Question

At 298 K the standard free energy of formation of ${{H}_{2}}O\left( l \right)$ is 2565 kJ/mol, while that of its ionization to ${{H}^{+}}$ and $O{{H}^{-}}$ is 80 kJ/mol. What will be reversible emf at 298 K of the cell?
\[{{H}_{2}}(g,1bar)/{{H}^{+}}(1M)\parallel OH(1M)/{{O}_{2}}(g,1bar)\]
Fill your answer by multiplying it with 10.

Answer 1

Hint: By using Nernst equation we calculate the emf of the cell and it is as follows.
\[{{E}_{cell}}=E_{cell}^{o}-\dfrac{0.059}{n}\log \left( \dfrac{acid}{base} \right)\]
Where ${{E}_{cell}}$ = emf of the cell
$E_{cell}^{o}$ = emf of the standard cell
n = number of electrons transferred

Complete answer:
- In the question it is given the standard free energy of formation of ${{H}_{2}}O\left( l \right)$ and asked to find the emf of the reversible reaction.
- Standard free energy of the reaction is given from this we can calculate the change in free of the reaction.
- The formula is given below
\[\Delta G=\Delta {{G}^{o}}-T\Delta S\]
$\Delta G$ = Gibbs free energy
$\Delta {{G}^{o}}$ = Standard free energy = 2565 kJ/mol
T = Temperature
$\Delta S$ = Ionization energy = 80 kJ/mol
- Substitute all the known values in the above to get Gibbs free energy of the reaction.
\[\begin{align}
  & \Delta G=\Delta {{G}^{o}}-T\Delta S \\
 & =2565-(298\times 80) \\
 & =-21,275 \\
\end{align}\]
- To get the standard cell emf substitute the free energy in the below formula.
\[\begin{align}
  & \Delta G=-nFE_{cell}^{o} \\
 & -21275=-2\times 96500\times E_{cell}^{o} \\
 & E_{cell}^{o}=0.1102 \\
\end{align}\]
- We got the value of standard cell emf and we need total cell emf.
\[{{E}_{cell}}=E_{cell}^{o}-\dfrac{0.059}{n}\log \left( \dfrac{acid}{base} \right)\]
Where ${{E}_{cell}}$ = emf of the cell
$E_{cell}^{o}$ = emf of the standard cell = 0.1102 V
n = number of electrons transferred = 2
Concentration of the acid = 1M
Concentration of the base = 1M
- So, we have to substitute all the known values in the Nernst equation and it is as follows.
\[\begin{align}
  & {{E}_{cell}}=E_{cell}^{o}-\dfrac{0.059}{n}\log \left( \dfrac{acid}{base} \right) \\
 & =0.1102-\dfrac{0.059}{2}\log \left( \dfrac{1M}{1M} \right) \\
 & =0.1102-\dfrac{0.059}{2}(0) \\
 & =0.1102V \\
\end{align}\]
- The emf of the cell is 0.1102 V.
- But in the question it is given that multiply the answer with 10.
- Therefore
Emf of the cell = (0.1102) (10) = 1.102 V.

Note:
First we have to calculate the Gibbs free energy from the given reaction conditions. Then after that we have to calculate the standard emf of the cell and later by substituting the standard emf of the cell in Nernst equation we will the emf of the cell.