Question

3.单选题（10分）已知φ-（pb²⁺/pb）=-0.1266v，kₛₚ-（pbcl₂）=1.7×10⁻⁵，则φ-（pbcl₂/pb）=（）v。a -0.268b 0.015c -0.409d 0.268

Explanation:

Step1: Recall Nernst equation

The Nernst equation for the half - reaction \(Pb^{2 +}+2e^- ightleftharpoons Pb\) is \(\varphi=\varphi^{\ominus}+\frac{0.0592}{n}\lg c(Pb^{2 +})\) (at \(25^{\circ}C\)), where \(n = 2\) for the reaction \(Pb^{2+}+2e^- ightarrow Pb\), and \(\varphi^{\ominus}(Pb^{2+}/Pb)=- 0.1266\ V\).

For the \(PbCl_2/Pb\) couple, the reaction is \(PbCl_2 + 2e^- ightleftharpoons Pb + 2Cl^-\). The solubility product \(K_{sp}(PbCl_2)=c(Pb^{2+})\cdot c^{2}(Cl^-)\), so \(c(Pb^{2+})=\frac{K_{sp}(PbCl_2)}{c^{2}(Cl^-)}\). When \(c(Cl^ -)=1\ mol/L\) (standard state for the \(PbCl_2/Pb\) couple), \(c(Pb^{2+}) = K_{sp}(PbCl_2)\).

Step2: Substitute \(c(Pb^{2+})\) into Nernst equation

Substitute \(c(Pb^{2+})=K_{sp}(PbCl_2)\) into the Nernst equation for the \(Pb^{2+}/Pb\) couple. The Nernst equation for \(\varphi(PbCl_2/Pb)\) (when \(c(Cl^ -) = 1\ mol/L\)) is:

\(\varphi(PbCl_2/Pb)=\varphi^{\ominus}(Pb^{2+}/Pb)+\frac{0.0592}{2}\lg K_{sp}(PbCl_2)\)

We know that \(\varphi^{\ominus}(Pb^{2+}/Pb)=- 0.1266\ V\), \(K_{sp}(PbCl_2)=1.7\times10^{-5}\)

First, calculate \(\lg K_{sp}(PbCl_2)=\lg(1.7\times 10^{-5})=\lg1.7+\lg10^{-5}\approx0.23 - 5=- 4.77\)

Then, \(\frac{0.0592}{2}\lg K_{sp}(PbCl_2)=\frac{0.0592}{2}\times(-4.77)\approx0.0296\times(-4.77)\approx - 0.141\) (approximate calculation, more accurately: \(\frac{0.0592}{2}\times\lg(1.7\times10^{-5})=\frac{0.0592}{2}\times(\lg1.7 - 5)\))

\(\lg1.7\approx0.2304\), so \(\frac{0.0592}{2}\times(0.2304 - 5)=\frac{0.0592}{2}\times(- 4.7696)=0.0296\times(-4.7696)\approx - 0.1412\)

Now, \(\varphi(PbCl_2/Pb)=-0.1266-0.1412\)? Wait, no, wait. Wait, the reaction for \(PbCl_2/Pb\) is \(PbCl_2 + 2e^- ightleftharpoons Pb + 2Cl^-\), and the relationship between \(\varphi(PbCl_2/Pb)\) and \(\varphi(Pb^{2+}/Pb)\) is derived from the fact that for the \(PbCl_2/Pb\) couple, the effective \(Pb^{2+}\) concentration is determined by \(K_{sp}\). Let's do the calculation more accurately.

\(\varphi=\varphi^{\ominus}(Pb^{2+}/Pb)+\frac{0.0592}{n}\lg c(Pb^{2+})\)

For \(PbCl_2\), \(K_{sp}=c(Pb^{2+})c^{2}(Cl^-)\), when \(c(Cl^ -) = 1\ M\), \(c(Pb^{2+})=K_{sp}\)

So \(\varphi(PbCl_2/Pb)=\varphi^{\ominus}(Pb^{2+}/Pb)+\frac{0.0592}{2}\lg K_{sp}\)

Substitute the values:

\(\varphi^{\ominus}(Pb^{2+}/Pb)=- 0.1266\ V\), \(K_{sp}=1.7\times10^{-5}\)

\(\lg(1.7\times10^{-5})=\ln(1.7\times10^{-5})/\ln(10)\approx(-9.779)/2.3026\approx - 4.247\) (Wait, earlier I made a mistake in \(\lg\) calculation. \(\lg(1.7\times10^{-5})=\lg1.7+\lg10^{-5}=0.2304 - 5=-4.7696\), correct. So \(\frac{0.0592}{2}\times(-4.7696)=0.0296\times(-4.7696)\approx - 0.1412\)

Then \(\varphi(PbCl_2/Pb)=- 0.1266-0.1412\)? No, wait, the Nernst equation is \(\varphi=\varphi^{\ominus}+\frac{0.0592}{n}\lg c(oxidized)\), for \(Pb^{2+}+2e^- ightarrow Pb\), the oxidized species is \(Pb^{2+}\), so when \(c(Pb^{2+})\) decreases (because \(Pb^{2+}\) is precipitated as \(PbCl_2\)), the potential should decrease? Wait, no, let's re - derive.

The reaction for \(PbCl_2/Pb\) is \(PbCl_2(s)+2e^- ightleftharpoons Pb(s)+2Cl^-(aq)\)

The standard potential for this reaction can be related to the \(Pb^{2+}/Pb\) potential. The equilibrium constant for the dissolution of \(PbCl_2\) is \(K_{sp}=c(Pb^{2+})c^{2}(Cl^-)\), so \(c(Pb^{2+}) = K_{sp}/c^{2}(Cl^-)\)

For the \(Pb^{2+}/Pb\) couple: \(\varphi=\varphi^{\ominus}(Pb^{2+}/Pb)+\frac{0.0592}{2}\lg c(Pb^{2+})\)

Substitute \(c(Pb^{2+})=K_{sp}/c^{2}(Cl^-)\) into it:

\(\varphi=\varphi^{\ominus}(Pb^{2+}/Pb)+\frac{0.0592}{2}\lg\frac{K_{sp}}{c^{2}(Cl^-)}\)

\(=\varphi^{\ominus}(Pb^{2+}/Pb)+\frac{0.0592}{2}\lg K_{sp}-\frac{0.0592}{2}\lg c^{2}(Cl^-)\)

\(=\varphi^{\ominus}(Pb^{2+}/Pb)+\frac{0.0592}{2}\lg K_{sp}-0.0592\lg c(Cl^-)\)

When \(c(Cl^ -)=1\ M\) (standard state for the \(PbCl_2/Pb\) couple), \(\lg c(Cl^ -) = 0\), so:

\(\varphi^{\ominus}(PbCl_2/Pb)=\varphi^{\ominus}(Pb^{2+}/Pb)+\frac{0.0592}{2}\lg K_{sp}\)

Now, calculate \(\frac{0.0592}{2}\lg(1.7\times10^{-5})\):

\(\lg(1.7\times10^{-5})=-4.7696\)

\(\frac{0.0592}{2}\times(-4.7696)=0.0296\times(-4.7696)\approx - 0.1412\)

Then \(\varphi^{\ominus}(PbCl_2/Pb)=- 0.1266-0.1412\)? No, that can't be right. Wait, I think I messed up the sign. Wait, the reaction \(PbCl_2 + 2e^- ightarrow Pb + 2Cl^-\) is a reduction reaction. The \(Pb^{2+}/Pb\) reaction is \(Pb^{2+}+2e^- ightarrow Pb\). If \(Pb^{2+}\) is less available (because it's precipitated as \(PbCl_2\)), the potential should be less positive (or more negative) than \(\varphi^{\ominus}(Pb^{2+}/Pb)\)? Wait, no, \(\varphi^{\ominus}(Pb^{2+}/Pb)\) is negative. Let's do the calculation with more precise numbers.

\(K_{sp}=1.7\times10^{-5}\), \(\lg K_{sp}=\lg1.7+\lg10^{-5}=0.2304 - 5=-4.7696\)

\(\frac{0.05916}{2}\times\lg K_{sp}\) (using \(0.05916\) instead of \(0.0592\) for more accuracy)

\(\frac{0.05916}{2}\times(-4.7696)=0.02958\times(-4.7696)\approx - 0.141\)

Then \(\varphi(PbCl_2/Pb)=- 0.1266-0.141\approx - 0.2676\ V\approx - 0.268\ V\)? Wait, no, that's option A. But wait, maybe I made a mistake in the direction. Wait, the correct formula:

The standard potential for \(PbCl_2/Pb\) is given by:

\(\varphi^{\ominus}(PbCl_2/Pb)=\varphi^{\ominus}(Pb^{2+}/Pb)+\frac{0.05916}{n}\lg K_{sp}(PbCl_2)\) when \(c(Cl^ -) = 1\ M\)

Wait, let's check the formula again. The Nernst equation for \(Pb^{2+}+2e^- ightleftharpoons Pb\) is \(\varphi=\varphi^{\ominus}+\frac{0.05916}{2}\lg c(Pb^{2+})\)

For \(PbCl_2\), \(c(Pb^{2+})=\frac{K_{sp}}{c^{2}(Cl^-)}\), so when \(c(Cl^ -)=1\ M\), \(c(Pb^{2+}) = K_{sp}\)

So \(\varphi=\varphi^{\ominus}+\frac{0.05916}{2}\lg K_{sp}\)

Substitute \(\varphi^{\ominus}=- 0.1266\ V\), \(K_{sp}=1.7\times10^{-5}\)

\(\lg K_{sp}=-4.7696\)

\(\frac{0.05916}{2}\times(-4.7696)=0.02958\times(-4.7696)\approx - 0.141\)

\(\varphi=-0.1266-0.141\approx - 0.2676\ V\approx - 0.268\ V\), which is option A.

Answer:

A. -0.268