Question

$FAD$ is a redox-active molecule which takes part in many important biological reactions. The redox potential of $FAD$ at $pH7.0$ is given below.$FAD+2H^{+}+2e^{-}\rightleftharpoons FAD{H}_2$; $E_{FAD{H}_2/FAD}=0.18V$. Calculate the redox potential when the media is acidified to $pH=0$.

• A. $0V$
• B. $-0.24V$
• C. $-0.12V$
• D. None of the above

Answer 1

The correct option is A $0V$

$FAD+2H^{+}+2e^{-}\rightleftharpoons FAD{H}_2$

Here $n=2$

$E_{cell}=E_{cell}^{\circ }-\frac{0.059}{2}\log \frac{\left[FAD{H}_2\right]}{\left[H^{+}{]}^2\right[FAD]}$

At $pH=7$ i.e. $\left[H^{+}\right]={10}^{-7}{E}_{cell}=0.18$

At E be $E_{cell}$ at $pH=0$ or $\left[H^{+}\right]=1$

$0.18=E^{\circ }-\frac{0.059}{2}\log \frac{\left[FAD{H}_2\right]}{{10}^{-14}\times \left[FAD\right]}$

$E=E^{\circ }-\frac{0.059}{2}\log \frac{\left[FAD{H}_2\right]}{\left[FAD\right]}$

$\therefore 0.18-E=-\frac{0.059}{2}\log \frac{1}{{10}^{-14}}=\frac{0.059\times 14}{2}$

$\therefore 0.18-E=0.413$

$\Rightarrow E=-0.24V$