Respuesta :
The chemical reaction is given as:
[tex]2NaDP^{+} +2H_{2}O\rightarrow 2NaDPH+2H^{+}+O_{2}[/tex]
Here, oxygen is oxidised and [tex]NaDP^{+}[/tex] is reduced. Thus, redox reaction occurs.
For cell reaction, [tex]\Delta G^{o} = -nFE^{o}_{cell}[/tex] (2)
where, [tex]\Delta G^{o} [/tex] = standard state free energy
n= number of electrons
F= Faraday constant ([tex]96485.33 C/mol[/tex])
[tex]E^{o}_{cell}[/tex] = cell potential
Substitute the value of number of electrons i.e. 2, Faraday constant and cell potential in the formula to determine the value of [tex]\Delta G^{o} [/tex].
Now, calculate the value of cell potential
[tex]E^{o}_{cell} = E^{o}_{cathode}- E^{o}_{anode}[/tex] (1)
[tex]E^{o}_{cathode}[/tex] = [tex]-0.324 V[/tex] (standard reduction potential of [tex]NaDP^{+}[/tex])
[tex]E^{o}_{anode}[/tex] = [tex]1.23 V[/tex] (standard reduction potential of [tex]O_{2}[/tex])
Put the above values in formula (1), we get:
[tex]E^{o}_{cell} = -0.324 V-1.23 V[/tex]
= [tex]-1.554 V[/tex]
Now, substitute above value in formula (2)
[tex]\Delta G^{o} = -2\times 96485.33 C/mol \times(-1.554 V) [/tex]
= [tex]299876.40564 CV/mol[/tex]
Since, one coulomb volt is equal to one joule.
Thus, value of [tex]\Delta G^{o}[/tex] is equal to [tex]299876.40564 J/mol[/tex] or [tex]299.87640564 kJ/mol[/tex]