Question

Standard free energies of formation (in kJ/mol) at 298K are $ - 237.2, - 394.4$ and $ - 8.2$ for ${H_2}O(I),C{O_2}(g)$ and pentane(g) respectively. The value of $E_{cell}^0$ for the pentane-oxygen fuel cell is
A.$0.0968V$
B.$1.968V$
C.$2,0968V$
D.$1.0968V$

Answer 1

Hint: The Gibbs energy is a thermodynamic value which gives the reversal of work performed by a thermodynamic system at a constant pressure and pressure. For the given reaction we use the formula ${\Delta _r}G = - nF{E_{\left( {cell} \right)}}$.

Complete step by step answer:
Cell potential is the potential difference between the two electrodes of a galvanic cell. It represents the electrode potential of the cation and the anion. Thus, the difference between the electrode potential of the cation and the anion is cell electromotive force (emf), when no current is drawn through the cell. The anode convention is taken on the left side and the cathode is taken on the right. The emf of the cell is positive and is calculated as the difference between the potential of the half cell on the right side and the potential of the half cell on the left side.
${E_{cell}} = {E_{right}} - {E_{left}}$

The Gibbs energy is a thermodynamic potential value which gives us the reversal of work performed by a thermodynamic system at a constant temperature and pressure. Thus, if the emf of a cell is represented by $E$ and the amount of charge passed is represented by $nF$, then the Gibbs energy for the reaction is given by
${\Delta _r}G = - nF{E_{\left( {cell} \right)}}$

The calculation of Gibbs energy also helps to calculate the equilibrium constant. The Gibbs energy reaction increases the entropy of the surroundings by giving out heat energy to the surroundings. Metals like platinum or gold are used as inert electrodes.
For the given question we use the formula ${\Delta _r}G = - nF{E_{\left( {cell} \right)}}$
Pentane-${C_5}{H_{12}}$
We know that,
${C_5}{H_{12}} + 8{O_2} \to 5C{O_2} + 6{H_2}O$
We apply the Gibbs energy formula
$\Delta {G^0}(298K) = 5( - 394.4) + 6( - 237.2) - ( - 8.2)$
$ \Rightarrow - 3387 \times {10^3}J/mole$
${\Delta _r}G = - nF{E_{\left( {cell} \right)}}$
$ \Rightarrow n = 5 \times 4 + 6 \times 2 = 32$
On applying the formula we get,
$\Delta E_{cell}^0 = \dfrac{{(3387 \times {{10}^3})}}{{\left( {32 \times 96500} \right)}}$
$ = 1.0968V$

Thus, we conclude that the correct option is (D).

Note:
The potential of an individual half cell cannot be calculated thus, we can only calculate the difference between the potential of the cathode electrode and the anode electrode. The hydrogen electrode is considered as the standard hydrogen electrode.
${H^ + }(aq) + {e^ - } \to \dfrac{1}{2}{H_2}(g)$