Radicals, e.g. ${\mathrm{A}}^{--}$${\mathrm{A}}^{--}$A^(--)\mathrm{A}^{--}may be present in equilibrium with oxidant, A and reductant, ${\mathrm{A}}^{2-}$${\mathrm{A}}^{2-}$A^(2-)\mathrm{A}^{2-} or their protonated conjugates:
自由基，例如可能与 ${\mathrm{A}}^{--}$${\mathrm{A}}^{--}$A^(--)\mathrm{A}^{--} 氧化剂 A 和还原剂或其 ${\mathrm{A}}^{2-}$${\mathrm{A}}^{2-}$A^(2-)\mathrm{A}^{2-} 质子化共轭物平衡存在：

and a radical formation constant can be defined:
并且可以定义一个自由基形成常数：

The value of $K_f$$K_f$K_(f)K_{f} is obviously a measure of the steady-state concentrations of radicals, ${\mathrm{A}}^{--}$${\mathrm{A}}^{--}$A^(--)\mathrm{A}^{--}obtained on mixing oxidant $A$$A$AA with reductant, ${\mathrm{A}}^{2-}$${\mathrm{A}}^{2-}$A^(2-)\mathrm{A}^{2-}. When experimental conditions result in sufficiently high concentrations of radicals to be measured, estimates of $K_{\mathrm{f}}$$K_{\mathrm{f}}$K_(f)K_{\mathrm{f}} can be used in conjuction with the two-electron potentials, $E^{\circ }\left(\mathrm{A}/{\mathrm{A}}^2\right)$$E^{\circ }\left(\mathrm{A}/{\mathrm{A}}^2\right)$E^(@)(A//A^(2))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{2}\right) or $E^{\circ }(\mathrm{A}$$E^{\circ }(\mathrm{A}$E^(@)(AE^{\circ}(\mathrm{A}, $2{\mathrm{H}}^{+}/{\mathrm{AH}}_2)$$\left.2{\mathrm{H}}^{+}/{\mathrm{AH}}_2\right)${:2H^(+)//AH_(2))\left.2 \mathrm{H}^{+} / \mathrm{AH}_{2}\right) to obtain estimates of the one-electron couples, $E^{\circ }\left(\mathrm{A}/{\mathrm{A}}^{\bullet }\right)$$E^{\circ }\left(\mathrm{A}/{\mathrm{A}}^{\bullet }\right)$E^(@)(A//A^(∙))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right), etc.
的值 $K_f$$K_f$K_(f)K_{f} 显然是自由基稳态浓度的量度， ${\mathrm{A}}^{--}$${\mathrm{A}}^{--}$A^(--)\mathrm{A}^{--} 通过将氧化剂 $A$$A$AA 与还原剂 混合而得到。 ${\mathrm{A}}^{2-}$${\mathrm{A}}^{2-}$A^(2-)\mathrm{A}^{2-} 当实验条件导致需要测量足够高浓度的自由基时，可以将 的 $K_{\mathrm{f}}$$K_{\mathrm{f}}$K_(f)K_{\mathrm{f}} 估计值与双电子电位结合使用， $E^{\circ }\left(\mathrm{A}/{\mathrm{A}}^2\right)$$E^{\circ }\left(\mathrm{A}/{\mathrm{A}}^2\right)$E^(@)(A//A^(2))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{2}\right) 或者 $E^{\circ }(\mathrm{A}$$E^{\circ }(\mathrm{A}$E^(@)(AE^{\circ}(\mathrm{A} $2{\mathrm{H}}^{+}/{\mathrm{AH}}_2)$$\left.2{\mathrm{H}}^{+}/{\mathrm{AH}}_2\right)${:2H^(+)//AH_(2))\left.2 \mathrm{H}^{+} / \mathrm{AH}_{2}\right) 来获得单电子对的估计值， $E^{\circ }\left(\mathrm{A}/{\mathrm{A}}^{\bullet }\right)$$E^{\circ }\left(\mathrm{A}/{\mathrm{A}}^{\bullet }\right)$E^(@)(A//A^(∙))E^{\circ}\left(\mathrm{A} / \mathrm{A}^{\bullet}\right) 等等。

Reaction 34 (above) can be obtained by subtracting $33a$$33a$33 a33 a from 32a:
反应 34（上图）可以通过 $33a$$33a$33 a33 a 从 32a 中减去得到：

32a  32 个

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：

If we add reaction $32a$$32a$32 a32 a to reaction $33a$$33a$33 a33 a we obtain reaction $31a$$31a$31 a31 a. Noting that $n=2$$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 $33a$$33a$33 a33 a 中加入反应 $32a$$32a$32 a32 a ，我们会得到反应 $31a$$31a$31 a31 a 。请注意， $n=2$$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：

(cf. Eq. (6)). Adding Eqs. (59) and (60) yields:
（参见方程 （6））。添加 eqs。（59） 和 （60） 得出：

while subtraction gives:  而 Subtraction 得到：

Using Eq. (28) with potentials in mV and $T\approx 298\mathrm{K}$$T\approx 298\mathrm{K}$T~~298KT \approx 298 \mathrm{~K} :
使用方程 （28），电位单位为 mV 和 $T\approx 298\mathrm{K}$$T\approx 298\mathrm{K}$T~~298KT \approx 298 \mathrm{~K} ：

where $K_{\mathrm{r}1},K_{\mathrm{r}2}$$K_{\mathrm{r}1},K_{\mathrm{r}2}$K_(r1),K_(r2)K_{\mathrm{r} 1}, K_{\mathrm{r} 2} are the dissociation constants for ${\mathrm{AH}}_2$${\mathrm{AH}}_2$AH_(2)\mathrm{AH}_{2} and ${\mathrm{AH}}^{-}$${\mathrm{AH}}^{-}$AH^(-)\mathrm{AH}^{-}respectively as defined in Eqs. (18) and (19).
其中 $K_{\mathrm{r}1},K_{\mathrm{r}2}$$K_{\mathrm{r}1},K_{\mathrm{r}2}$K_(r1),K_(r2)K_{\mathrm{r} 1}, K_{\mathrm{r} 2} 是 ${\mathrm{AH}}_2$${\mathrm{AH}}_2$AH_(2)\mathrm{AH}_{2} 和 ${\mathrm{AH}}^{-}$${\mathrm{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_{\mathrm{f}}$$K_{\mathrm{f}}$K_(f)K_{\mathrm{f}} directly, e.g. because very high pH values may be required to ionize completely the reductant to ${\mathrm{A}}^{2-}$${\mathrm{A}}^{2-}$A^(2-)\mathrm{A}^{2-}. It is much more convenient to define an apparent formation constant, $K_{fi}$$K_{fi}$K_(fi)K_{f i} at an experimentally accessible $\mathrm{pH},i$$\mathrm{pH},i$pH,i\mathrm{pH}, i :
它可能很难直接测量 $K_{\mathrm{f}}$$K_{\mathrm{f}}$K_(f)K_{\mathrm{f}} ，例如，因为可能需要非常高的 pH 值才能将还原剂完全电离到 ${\mathrm{A}}^{2-}$${\mathrm{A}}^{2-}$A^(2-)\mathrm{A}^{2-} 。在实验可访问 $K_{fi}$$K_{fi}$K_(fi)K_{f i} $\mathrm{pH},i$$\mathrm{pH},i$pH,i\mathrm{pH}, i 的 ：

We follow previous symbolism and define $S_0$$S_0$S_(0)S_{0} and $S_{\mathrm{r}}$$S_{\mathrm{r}}$S_(r)S_{\mathrm{r}} as the sums of the oxidant (only A) and reductant $({\mathrm{AH}}_2+{\mathrm{AH}}^{-}$$\left({\mathrm{AH}}_2+{\mathrm{AH}}^{-}\right.$(AH_(2)+AH^(-):}\left(\mathrm{AH}_{2}+\mathrm{AH}^{-}\right. $+{\mathrm{A}}^2$$+{\mathrm{A}}^2$+A^(2)+\mathrm{A}^{2} ] respectively, as before, and use $S_{\mathrm{s}}$$S_{\mathrm{s}}$S_(s)S_{\mathrm{s}} to represent the sum of the radical intermediate species. The subscript s is convenient because the radical will be a semiquinone in many examples. It is easily shown, using the approach already used in Sec. 4.3, that:
我们遵循前面的象征主义，像以前一样分别将 和 定义为 $S_0$$S_0$S_(0)S_{0} 氧化剂（只有 A）和还原剂 $({\mathrm{AH}}_2+{\mathrm{AH}}^{-}$$\left({\mathrm{AH}}_2+{\mathrm{AH}}^{-}\right.$(AH_(2)+AH^(-):}\left(\mathrm{AH}_{2}+\mathrm{AH}^{-}\right. $+{\mathrm{A}}^2$$+{\mathrm{A}}^2$+A^(2)+\mathrm{A}^{2} ] 的总和，并用于 $S_{\mathrm{s}}$$S_{\mathrm{s}}$S_(s)S_{\mathrm{s}} 表示自由基中间物种的总 $S_{\mathrm{r}}$$S_{\mathrm{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：

where $K_{\mathrm{r}1},K_{\mathrm{r}2}$$K_{\mathrm{r}1},K_{\mathrm{r}2}$K_(r1),K_(r2)K_{\mathrm{r} 1}, K_{\mathrm{r} 2} are defined in Eqs. (18) and (19) as before and $K_{\mathrm{s}}=K_{29}$$K_{\mathrm{s}}=K_{29}$K_(s)=K_(29)K_{\mathrm{s}}=K_{29}.
其中 $K_{\mathrm{r}1},K_{\mathrm{r}2}$$K_{\mathrm{r}1},K_{\mathrm{r}2}$K_(r1),K_(r2)K_{\mathrm{r} 1}, K_{\mathrm{r} 2} 定义在 Eqs 中。（18） 和 （19） 如前所述 和 $K_{\mathrm{s}}=K_{29}$$K_{\mathrm{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_{\mathrm{f}}$$K_{\mathrm{f}}$K_(f)K_{\mathrm{f}} or $K_{\mathrm{fi}}$$K_{\mathrm{fi}}$K_(fi)K_{\mathrm{fi}} at some ionic strength, $I$$I$II. Using $K_{\mathrm{f}}^{\prime },K_{\mathrm{f}i}^{\prime }$$K_{\mathrm{f}}^{\prime },K_{\mathrm{f}i}^{\prime }$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 $\left({\mathrm{H}}^{+}\right)$$\left({\mathrm{H}}^{+}\right)$(H^(+))\left(\mathrm{H}^{+}\right), together with the mid-point potentials $E_{\text{mi}}$$E_{\text{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_{\mathrm{f}}$$K_{\mathrm{f}}$K_(f)K_{\mathrm{f}} $K_{\mathrm{fi}}$$K_{\mathrm{fi}}$K_(fi)K_{\mathrm{fi}} 某个离子强度 的估计值。 $I$$I$II 如前所述，用 $K_{\mathrm{f}}^{\prime },K_{\mathrm{f}i}^{\prime }$$K_{\mathrm{f}}^{\prime },K_{\mathrm{f}i}^{\prime }$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 $\left({\mathrm{H}}^{+}\right)$$\left({\mathrm{H}}^{+}\right)$(H^(+))\left(\mathrm{H}^{+}\right) 之外，再加上在相同离子强度 $E_{\text{mi}}$$E_{\text{mi }}$E_("mi ")E_{\text {mi }} 下测得的中点电位，可以证明：

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。

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}$${}^{44,45}$^(44,45){ }^{44,45} to yield values at 298 K give: $\mathrm{p}{K}_{\mathrm{r}1}^{\prime }=11.24,\mathrm{p}{K}_{\mathrm{r}2}^{\prime }=12.83$$\mathrm{p}{K}_{\mathrm{r}1}^{\prime }=11.24,\mathrm{p}{K}_{\mathrm{r}2}^{\prime }=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 $\mathrm{p}{K}_{\mathrm{f}}^{\prime }=0.11$$\mathrm{p}{K}_{\mathrm{f}}^{\prime }=0.11$pK_(f)^(')=0.11\mathrm{p} K_{\mathrm{f}}^{\prime}=0.11 at $I=0.65$$I=0.65$I=0.65I=0.65. Conant and Fieser ${}^{47}$${}^{47}$^(47){ }^{47} indicate $E^{\circ }(\mathrm{DQ},2{\mathrm{H}}^{+}/{\mathrm{DQH}}_2)=480\mathrm{mV}$$E^{\circ }\left(\mathrm{DQ},2{\mathrm{H}}^{+}/{\mathrm{DQH}}_2\right)=480\mathrm{mV}$E^(@)(DQ,2H^(+)//DQH_(2))=480mVE^{\circ}\left(\mathrm{DQ}, 2 \mathrm{H}^{+} / \mathrm{DQH}_{2}\right)=480 \mathrm{mV} (but used $50\mathrm{\%}$$50\mathrm{\%}$50%50 \% ethanol). Equation (63) then yields an estimate of $E^{\circ }\left({\mathrm{DQ}}^{\prime }{\mathrm{DQ}}^{\bullet }\right)=-236\mathrm{mV}$$E^{\circ }\left({\mathrm{DQ}}^{\prime }{\mathrm{DQ}}^{\bullet }\right)=-236\mathrm{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}$${}^{48}$^(48){ }^{48} estimated $E_{\mathrm{m}i}$$E_{\mathrm{m}i}$E_(mi)E_{\mathrm{m} i} (duroquinone/durohydroquinone) using $20\mathrm{\%}$$20\mathrm{\%}$20%20 \% pyridine in water at 303 K , for $\mathrm{pH}(i)=7.4$$\mathrm{pH}(i)=7.4$pH(i)=7.4\mathrm{pH}(i)=7.4 to 13.5 ; a value of $E_{\mathrm{m}7}=41\mathrm{mV}$$E_{\mathrm{m}7}=41\mathrm{mV}$E_(m7)=41mVE_{\mathrm{m} 7}=41 \mathrm{mV} is interpolated. Baxendale and Hardy’s data, ${}^{44,45}$${}^{44,45}$^(44,45){ }^{44,45} and $\mathrm{p}{K}_{\mathrm{s}}^{\prime }=5.1$$\mathrm{p}{K}_{\mathrm{s}}^{\prime }=5.1$pK_(s)^(')=5.1\mathrm{p} K_{\mathrm{s}}^{\prime}=5.1 from pulse radiolysis, ${}^3$${}^3$^(3){ }^{3} yields $K_{\mathrm{f}7}^{\prime }=1.1\times {10}^{-10}$$K_{\mathrm{f}7}^{\prime }=1.1\times {10}^{-10}$K_(f7)^(')=1.1 xx10^(-10)K_{\mathrm{f} 7}^{\prime}=1.1 \times 10^{-10}. Using Eq. (67), $E_7\left(\mathrm{DQ}/{\mathrm{DQ}}^{\bullet }\right)=$$E_7\left(\mathrm{DQ}/{\mathrm{DQ}}^{\bullet }\right)=$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： $\mathrm{p}{K}_{\mathrm{r}1}^{\prime }=11.24,\mathrm{p}{K}_{\mathrm{r}2}^{\prime }=12.83$$\mathrm{p}{K}_{\mathrm{r}1}^{\prime }=11.24,\mathrm{p}{K}_{\mathrm{r}2}^{\prime }=12.83$pK_(r1)^(')=11.24,pK_(r2)^(')=12.83\mathrm{p} K_{\mathrm{r} 1}^{\prime}=11.24, \mathrm{p} K_{\mathrm{r} 2}^{\prime}=12.83 和 $\mathrm{p}{K}_{\mathrm{f}}^{\prime }=0.11$$\mathrm{p}{K}_{\mathrm{f}}^{\prime }=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 ${}^{47}$