Question

A fuel cell involves combustion of the butane at 1 atm and 298 K,

$C_{4}H_{10}\left(\right.g\left.\right)+\frac{13}{2}O_{2}\left(\right.g\left.\right) \rightarrow 4CO_{2}\left(\right.g\left.\right)+5H_{2}O\left(\right.l\left.\right)$

$\Delta G^\circ =-2746 \, kJ/mol$

What is $E^\circ $ of a cell?

• A. +4.74 V
• B. +0.547 V
• C. +1.09 V
• D. +4.37 V

Answer 1

Answer: (C)

In the reaction

$\overset{- 10 + 10}{C_{4} H_{10} \left(g\right)}+\frac{13}{2}O_{2}\left(g\right) \rightarrow \overset{+ 4 - 4}{4 C O_{2} \left(g\right)}+5H_{2}O\left(l\right)$

Change in oxidation number of carbon $=+16-\left(\right.-10\left.\right)=+26$

So number of electrons involved in cell process will be 26.

$E^\circ =\frac{- \Delta G ^\circ }{n F}=\frac{\left(\right. - 2746 \left.\right) \times 1000}{26 \times 96500}$

$=+1.09 \, V$