Question

A fuel cell develops an electrical potential from the combustion of butane at 1 bar and 298 K.
\[{{\text{C}}_{4}}{{\text{H}}_{10}}_{\left( \text{g} \right)}\text{ + 6}\text{.5}{{\text{O}}_{2(g)}}\text{ }\to \text{ 4C}{{\text{O}}_{2(g)}}\text{ + 5}{{\text{H}}_{2}}{{\text{O}}_{\left( \text{g} \right)}};\text{ }{{\Delta }_{r}}G{}^\circ \text{ = -2746 kJ/mol}\]
What is \[\text{E}{}^\circ \] of a cell?
A. 4.74 V
B. 0.547 V
C. 4.37 V
D. 1.00 V

Answer 1

Hint: \[\text{E}{}^\circ \] is the standard electrode potential of a cell whereas $\vartriangle \text{G}{}^\circ $is the standard Gibbs Free energy. The relation between standard electrode potential and standard Gibbs free energy is $\vartriangle \text{G}{}^\circ \text{ = nFE}{}^\circ $ where n is the no. of electrons, F is the Faraday's constant.

Complete answer:
-Firstly, we know that we have to calculate the standard electrode potential of a cell by using the formula:
 $\vartriangle \text{G}{}^\circ \text{ = nFE}{}^\circ $ ….(1)
-It is given that the value of $\vartriangle \text{G}{}^\circ $is -2746 kJ/mol and F is the constant whose value is constant in every condition.
-So, we have to find the value of n i.e. number of the electron that participates in the reaction.
-The reaction is:
${{\text{C}}_{4}}{{\text{H}}_{10}}\text{ + 6}\text{.5}{{\text{O}}_{2}}\text{ }\to \text{ 4C}{{\text{O}}_{2}}\text{ + 5}{{\text{H}}_{2}}\text{O}$
-Here, we can see that the change in the oxidation state of the oxygen i.e. form 0 to -2 because the oxidation state of oxygen in ${{\text{O}}_{2}}$ is 0. After all, according to rules of oxidation state, the single element has the oxidation state of zero.
-Whereas in carbon dioxide, the oxidation state of carbon dioxide is:
$\begin{align}
  & \text{4 + 2x = 0} \\
 & \text{x = -2} \\
\end{align}$
-In water, the oxidation state of oxygen is:
$\begin{align}
  & \text{2 + x = 0} \\
 & \text{x = -2} \\
\end{align}$
-So, now by applying all values in equation (1), we will get:
$\text{-2746 = 26 }\times \text{ 96500 }\times \text{ E}{{{}^\circ }_{\text{Cell}}}$
$\text{E}{{{}^\circ }_{\text{Cell}}}\text{ }=\text{ }\dfrac{-2746}{2509}\text{ = 1}\text{.09V}$
So, the correct answer is “Option D”.


Note: If the value $\vartriangle \text{G}{}^\circ $ is negative, then the reaction is spontaneous which means that the reaction is possible in the forward direction that is from reactant to product. Whereas if the value of $\vartriangle \text{G}{}^\circ $is positive then the reaction will be spontaneous but it proceeds in the opposite direction that is from product to reactant.