Question

The standard free energies of formation of $H_2{O}_{\left(l\right)},{CO}_{2\left(g\right)}$ an n-pentane are $-237.2kJ,$ $-394.4kJ,$ $-8.2kJ$. Then the value of $E^0$ for pentane-oxygen fuel cell is:

• A. $1.1V$
• B. $0.1V$
• C. $2.0V$
• D. $2.3V$

Answer 1

The correct option is A $1.1V$

$C_5{H}_{12}+8O_2\to 5{CO}_2+6H_{2}O$

change in gibb's free energy of equation,

$△G=6\times △G_{H_{2}O}+5△G{CO}_2-{△G}_{npentane}$

$=6\times (-237.2)+5\times (-394.4)-(-8.2)$

$△G=-1423.2-1972+8.2$

$△G=-3387KJ/mol$

we know, $△G=-nF{E}^0$

$-3387=-32\times 96500\times E^0$

$E^0=1.1V$

for the reaction, there is $32e^{-}$ transfer.