Question

If $\Delta {G^0}$ value of fuel cell using ${C_4}{H_{10}}$ and ${O_2}$ is $- 2.6 \times {10^6}J$, then ${E^0}$ value of cell is
(1) 2.05 V
(2) 1.04 V
(3) 1.85 V
(4) 2.86 V

Answer 1

In a Galvanic cell, the Gibbs free energy is related to the electromotive force of the cell, faraday constant, and a number of electrons. There are two types of redox reactions taking place which helps the cell to generate an electrical potential and determine the sign of the Gibbs free energy.

Complete step by step answer: Given,
The $\Delta {G^0}$ value is$- 2.6 \times {10^6}J$.
The reaction between the butane and oxygen is shown below.
${C_4}{H_{10}} + 6.5{O_2}(g) \to 4C{O_2}(g) + 5{H_2}O(l)$
In this reaction, one mole of butane reacts with 6.5 moles of oxygen to form 4 mole of carbon dioxide and five moles of water.
The oxidation state of oxygen changes from 0 present on the left side of the reaction to -2 present on the right side of the reaction.
$n = 2 \times 13$
$\Rightarrow n = 26$
The relation between the Gibbs free energy and the electromotive force of the cell is shown below.
$\Delta {G^0} = - nF{E^0}cell$
Where,
$\Delta {G^0}$ is Gibbs free energy
n is the number of electrons.
F is Faraday’s constant.
${E^0}cell$ is the electromotive force of the cell.
The value of Faraday’s constant is 96500.
To calculate the ${E^0}cell$, substitute the values of Gibbs free energy, Faraday’s constant, and a number of electrons in the above equation.
$- 2.6 \times {10^6} = - 26 \times 96500 \times {E^0}cell$
$\Rightarrow {E^0}cell = \dfrac{{2.6 \times {{10}^6}}}{{26 \times 96500}}$
$\Rightarrow {E^0}cell = 1.04V$
Thus, the value of ${E^0}cell$ is 1.04V.
Therefore, the correct option is 2.

Note:
When the value of ${E^0}cell$ is greater than zero then the reaction taking place is spontaneous and the sign of Gibbs free energy is negative and when the value of ${E^0}cell$ is less than zero then the reaction is nonspontaneous and the sign of Gibbs free energy is positive.