Question

Calculate $\mathit{\Delta }{\mathit{\epsilon }}^{\mathbf{0}\mathbf{\text{'}}}$and $\mathit{\Delta }{\mathit{G}}^{\mathbf{0}\mathbf{\text{'}}}$for the light reactions in plants, that is, the four-electron oxidation of H₂O by NADP⁺.

Short answer

The values of${\mathrm{\Delta G}}^{0\text{'}}=438\mathrm{kJ}/\mathrm{mol}$ and ${\mathrm{\Delta \epsilon }}^{0\text{'}}=-1.135\mathrm{V}$.

Step-by-step solution

Obtaining values of Δε0'and ΔG0'

The values of ${\mathrm{\Delta \epsilon }}^{0\text{'}}$and ${\mathrm{\Delta G}}^{0\text{'}}$are obtained from the figure of the Z-scheme photosynthesis.

Calculating the number of electrons

The value of ${\mathrm{\Delta \epsilon }}^{0\text{'}}$is found to be;

$\begin{array}{c}{\mathrm{\Delta \epsilon }}^{0\text{'}}=-0.320\mathrm{V}-0.815\mathrm{V}\\ =-1.135\mathrm{V}\end{array}$

From the following reaction, the number of electrons is obtained;

$2{\mathrm{H}}_2\mathrm{O}\to 4{\mathrm{H}}^{+}+{\mathrm{O}}_2+4{\mathrm{e}}^{-}$

Therefore, the number of electrons was found to be 4.

Calculating the change of free energy

$\begin{array}{c}{\mathrm{\Delta G}}^{0\text{'}}=-{\mathrm{nF\Delta \epsilon }}^{0\text{'}}\\ {\mathrm{\Delta G}}^{0\text{'}}=\left(-4\right)\times \left(96,485\mathrm{C}/\mathrm{mol}\right)\times \left(-1.135\mathrm{V}\right)\\ =438000\mathrm{J}/\mathrm{mol}\\ =438\mathrm{kJ}/\mathrm{mol}\end{array}$

Hence, the value of ${\mathrm{\Delta G}}^{0\text{'}}=438\mathrm{kJ}/\mathrm{mol}$and ${\mathrm{\Delta \epsilon }}^{0\text{'}}=-1.135\mathrm{V}$.