Answer:
[tex]1.713\text{ V}[/tex]
Explanation:
Here, we want to calculate the standard reduction potential
Using the table of standard reduction potential, we have it that:
[tex]\begin{gathered} Pb^{2+}\text{ + 2e}^-\text{ }\rightarrow\text{ Pb E = -0.13 V} \\ Au^{3+}\text{ + 3e}^-\text{ }\rightarrow\text{ +1.498 V} \end{gathered}[/tex]
We have the value of E as:
[tex]1.498\text{ - \lparen-0.13\rparen= +1.628 V}[/tex]
Now, let us calculate the value of Q, we have that as:
[tex]Q\text{ = }\frac{[Pb^{2+}]\placeholder{⬚}^3}{[Au^{3+}]\placeholder{⬚}^2}\text{ = }\frac{0.00120^3}{0.871^2}\text{ = 2.278 }\times\text{ 10}^{-9}[/tex]
Mathematically:
[tex]E_{cell}\text{ = E}^o\text{ -\lparen}\frac{0.0591}{n})log\text{ Q}[/tex]
where n is the number of electrons transferred is 6
Substituting the values:
[tex]\begin{gathered} E_{cell}\text{ = +1.628 - \lparen}\frac{0.0591}{6})log\text{ 2.278 }\times\text{ 10}^{-9} \\ \\ E_{cell}=\text{ 1.713 V} \end{gathered}[/tex]
