The carbon dioxide gas-sensing electrode consists of a gas-permeable membrane, a <code>$0.010\mathrm{mol{\cdot}L^{-1}}$</code> <code>$\mathrm {NaHCO_{3}}$</code> solution, a pH electrode, and an external reference electrode. The external reference electrode is a standard calomel electrode with <code>$\varphi\left(\mathrm{Cl}^{-}|\mathrm{Hg}_{2}\mathrm{Cl}_{2}|\mathrm{Hg}\right)=0.2801\ \mathrm{V}$</code>. Calculate the electrode potential of the galvanic cell when the mass fraction of carbon dioxide in the atmosphere is <code>$611\ \mathrm{ppm\ by\ mass}$</code>. Given: Henry's law constant for <code>$\mathrm{CO}_{2}$</code> is <code>$K_{\mathrm{H}}=3.0\times10^{3}\ \mathrm{kPa{\cdot}L{\cdot}mol^{-1}}$</code>, the equilibrium constant for <code>$\mathrm{CO}_{2}+\mathrm{H}_{2}\mathrm{O}=\mathrm{H}_{2}\mathrm{CO}_{3}$</code> is <code>$K=1.70\times10^{-3}$</code>, and the acid dissociation constants of carbonic acid are <code>$\mathrm{p}K_{\mathrm{a1,2}}=6.35,10.33$</code>.
二氧化碳气敏电极由透气膜、$0.010\mathrm{mol{\cdot}L^{-1}}$的$\mathrm {NaHCO_{3}}$溶液、pH电极和外参比电极组成。外参比电极为标准甘汞电极，$\varphi\left(\mathrm{Cl}^{-}|\mathrm{Hg}_{2}\mathrm{Cl}_{2}|\mathrm{Hg}\right)=0.2801\ \mathrm{V}$。计算大气中二氧化碳质量分数为$611\ \mathrm{ppm\ by\ mass}$时原电池的电极电势。已知：$\mathrm{CO}_{2}$的亨利常数$K_{\mathrm{H}}=3.0\times10^{3}\ \mathrm{kPa{\cdot}L{\cdot}mol^{-1}}$，$\mathrm{CO}_{2}+\mathrm{H}_{2}\mathrm{O}=\mathrm{H}_{2}\mathrm{CO}_{3}$的$K=1.70\times10^{-3}$碳酸的电离常数$\mathrm{p}K_{\mathrm{a1,2}}=6.35,10.33$。
$0.254\ \mathrm{V}$
$0.397\ \mathrm{V}$
$0.543\ \mathrm{V}$
$0.689\ \mathrm{V}$
$0.731\ \mathrm{V}$
$0.826\ \mathrm{V}$
$1.031\ \mathrm{V}$
$1.102\ \mathrm{V}$
$0.254\ \mathrm{V}$
$0.397\ \mathrm{V}$
$0.543\ \mathrm{V}$
$0.689\ \mathrm{V}$
$0.731\ \mathrm{V}$
$0.826\ \mathrm{V}$
$1.031\ \mathrm{V}$
$1.102\ \mathrm{V}$
From $\omega_\mathrm{CO_2}=611\ \mathrm{ppm\ by\ mass}$, i.e., $\omega_\mathrm{CO_2}= 6.11\times10^{-4}$, and the molar mass and standard atmospheric pressure data $M_\mathrm{CO_2}=44\ \mathrm{g{\cdot}mol^{-1}},\bar{M}_\mathrm{air}=29\ \mathrm{g{\cdot}mol^{-1}}, 1\ \mathrm{atm}=101.325\ \mathrm{kPa},$ we calculate: $p_\mathrm{CO_2}=\omega_\mathrm{CO_2}\times\frac{\bar{M}_\mathrm{air}}{M_\mathrm{CO_2}}\times1\ \mathrm{atm}=0.0403\ \mathrm{kPa}.$

<Checkpoint>[$p_\mathrm{CO_2}=0.0403\ \mathrm{kPa}$](1)</Checkpoint>

From Henry's law, the concentration of $\mathrm{CO_2}$ at dissolution equilibrium in the solution is $\left[\mathrm{CO_2}\right]=\frac{p_\mathrm{CO_2}}{K_\mathrm{H}}=1.36\times10^{-5}\ \mathrm{mol{\cdot}L^{-1}}$.

<Checkpoint>[$\left[\mathrm{CO_2}\right]=1.36\times10^{-5}\ \mathrm{mol{\cdot}L^{-1}}$](1)</Checkpoint>

In this system, $\left[\mathrm{HCO}_3^-\right]$ is much larger than $\left[\mathrm{H_2CO_3}\right]$, so it can be approximately considered that $\left[\mathrm{HCO_3^-}\right]=0.010\ \mathrm{mol{\cdot}L^{-1}}$ remains unchanged.

<Checkpoint>[$\left[\mathrm{HCO_3^-}\right]=0.010\ \mathrm{mol{\cdot}L^{-1}}$ remains unchanged](1)</Checkpoint>

From ionization equilibrium, $\left[\mathrm{H^+}\right]=\frac{K_\mathrm{a1}\left[\mathrm{CO_2}\right]}{\left[\mathrm{HCO_3^-}\right]}=\frac{10^{-6.35}\times1.36\times10^{-5}}{0.010}=6.07\times10^{-10}\ \mathrm{mol{\cdot}L^{-1}}$, i.e., $\mathrm{pH}=9.22$.

<Checkpoint>[$\mathrm{pH}=9.22$](1)</Checkpoint>

The standard calomel electrode is the anode, and the cathode is where water is reduced to produce hydrogen. $\varphi_\mathrm{anode}=0.2801\ \mathrm{V}$, $\varphi_\mathrm{cathode}=0.0592\lg\left[\mathrm{H^+}\right]=-0.546\ \mathrm{V}$, so $E=\left(0.2801+0.546\right)=0.826\ \mathrm{V}$.

<Checkpoint>[$E=0.826\ \mathrm{V}$](1)</Checkpoint>
由$\omega_\mathrm{CO_2}=611\ \mathrm{ppm\ by\ mass}$，即$\omega_\mathrm{CO_2}= 6.11\times10^{-4}$，以及摩尔质量和标准大气压数据$M_\mathrm{CO_2}=44\ \mathrm{g{\cdot}mol^{-1}},\bar{M}_\mathrm{air}=29\ \mathrm{g{\cdot}mol^{-1}}, 1\ \mathrm{atm}=101.325\ \mathrm{kPa},$计算得到：$p_\mathrm{CO_2}=\omega_\mathrm{CO_2}\times\frac{\bar{M}_\mathrm{air}}{M_\mathrm{CO_2}}\times1\ \mathrm{atm}=0.0403\ \mathrm{kPa}.$

<Checkpoint>[$p_\mathrm{CO_2}=0.0403\ \mathrm{kPa}$](1)</Checkpoint>

由亨利定律，溶液中溶解平衡时的$\mathrm{CO_2}$的浓度为$\left[\mathrm{CO_2}\right]=\frac{p_\mathrm{CO_2}}{K_\mathrm{H}}=1.36\times10^{-5}\ \mathrm{mol{\cdot}L^{-1}}$。

<Checkpoint>[$\left[\mathrm{CO_2}\right]=1.36\times10^{-5}\ \mathrm{mol{\cdot}L^{-1}}$](1)</Checkpoint>

在该体系中，$\left[\mathrm{HCO}_3^-\right]$远大于$\left[\mathrm{H_2CO_3}\right]$，故可近似认为$\left[\mathrm{HCO_3^-}\right]=0.010\ \mathrm{mol{\cdot}L^{-1}}$不变。

<Checkpoint>[$\left[\mathrm{HCO_3^-}\right]=0.010\ \mathrm{mol{\cdot}L^{-1}}$不变](1)</Checkpoint>

由电离平衡，$\left[\mathrm{H^+}\right]=\frac{K_\mathrm{a1}\left[\mathrm{CO_2}\right]}{\left[\mathrm{HCO_3^-}\right]}=\frac{10^{-6.35}\times1.36\times10^{-5}}{0.010}=6.07\times10^{-10}\ \mathrm{mol{\cdot}L^{-1}}$，即$\mathrm{pH}=9.22$。

<Checkpoint>[$\mathrm{pH}=9.22$](1)</Checkpoint>

标准甘汞电极为阳极，阴极为水被还原出氢气。$\varphi_\mathrm{anode}=0.2801\ \mathrm{V}$，$\varphi_\mathrm{cathode}=0.0592\lg\left[\mathrm{H^+}\right]=-0.546\ \mathrm{V}$，故$E=\left(0.2801+0.546\right)=0.826\ \mathrm{V}$。

<Checkpoint>[$E=0.826\ \mathrm{V}$](1)</Checkpoint>