathrm{r} 1}^{\prime}=4.21, \mathrm{p} K_{\mathrm{r} 2}^{\prime}=11.52 （代表性文献值） p K s = − 0.45 . 50 A p K s = − 0.45 . 50 A pK_(s)=-0.45.^(50)A\mathrm{p} K_{\mathrm{s}}=-0.45 .{ }^{50} \mathrm{~A} 获得 的 K f = 1.2 × 10 − 3 K f = 1.2 × 10 − 3 K_(f)=1.2 xx10^(-3)K_{\mathrm{f}}=1.2 \times 10^{-3} 估计值，双电子还原的 值 E m 0 = 400 mV E m 0 = 400 mV E_(m0)=400mVE_{\mathrm{m} 0}=400 \mathrm{mV} （参见 Clark， 22 22 ^(22){ }^{22} 第 470 页）将接近方程（24）中的 E ∘ ( A , 2 H + / AH 2 ) E ∘ A , 2 H + / AH 2 E^(@)(A,2H^(+)//AH_(2))E^{\circ}\left(\mathrm{A}, 2 \mathrm{H}^{+} / \mathrm{AH}_{2}\right) 。方程 （64） 抗 E ∘ ( A − / A 2 ) ≈ 19 mV E ∘ A − / A 2 ≈ 19 mV E^(@)(A^(-)//A^(2))~~19mVE^{\circ}\left(\mathrm{A}^{-} / \mathrm{A}^{2}\right) \approx 19 \mathrm{mV} 坏血酸的 A 2 − = A 2 − = A^(2-)=\mathrm{A}^{2-}= 产量。Steenken 和 Neta 8 8 ^(8){ }^{8} 使用脉冲辐射分解氧化还原平衡方法，估计 E 13.5 ( A ∙ / A 2 ) = E 13.5 A ∙ / A 2 = E_(13.5)(A^(∙)//A^(2))=E_{13.5}\left(\mathrm{~A}^{\bullet} / \mathrm{A}^{2}\right)= 为 15 mV。这完全在独立计算的不确定性范围内。（因为 p K r 2 ′ ≈ 11.5 , E 13.5 ( A − / A 2 ) ≈ p K r 2 ′ ≈ 11.5 , E 13.5 A − / A 2 ≈ pK_(r2)^(')~~11.5,E_(13.5)(A^(-)//A^(2))~~\mathrm{p} K_{\mathrm{r} 2}^{\prime} \approx 11.5, E_{13.5}\left(\mathrm{~A}^{-} / \mathrm{A}^{2}\right) \approx E ∘ ( A − / A 2 ) E ∘ A − / A 2 E^(@)(A^(-)//A^(2))E^{\circ}\left(\mathrm{A}^{-} / \mathrm{A}^{2}\right) ）。

5.4. Uncertainties in the Calculations
5.4. 计算中的不确定性

As an example, consider the calculation for E ∘ ( A / A ∙ ) E ∘ A / A ∙ E^(@)(A//A^(∙))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right) for A = A = A=\mathrm{A}= simple quinones. Clark’s tables 22 22 ^(22){ }^{22} of values of E 0 E 0 E_(0)E_{0} for the two-electron reduction of many quinones indicate random uncertainties of 5 − 15 mV 5 − 15 mV 5-15mV5-15 \mathrm{mV}, the higher values including measurements in partly non-aqueous solvents. In these cases E 0 E 0 E_(0)E_{0} approximates to E ∘ ( A , 2 H + / AH 2 ) E ∘ A , 2 H + / AH 2 E^(@)(A,2H^(+)//AH_(2))E^{\circ}\left(\mathrm{A}, 2 \mathrm{H}^{+} / \mathrm{AH}_{2}\right). To calculate the uncertainty in the estimate of E ∘ ( A / A ∙ ) E ∘ A / A ∙ E^(@)(A//A^(∙))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right), for example, we also need to consider the uncertainty in the
例如，考虑简单醌的计算 E ∘ ( A / A ∙ ) E ∘ A / A ∙ E^(@)(A//A^(∙))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right) A = A = A=\mathrm{A}= 。许多醌的双电子还原的 的 Clark 值 E 0 E 0 E_(0)E_{0} 表 22 22 ^(22){ }^{22} 表明 5 − 15 mV 5 − 15 mV 5-15mV5-15 \mathrm{mV} 的随机不确定性，较高的值包括在部分非水性溶剂中的测量值。在这些情况下， E 0 E 0 E_(0)E_{0} 近似值为 E ∘ ( A , 2 H + / AH 2 ) E ∘ A , 2 H + / AH 2 E^(@)(A,2H^(+)//AH_(2))E^{\circ}\left(\mathrm{A}, 2 \mathrm{H}^{+} / \mathrm{AH}_{2}\right) 。例如，要计算 估计 E ∘ ( A / A ∙ ) E ∘ A / A ∙ E^(@)(A//A^(∙))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right) 中的不确定性，我们还需要考虑 中的不确定性
sum: p K r 1 + p K r 2 + p K f p K r 1 + p K r 2 + p K f pK_(r1)+pK_(r2)+pK_(f)\mathrm{p} K_{\mathrm{r} 1}+\mathrm{p} K_{\mathrm{r} 2}+\mathrm{p} K_{\mathrm{f}}, as indicated in Eq. (63). Estimates 44 − 46 44 − 46 ^(44-46){ }^{44-46} of p K r 1 ′ , p K r 2 ′ p K r 1 ′ , p K r 2 ′ pK_(r1)^('),pK_(r2)^(')\mathrm{p} K_{\mathrm{r} 1}^{\prime}, \mathrm{p} K_{\mathrm{r} 2}^{\prime} and p K f ′ p K f ′ pK_(f)^(')\mathrm{p} K_{\mathrm{f}}^{\prime} refer to ionic strengths of 0.65 or 0.375 , and the substitution of these practical constants for the thermodynamic constants required in Eq. (63) introduces systematic errors.
sum： p K r 1 + p K r 2 + p K f p K r 1 + p K r 2 + p K f pK_(r1)+pK_(r2)+pK_(f)\mathrm{p} K_{\mathrm{r} 1}+\mathrm{p} K_{\mathrm{r} 2}+\mathrm{p} K_{\mathrm{f}} ，如方程 （63） 所示。离子强度的估计 44 − 46 44 − 46 ^(44-46){ }^{44-46} 值 p K r 1 ′ , p K r 2 ′ p K r 1 ′ , p K r 2 ′ pK_(r1)^('),pK_(r2)^(')\mathrm{p} K_{\mathrm{r} 1}^{\prime}, \mathrm{p} K_{\mathrm{r} 2}^{\prime} 和 p K f ′ p K f ′ pK_(f)^(')\mathrm{p} K_{\mathrm{f}}^{\prime} 指离子强度为 0.65 或 0.375 ，用这些实际常数代替方程（63）中要求的热力学常数会引入系统误差。
Perrin et al. 51 51 ^(51){ }^{51} derived a formula to correct practical ionization constants. For dissociation of the weak acid HA ( n − 1 ) − HA ( n − 1 ) − HA^((n-1)-)\mathrm{HA}^{(\mathrm{n}-1)-} :
Perrin 等人 51 51 ^(51){ }^{51} 得出了一个公式来校正实际电离常数。对于弱酸 HA ( n − 1 ) − HA ( n − 1 ) − HA^((n-1)-)\mathrm{HA}^{(\mathrm{n}-1)-} 的解离：
35
HA ( n − 1 ) − ⇄ A n − + H + pK ≈ p K ′ + [ ( 2 n − 1 ) / 2 ] f ( I ) HA ( n − 1 ) − ⇄ A n − + H + pK ≈ p K ′ + [ ( 2 n − 1 ) / 2 ] f ( I ) {:[HA^((n-1)-)⇄A^(n-)+H^(+)],[pK~~pK^(')+[(2n-1)//2]f(I)]:}\begin{aligned} \mathrm{HA}^{(\mathrm{n}-1)-} & \rightleftarrows \mathrm{A}^{\mathrm{n}-}+\mathrm{H}^{+} \\ \mathrm{pK} & \approx \mathrm{p} K^{\prime}+[(2 n-1) / 2] f(I) \end{aligned}
We have adapted his formula to use the ionic strength function, f ( I ) f ( I ) f(I)f(I) previously defined:
我们调整了他的公式以使用 f ( I ) f ( I ) f(I)f(I) 前面定义的离子强度函数：
f ( I ) ≈ 1.02 ( I 1 / 2 [ 1 + I 1 / 2 ] − 1 − 0.2 I ) f ( I ) ≈ 1.02 I 1 / 2 1 + I 1 / 2 − 1 − 0.2 I f(I)~~1.02(I^(1//2)[1+I^(1//2)]^(-1)-0.2 I)f(I) \approx 1.02\left(I^{1 / 2}\left[1+I^{1 / 2}\right]^{-1}-0.2 I\right)
At high ionic strengths, I ≈ 0.4 − 0.6 I ≈ 0.4 − 0.6 I~~0.4-0.6I \approx 0.4-0.6, reliable use of Eq. (70) is doubtful. However, we see that for uncharged quinones (e.g. duroquinone), p K r 1 ′ p K r 1 ′ pK_(r1)^(')\mathrm{p} K_{\mathrm{r} 1}^{\prime} and p K r 2 ′ p K r 2 ′ pK_(r2)^(')\mathrm{p} K_{\mathrm{r} 2}^{\prime} may underestimate the thermodynamic values by ca. 0.1-0.2 and 0.5 respectively. It can be shown that
在高离子强度下， I ≈ 0.4 − 0.6 I ≈ 0.4 − 0.6 I~~0.4-0.6I \approx 0.4-0.6 方程 （70） 的可靠使用是值得怀疑的。然而，我们看到对于不带电荷的醌（例如杜醌）， p K r 1 ′ p K r 1 ′ pK_(r1)^(')\mathrm{p} K_{\mathrm{r} 1}^{\prime} p K r 2 ′ p K r 2 ′ pK_(r2)^(')\mathrm{p} K_{\mathrm{r} 2}^{\prime} 并且可能分别低估了热力学值约 0.1-0.2 和 0.5。可以证明
p K f ≈ p K f ′ − f ( I ) p K f ≈ p K f ′ − f ( I ) pK_(f)~~pK_(f)^(')-f(I)\mathrm{p} K_{\mathrm{f}} \approx \mathrm{p} K_{\mathrm{f}}^{\prime}-f(I)
for uncharged oxidants A , i.e. for uncharged quinones. The semiquinone formation constant decreases with increasing I I II so that p K f ′ p K f ′ pK_(f)^(')\mathrm{p} K_{\mathrm{f}}^{\prime} overestimates p K f p K f pK_(f)\mathrm{p} K_{\mathrm{f}} by c a .0 .3 c a .0 .3 ca.0.3c a .0 .3 at I ≈ 0.4 − 0.6 I ≈ 0.4 − 0.6 I~~0.4-0.6I \approx 0.4-0.6. There is thus partial canceling-out of these systematic errors in the application of Eq. (63). The systematic error introduced into the calculation of E ∘ ( A / A ∙ ) E ∘ A / A ∙ E^(@)(A//A^(∙))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right) will still amount to the estimate being ca. 10 mV more positive than the true value.
对于不带电荷的氧化剂 A ，即对于不带电荷的醌。半醌形成常数随着增加 I I II 而减小， p K f ′ p K f ′ pK_(f)^(')\mathrm{p} K_{\mathrm{f}}^{\prime} 因此高估 p K f p K f pK_(f)\mathrm{p} K_{\mathrm{f}} c a .0 .3 c a .0 .3 ca.0.3c a .0 .3 了 。 I ≈ 0.4 − 0.6 I ≈ 0.4 − 0.6 I~~0.4-0.6I \approx 0.4-0.6 因此，在方程 （63） 的应用中，这些系统性错误被部分抵消了。计算中引入的系统误差仍 E ∘ ( A / A ∙ ) E ∘ A / A ∙ E^(@)(A//A^(∙))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right) 将导致估计值比真实值高出约 10 mV。
Even for these simple quinones, generally only one estimate 44 − 46 44 − 46 ^(44-46){ }^{44-46} of the ionization and formation constants required is available. Even discounting random errors in their determination, the calculations of one-electron reduction potential as described in this section must involve uncertainties of at least 10 − 20 mV 10 − 20 mV 10-20mV10-20 \mathrm{mV} is general. Similar consideration may be given to other applications of the formulae derived.
即使对于这些简单的醌，通常也只有一个所需的电离和形成常数的估计 44 − 46 44 − 46 ^(44-46){ }^{44-46} 值可用。即使不考虑测定中的随机误差，本节中描述的单电子还原势的计算也必须涉及至少 10 − 20 mV 10 − 20 mV 10-20mV10-20 \mathrm{mV} 是一般性的不确定性。对所推导的公式的其他应用也可以给予类似的考虑。
These illustrations may be used, in turn, to refine calculations of standard potentials using experimental measurements of ionization and formation constants. Thus the literature data 44 , 46 44 , 46 ^(44,46){ }^{44,46} for 1,4-benzoquinone may be corrected to yield estimates of the thermodynamic constants: p K r 1 , p K r 2 , p K s p K r 1 , p K r 2 , p K s pK_(r1),pK_(r2),pK_(s)\mathrm{p} K_{\mathrm{r} 1}, \mathrm{p} K_{\mathrm{r} 2}, \mathrm{p} K_{\mathrm{s}} and p K f p K f pK_(f)\mathrm{p} K_{\mathrm{f}} of 10.0 , 11.9 , 4.0 10.0 , 11.9 , 4.0 10.0,11.9,4.010.0,11.9,4.0 and -0.92 respectively. Using the well-established 22 E ∘ ( Q , 2 H + / QH 2 ) = 699 mV 22 E ∘ Q , 2 H + / QH 2 = 699 mV ^(22)E^(@)(Q,2H^(+)//QH_(2))=699mV{ }^{22} E^{\circ}\left(\mathrm{Q}, 2 \mathrm{H}^{+} / \mathrm{QH}_{2}\right)=699 \mathrm{mV} yields estimates of E ∘ ( Q , Q ∙ ) = 78 mV E ∘ Q , Q ∙ = 78 mV E^(@)(Q,Q^(∙))=78mVE^{\circ}\left(\mathrm{Q}, \mathrm{Q}^{\bullet}\right)=78 \mathrm{mV} and E ∘ ( Q ⋅ − , Q 2 − ) = 24 mV E ∘ Q ⋅ − , Q 2 − = 24 mV E^(@)(Q^(*-),Q^(2-))=24mVE^{\circ}\left(\mathrm{Q}^{\cdot-}, \mathrm{Q}^{2-}\right)=24 \mathrm{mV}, the former some 20 mV lower than previous estimates. 42 42 ^(42){ }^{42} In fact, such corrections are not so straight-forward, since Baxendale and Hardy 44 44 ^(44){ }^{44} included some activity coefficients (of the buffers used) in defining K r 1 ′ , K r 2 ′ K r 1 ′ , K r 2 ′ K_(r1)^('),K_(r2)^(')K_{\mathrm{r} 1}^{\prime}, K_{\mathrm{r} 2}^{\prime}. The simple application of Eqs. (70) or (71) may be inappropriate in some instances.
反过来，这些插图可用于使用电离和形成常数的实验测量来改进标准电位的计算。因此，可以更正 1,4-苯醌的文献数据 44 , 46 44 , 46 ^(44,46){ }^{44,46} 以产生热力学常数的估计值： p K r 1 , p K r 2 , p K s p K r 1 , p K r 2 , p K s pK_(r1),pK_(r2),pK_(s)\mathrm{p} K_{\mathrm{r} 1}, \mathrm{p} K_{\mathrm{r} 2}, \mathrm{p} K_{\mathrm{s}} 分别为 of p K f p K f pK_(f)\mathrm{p} K_{\mathrm{f}} 10.0 , 11.9 , 4.0 10.0 , 11.9 , 4.0 10.0,11.9,4.010.0,11.9,4.0 和 -0.92。使用公认的 22 E ∘ ( Q , 2 H + / QH 2 ) = 699 mV 22 E ∘ Q , 2 H + / QH 2 = 699 mV ^(22)E^(@)(Q,2H^(+)//QH_(2))=699mV{ }^{22} E^{\circ}\left(\mathrm{Q}, 2 \mathrm{H}^{+} / \mathrm{QH}_{2}\right)=699 \mathrm{mV} E ∘ ( Q , Q ∙ ) = 78 mV E ∘ Q , Q ∙ = 78 mV E^(@)(Q,Q^(∙))=78mVE^{\circ}\left(\mathrm{Q}, \mathrm{Q}^{\bullet}\right)=78 \mathrm{mV} 和 E ∘ ( Q ⋅ − , Q 2 − ) = 24 mV E ∘ Q ⋅ − , Q 2 − = 24 mV E^(@)(Q^(*-),Q^(2-))=24mVE^{\circ}\left(\mathrm{Q}^{\cdot-}, \mathrm{Q}^{2-}\right)=24 \mathrm{mV} 的产量估计值，前者比以前的估计值低约 20 mV。 42 42 ^(42){ }^{42} 事实上，这样的更正并不是那么简单，因为 Baxendale 和 Hardy 44 44 ^(44){ }^{44} 在定义 K r 1 ′ , K r 2 ′ K r 1 ′ , K r 2 ′ K_(r1)^('),K_(r2)^(')K_{\mathrm{r} 1}^{\prime}, K_{\mathrm{r} 2}^{\prime} 时包括了一些活动系数（使用的缓冲区）。Eqs 的简单应用。（70） 或 （71） 在某些情况下可能不合适。
The choice of redox indicators B with which to establish and measure the position of the desired equilibrium 1 with the unknown A is influenced by several factors. Ideally, determinations of K 1 K 1 K_(1)K_{1} with two indicators - one higher than the unknown by (say) 50-100 mV , one lower - will lead to the most reliable value. In practice, the choice depends on solubilities, absorption spectra of reactants and products, p K a p K a pK_(a)\mathrm{p} K_{\mathrm{a}} 's, kinetic constraints, (especially the need for fast electron transfer, see above, Sec. 3.5) and ready availability with adequate purity.
选择氧化还原指示剂 B 来建立和测量所需平衡 1 与未知物 A 的位置受几个因素的影响。理想情况下，使用两个指标进行 K 1 K 1 K_(1)K_{1} 测定 - 一个比未知值高（比如）50-100 mV ，一个低 - 将导致最可靠的值。在实践中，选择取决于溶解度、反应物和产物的吸收光谱、 p K a p K a pK_(a)\mathrm{p} K_{\mathrm{a}} 动力学约束（尤其是需要快速电子转移，见上文第 3.5 节）以及具有足够纯度的现成可用性。

6.1. Oxygen  6.1. 氧气

Oxygen is an important reactant with many radicals, although electron-transfer rather than radical-addition is a pre-requisite and it is somewhat inconvenient to vary the concentration of oxygen over a wide range. It is useful to draw attention again to the standard definition: E ∘ ( O 2 ( 1 atm . ) / O 2 ∙ ) = − 325 mV E ∘ O 2 ( 1 atm . ) / O 2 ∙ = − 325 mV E^(@)(O_(2)(1(atm).)//O_(2)^(∙))=-325mVE^{\circ}\left(\mathrm{O}_{2}(1 \mathrm{~atm}.) / \mathrm{O}_{2}{ }^{\bullet}\right)=-325 \mathrm{mV} whereas E ( O 2 ( 1 mol dm − 3 ) / O 2 − ) E O 2 1 mol dm − 3 / O 2 − E(O_(2)(1(mol)dm^(-3))//O_(2)^(-))E\left(\mathrm{O}_{2}\left(1 \mathrm{~mol} \mathrm{dm}^{-3}\right) / \mathrm{O}_{2}{ }^{-}\right) = − 155 mV = − 155 mV =-155mV=-155 \mathrm{mV}.
氧是具有许多自由基的重要反应物，尽管电子转移而不是自由基加成是先决条件，并且在较大范围内改变氧浓度有些不方便。再次提请注意标准定义是有用的： E ∘ ( O 2 ( 1 atm . ) / O 2 ∙ ) = − 325 mV E ∘ O 2 ( 1 atm . ) / O 2 ∙ = − 325 mV E^(@)(O_(2)(1(atm).)//O_(2)^(∙))=-325mVE^{\circ}\left(\mathrm{O}_{2}(1 \mathrm{~atm}.) / \mathrm{O}_{2}{ }^{\bullet}\right)=-325 \mathrm{mV} whereas E ( O 2 ( 1 mol dm − 3 ) / O 2 − ) E O 2 1 mol dm − 3 / O 2 − E(O_(2)(1(mol)dm^(-3))//O_(2)^(-))E\left(\mathrm{O}_{2}\left(1 \mathrm{~mol} \mathrm{dm}^{-3}\right) / \mathrm{O}_{2}{ }^{-}\right) = − 155 mV = − 155 mV =-155mV=-155 \mathrm{mV} .

6.2. Quinones  6.2. 醌

Reduction potentials for the couples A / A − − A / A − − A//A^(--)A / A^{--}and A − / A 2 − A − / A 2 − A^(-)//A^(2-)\mathrm{A}^{-} / \mathrm{A}^{2-} for A = A = A=\mathrm{A}= quinones may be calculated 4 , 5 , 6 , 42 4 , 5 , 6 , 42 ^(4,5,6,42){ }^{4,5,6,42} from the ionization constants of AH 2 AH 2 AH_(2)\mathrm{AH}_{2} and the semiquinone formation constants, as described above (Sec. 5). Completely independent estimates of E ∘ ( A / A ⋅ ) E ∘ A / A ⋅ E^(@)(A//A^(*))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\cdot}\right) for A = A = A=\mathrm{A}= duroquinone are provided by the measurements of Δ E 1 Δ E 1 DeltaE_(1)\Delta E_{1} corrected to I = I = I=I= 0 for A = A = A=\mathrm{A}= duroquinone and B = 1 , 1 ′ B = 1 , 1 ′ B=1,1^(')\mathrm{B}=1,1^{\prime}-dibenzyl-4, 4 ′ 4 ′ 4^(')4^{\prime}-bipyridinium dication. 32 32 ^(32){ }^{32} Values of Δ E 1 Δ E 1 DeltaE_(1)\Delta E_{1} of 110 ± 4 32 , 113 ± 4 52 110 ± 4 32 , 113 ± 4 52 110+-4^(32),113+-4^(52)110 \pm 4^{32}, 113 \pm 4^{52}, and 107 ± 3 53 mV 107 ± 3 53 mV 107+-3^(53)mV107 \pm 3^{53} \mathrm{mV} together with E ∘ ( B / B ∙ ) = − 354 mV E ∘ B / B ∙ = − 354 mV E^(@)(B//B^(∙))=-354mVE^{\circ}\left(\mathrm{B} / \mathrm{B}^{\bullet}\right)=-354 \mathrm{mV} (but see below, Sec. 6.3) yield E ∘ ( A / A ∙ ) = − 244 mV E ∘ A / A ∙ = − 244 mV E^(@)(A//A^(∙))=-244mVE^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right)=-244 \mathrm{mV} for
如上所述，对 A / A − − A / A − − A//A^(--)A / A^{--} 和 A − / A 2 − A − / A 2 − A^(-)//A^(2-)\mathrm{A}^{-} / \mathrm{A}^{2-} A = A = A=\mathrm{A}= 醌的还原电位可以根据电离常数 AH 2 AH 2 AH_(2)\mathrm{AH}_{2} 和半醌形成常数计算 4 , 5 , 6 , 42 4 , 5 , 6 , 42 ^(4,5,6,42){ }^{4,5,6,42} （第 5 节）。杜罗醌 E ∘ ( A / A ⋅ ) E ∘ A / A ⋅ E^(@)(A//A^(*))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\cdot}\right) 的 A = A = A=\mathrm{A}= 完全独立估计值由 Δ E 1 Δ E 1 DeltaE_(1)\Delta E_{1} 杜罗醌和 B = 1 , 1 ′ B = 1 , 1 ′ B=1,1^(')\mathrm{B}=1,1^{\prime} -二苄基-4， 4 ′ 4 ′ 4^(')4^{\prime} -联吡啶指示校正为 I = I = I=I= 0 A = A = A=\mathrm{A}= 的测量值提供。 32 32 ^(32){ }^{32} Δ E 1 Δ E 1 DeltaE_(1)\Delta E_{1} 的值 110 ± 4 32 , 113 ± 4 52 110 ± 4 32 , 113 ± 4 52 110+-4^(32),113+-4^(52)110 \pm 4^{32}, 113 \pm 4^{52} ，并且与 E ∘ ( B / B ∙ ) = − 354 mV E ∘ B / B ∙ = − 354 mV E^(@)(B//B^(∙))=-354mVE^{\circ}\left(\mathrm{B} / \mathrm{B}^{\bullet}\right)=-354 \mathrm{mV} （但见下文，第 6.3 节） 107 ± 3 53 mV 107 ± 3 53 mV 107+-3^(53)mV107 \pm 3^{53} \mathrm{mV} 一起产生 E ∘ ( A / A ∙ ) = − 244 mV E ∘ A / A ∙ = − 244 mV E^(@)(A//A^(∙))=-244mVE^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right)=-244 \mathrm{mV}