Question

$Pt({{H}_{2}})({{p}_{1}})/{{H}^{+}}(1M)({{H}_{2}})({{p}_{2}}),Pt$cell reaction will be exergonic if:
[A]$({{p}_{1}})=({{p}_{2}})$
[B]$({{p}_{1}})>({{p}_{2}})$
[C]$({{p}_{2}})>({{p}_{1}})$
[D]${{p}_{1}}=1atm$

Answer 1

Hint: For an exergonic reaction, release in free energy is negative. We can solve this by using the Nernst equation and solving it for the electrode provided. Nernst equation gives the relation between standard electrode potential and cell potential using the cathode and anode half reactions.
  Complete step by step solution:
A reaction is exergonic if the net change in free energy is negative. There will be a net release of free energy.
The above condition can be written as- $\Delta G={{G}_{product}}-{{G}_{reac\tan t}}<0$
I.e.$\Delta G$ should be negative.
As we know, for a spontaneous reaction,$\Delta G$is negative and $\Delta E$ will be positive.
Here, the given cell reaction$Pt({{H}_{2}})({{p}_{1}})/{{H}^{+}}(1M)({{H}_{2}})({{p}_{2}})$ we have a Platinum-Hydrogen electrode.

We know, Nernst equation is-
$\begin{align}
  & {{E}_{cell}}={{E}^{\circ }}-\frac{0.059}{n}\log Q \\
 & Q=\frac{[{{M}_{red}}]}{[{{M}_{ox}}]} \\
\end{align}$
Where, ${{E}_{cell}}$= Cell potential
${{E}^{\circ }}$=Standard cell potential
n= number of electrons transferred
Q= reaction quotient, which is given by the
 (Concentration of reduced metal ion)$\div $(Concentration of oxidised metal ion)
Cell reaction for $Pt({{H}_{2}})({{p}_{1}})/{{H}^{+}}(1M)({{H}_{2}})({{p}_{2}})$ will be-
$\begin{align}
  & {{H}_{2}}({{p}_{1}})\to {{H}^{+}} \\
 & {{H}^{+}}\to {{H}_{2}}({{p}_{2}}) \\
\end{align}$
Now, we can write the Nernst equation for $Pt({{H}_{2}})({{p}_{1}})/{{H}^{+}}(1M)({{H}_{2}})({{p}_{2}})$ as-
     ${{E}_{cell}}=0-\frac{0.059}{n}\log \frac{[{{H}^{+}}]({{p}_{2}})}{[{{H}^{+}}]({{p}_{1}})}$
$\because [{{H}^{+}}]=1$
     \[\begin{align}
  & \therefore {{E}_{cell}}=-\frac{0.059}{n}\log \frac{({{p}_{2}})}{({{p}_{1}})} \\
 & or,{{E}_{cell}}=\frac{0.059}{n}\log \frac{({{p}_{1}})}{({{p}_{2}})} \\
\end{align}\]
To make ${{E}_{cell}}$positive, we interchanged the positions of ${{p}_{1}}and{{p}_{2}}$
${{E}_{cell}}$ will be positive when $\frac{{{p}_{1}}}{{{p}_{2}}}$ is positive and,
 $\frac{{{p}_{1}}}{{{p}_{2}}}$ will be positive when $\log \frac{{{p}_{1}}}{{{p}_{2}}}$ is positive
Therefore, we can write that-
 $\begin{align}
  & \frac{{{p}_{1}}}{{{p}_{2}}}>1 \\
 & or,{{p}_{1}}>{{p}_{2}} \\
\end{align}$

Therefore, the reaction to be exergonic and for $\Delta G$ to be negative and ${{E}_{cell}}$ to be positive, ${{p}_{1}}$ must be greater than ${{p}_{2}}$.

Therefore, the correct answer is option [B] $({{p}_{1}})>({{p}_{2}})$.

Note: It is important to remember how free energy is related to the cell potential. It is also important to write down the cells oxidation and reduction reactions correctly. If we write the oxidation and reduction half reactions incorrectly, we will get the opposite answer,$({{p}_{2}})>({{p}_{1}})$ which is incorrect. Therefore, the cell notations should be written correctly here. Also, we must remember the Nernst Equation to solve such questions.