Question

Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that [Pb2+] = 0.00120 M and[Au³+] = 0.871 M. Standard reduction potentials can be found in this table.3 Pb(s) + 2 Au³+ (aq) 3 Pb²+ (aq) + 2 Au(s)E =

Answer 1

`1.713\text{ V}`

Step-by-step explanation:

Here, we want to calculate the standard reduction potential

Using the table of standard reduction potential, we have it that:

`\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}`

We have the value of E as:

`1.498\text{ - \lparen-0.13\rparen= +1.628 V}`

Now, let us calculate the value of Q, we have that as:

`Q\text{ = }\frac{[Pb^(2+)]\placeholder{⬚}^3}{[Au^(3+)]\placeholder{⬚}^2}\text{ = }(0.00120^3)/(0.871^2)\text{ = 2.278 }*\text{ 10}^(-9)`

Mathematically:

`E_(cell)\text{ = E}^o\text{ -\lparen}(0.0591)/(n))log\text{ Q}`

where n is the number of electrons transferred is 6

Substituting the values:

`\begin{gathered} E_(cell)\text{ = +1.628 - \lparen}(0.0591)/(6))log\text{ 2.278 }*\text{ 10}^(-9) \\ \\ E_(cell)=\text{ 1.713 V} \end{gathered}`