Question

Glucose is the major fuel for most living cells. The oxidative breakdown of glucose by our body to produce energy is called respiration. The reaction for the complete combustion of glucose is
$C_6{H}_{12}{O}_6\left(s\right)+6O_2\left(g\right)\to 6C{O}_2\left(g\right)+6H_{2}O\left(l\right)$
If this combustion reaction could be harnessed as a galvanic cell, calculate the theoretical voltage that could be produced at standard conditions. (Hint: Use $\Delta {G^o}_f$values from Appendix 4.)

Short answer

The value for theoretical voltage is 1.34 V when the combustion reaction is connected as galvanic cell.

Step-by-step solution

Combustion reaction and galvanic cell

The substance that reacts with oxygen to release light and heat in the form of energy called combustion reaction. The electrochemical cell where the use of chemical energy from redox reaction and it converts as electrical energy called galvanic cell.

Calculation for Gibbs free energy

The formation of substance from elements associated with change in free energy are stable to form 1 atm pressure with 298 K called standard free energy. The Gibbs free energy can be calculated as:

$$\begin{gathered} \Delta {G^o}_{\left(reaction\right)}=\sum \Delta {G^o}_{f \left(products\right)}-\sum \Delta {G^o}_{f \left(reactants\right)} \\ =[6\times \Delta {G^o}_{f\left(H_{2}O\right)}+6\times \Delta {G^o}_{f\left(C{O}_2\right)}]-[\Delta {G^o}_{f \left(C_6{H}_{12}{O}_6\right)}+6\times \Delta {G^o}_{f \left(O_2\right)}] \\ =-1638.78-2366.16-(-910.56) \\ =-3094.38 kJ/mol \end{gathered}$$

Calculation of theoretical voltage value

The free energy change can be calculated when the standard state cell potential is known. The equation can be calculated as:

$\Delta G^o=-nF{E}^o$

On substituting the values,

$$\begin{gathered} -3094.38 kJ/mol=-24\times 96500\times E^o \\ {E}^o=\frac{-3094380}{-24\times 96500} \\ {E}^o=1.34 V \end{gathered}$$