Question

What is the value of δg°' (or, to put it another way, the cost) when 2nadp+ and 2h2o are converted to 2nadph plus 2h+ plus o2?

Answer 1

The chemical reaction is given as:

$2NaDP^{+} +2H_{2}O\rightarrow 2NaDPH+2H^{+}+O_{2}$

Here, oxygen is oxidised and $NaDP^{+}$ is reduced. Thus, redox reaction occurs.

For cell reaction, $\Delta G^{o} = -nFE^{o}_{cell}$          (2)

where, $\Delta G^{o}$ = standard state free energy

n= number of electrons

F= Faraday constant ($96485.33 C/mol$)

$E^{o}_{cell}$ = 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  $\Delta G^{o}$.

Now, calculate the value of cell potential

$E^{o}_{cell} = E^{o}_{cathode}- E^{o}_{anode}$           (1)

$E^{o}_{cathode}$ = $-0.324 V$ (standard reduction potential of $NaDP^{+}$)

$E^{o}_{anode}$ = $1.23 V$ (standard reduction potential of $O_{2}$)

Put the above values in formula (1), we get:

$E^{o}_{cell} = -0.324 V-1.23 V$

= $-1.554 V$

Now, substitute above value in formula (2)

$\Delta G^{o} = -2\times 96485.33 C/mol \times(-1.554 V)$  

= $299876.40564 CV/mol$

Since, one coulomb volt is equal to one joule.

Thus, value of $\Delta G^{o}$ is equal to $299876.40564 J/mol$ or $299.87640564 kJ/mol$

Answer 2

The answer is 104.9, i dont know how though