Question

At equimolar concentrations of $\text{ F}{{\text{e}}^{\text{2+}}}$ and$\text{ F}{{\text{e}}^{\text{3+}}}$, what must [Ag+] be so that the voltage of the galvanic cell made from the \[\text{ (A}{{\text{g}}^{\text{+}}}\text{ }\!\!|\!\!\text{ Ag) }\!\!~\!\!\text{ }\] and \[\text{ (F}{{\text{e}}^{\text{3+}}}\text{ }\!\!|\!\!\text{ F}{{\text{e}}^{\text{2+}}}\text{) }\!\!~\!\!\text{ }\] electrodes equals zero?
\[\text{ F}{{\text{e}}^{\text{2+}}}\text{+A}{{\text{g}}^{\text{+}}}\rightleftharpoons \text{F}{{\text{e}}^{\text{3+}}}\text{+Ag}\]
${{\text{E}}_{\text{A}{{\text{g}}^{\text{+}}}\text{/Ag}}}=0.7991$; ${{\text{E}}_{\text{F}{{\text{e}}^{\text{3+}}}\text{/F}{{\text{e}}^{\text{2+}}}}}=0.771$

Answer 1

Hint: The Nernst equation states a relation between the cell potential \[{{\text{E}}_{\text{cell}}}\]with the standard cell potential$\text{E}_{\text{cell}}^{\text{0}}$, temperature T, and the reaction quotient. The reaction quotient for the redox reaction is equal to the ratio of the concentration of reducing and oxidizing species.
The Nernst equation is as shown below,
${{\text{E}}_{\text{cell}}}\text{= E}_{\text{cell}}^{\text{0}}\text{ }-\text{ }\dfrac{0.059}{\text{n}}\text{ ln}\dfrac{\left[ \text{Red} \right]}{\left[ \text{Oxd} \right]}$
The $\text{ F}{{\text{e}}^{\text{2+}}}$ oxidized to $\text{ F}{{\text{e}}^{\text{3+}}}$ and $\text{ A}{{\text{g}}^{\text{+}}}$ reduces to the$\text{ Ag}$. Here, the number of electrons involves equal to one.

Complete step by step answer:
We know that reaction between $\text{ F}{{\text{e}}^{\text{2+}}}$ and $\text{ A}{{\text{g}}^{\text{+}}}$ . It can be written as:
\[\text{ F}{{\text{e}}^{\text{2+}}}\text{+A}{{\text{g}}^{\text{+}}}\rightleftharpoons \text{F}{{\text{e}}^{\text{3+}}}\text{+Ag}\]

Here, the$\text{ F}{{\text{e}}^{\text{2+}}}$undergoes the oxidation to form $\text{ F}{{\text{e}}^{\text{3+}}}$and $\text{ A}{{\text{g}}^{\text{+}}}$ reduces to the$\text{ Ag}$. We are given that, the concentration of $\text{ F}{{\text{e}}^{\text{2+}}}$ and $\text{ F}{{\text{e}}^{\text{3+}}}$ are same or equimolar. Therefore,
$\text{ }\left[ \text{ F}{{\text{e}}^{\text{2+}}} \right]=\left[ \text{ F}{{\text{e}}^{\text{3+}}} \right]$ and the voltage of the galvanic cell is zero that is \[{{\text{E}}_{\text{cell}}}\text{= 0}\]
We have to find the concentration of $\text{ A}{{\text{g}}^{\text{+}}}$ .

We will use the Nernst equation. the Nernst equation is given as follows,
${{\text{E}}_{\text{cell}}}\text{= E}_{\text{cell}}^{\text{0}}\text{ }-\text{ }\dfrac{\text{RT}}{\text{nF}}\text{ ln}\dfrac{\left[ \text{Red} \right]}{\left[ \text{Oxd} \right]}$
Where,
\[{{\text{E}}_{\text{cell}}}\] is the voltage of the cell
$\text{E}_{\text{cell}}^{\text{0}}$ is the standard potential of cell
R is the gas constant
T is the absolute temperature
n is the number of electrons involved in the redox reaction
F is the faraday's constant 96500 C.
$\left[ \text{Red} \right]$ is the concentration of reducing species
$\left[ \text{Oxd} \right]$ is the concentration of oxidized species

The equation can be modified for absolute temperature as room temperature we get,
${{\text{E}}_{\text{cell}}}\text{= E}_{\text{cell}}^{\text{0}}\text{ }-\text{ }\dfrac{0.059}{\text{n}}\text{ ln}\dfrac{\left[ \text{Red} \right]}{\left[ \text{Oxd} \right]}$

Now, let's substitute the values in the Nernst equation. We have,
${{\text{E}}_{\text{cell}}}\text{= E}_{\text{cell}}^{\text{0}}\text{ - }\dfrac{\text{0}\text{.059}}{\text{n}}\text{ ln}\dfrac{\left[ \text{F}{{\text{e}}^{\text{3+}}} \right]}{\left[ \text{F}{{\text{e}}^{\text{2+}}} \right]\left[ \text{A}{{\text{g}}^{\text{+}}} \right]}$

Let us calculate the $\text{E}_{\text{cell}}^{\text{0}}$.
$\begin{align}
  & \text{E}_{\text{cell}}^{\text{0}}\text{= }{{\text{E}}_{\text{A}{{\text{g}}^{\text{+}}}\text{/Ag}}}-{{\text{E}}_{\text{F}{{\text{e}}^{\text{3+}}}\text{/F}{{\text{e}}^{\text{2+}}}}} \\
 & \text{ = 0}\text{.7991 }-\text{ 0}\text{.771} \\
 & \text{ = 0}\text{.0281 V} \\
\end{align}$
Therefore, the $\text{E}_{\text{cell}}^{\text{0}}$ is $\text{0}\text{.0281 V}$

Substitute the values in the first equation. We have,
$\begin{align}
  & \text{0 = 0}\text{.0281 - }\dfrac{\text{0}\text{.059}}{1}\text{ lo}{{\text{g}}_{\text{10}}}\dfrac{\left[ \text{F}{{\text{e}}^{\text{3+}}} \right]}{\left[ \text{F}{{\text{e}}^{\text{2+}}} \right]\left[ \text{A}{{\text{g}}^{\text{+}}} \right]}\text{ }\because \text{ n = 1 }{{\text{e}}^{\text{-}}} \\
 & \text{since, }\left[ \text{F}{{\text{e}}^{\text{3+}}} \right]\text{=}\left[ \text{F}{{\text{e}}^{\text{2+}}} \right] \\
 & \text{we have ,} \\
 & \text{0 = 0}\text{.0281 - }\dfrac{\text{0}\text{.059}}{1}\text{ lo}{{\text{g}}_{\text{10}}}\dfrac{\left[ \text{F}{{{\text{}}}^{\text{2+}}} \right]}{\left[ \text{F}{{{\text{}}}^{\text{2+}}} \right]\left[ \text{A}{{\text{g}}^{\text{+}}} \right]} \\
 & \text{0 = 0}\text{.0281 - }\dfrac{\text{0}\text{.059}}{1}\text{ lo}{{\text{g}}_{\text{10}}}\dfrac{1}{\left[ \text{A}{{\text{g}}^{\text{+}}} \right]} \\
\end{align}$

We are interests to find out the $\left[ \text{A}{{\text{g}}^{\text{+}}} \right]$, rearrange with respect we have,
$\begin{align}
  & \text{ 0 = 0}\text{.0281 - }\dfrac{\text{0}\text{.059}}{1}\text{ lo}{{\text{g}}_{\text{10}}}\dfrac{1}{\left[ \text{A}{{\text{g}}^{\text{+}}} \right]} \\
 & \Rightarrow \text{0}\text{.0281 = 0}\text{.059 lo}{{\text{g}}_{\text{10}}}\dfrac{1}{\left[ \text{A}{{\text{g}}^{\text{+}}} \right]} \\
 & \Rightarrow \text{ lo}{{\text{g}}_{\text{10}}}\dfrac{1}{\left[ \text{A}{{\text{g}}^{\text{+}}} \right]}\text{ = }\dfrac{\text{0}\text{.0281}}{\text{0}\text{.059}} \\
 & \Rightarrow \text{ lo}{{\text{g}}_{\text{10}}}\dfrac{1}{\left[ \text{A}{{\text{g}}^{\text{+}}} \right]}\text{ = 0}\text{.476} \\
 & \text{Take antilog we have,} \\
 & \Rightarrow \text{ }\dfrac{1}{\left[ \text{A}{{\text{g}}^{\text{+}}} \right]}\text{ = 1}{{\text{0}}^{\text{0}\text{.476}}} \\
 & \Rightarrow \text{ }\dfrac{1}{\left[ \text{A}{{\text{g}}^{\text{+}}} \right]}\text{ = 2}\text{.992} \\
 & \Rightarrow \left[ \text{A}{{\text{g}}^{\text{+}}} \right]\text{ = }\dfrac{1}{\text{2}\text{.992}} \\
 & \therefore \left[ \text{A}{{\text{g}}^{\text{+}}} \right]=\text{ 0}\text{.334 M} \\
\end{align}$
Therefore, the concentration of $\left[ \text{A}{{\text{g}}^{\text{+}}} \right]$ is $\text{0}\text{.334 M}$.

Note: The Nernst equation can be used to determine the equilibrium constant of the reaction. The Nernst equation can be written as,
$\text{ E}_{\text{cell}}^{\text{0}}\text{ = }\dfrac{\text{RT}}{\text{nF}}\text{ ln K}$
At equilibrium constant, ${{\text{E}}_{\text{cell}}}$ is equal to zero and K is the equilibrium constant.