Oxidation Reduction Potentials: Nernst's Law

Exercise 3

            

Use arrows ↓ and ↑ !

We consider the following redox system: $Ag^{+}$ // $Ag$     $E^o$= $0.81 \;V$ What is his →           potential at $90^oC$ for $[Ag^{+}]$ $=$ $0.1 \frac{mol}{L}$ ?

$E$ $ =$ $ E^o+\frac{RT}{nF}ln[Ag^{+}]$

$n\; =\; 1$ because $Ag^{+}$ + 1 $e^- $ $\leftrightarrows$ $Ag$

$\frac{RT}{nF}\;$$ =$$\;\frac{8.3145\cdot (273.15+90)}{1\cdot96 485}$$ = $$0.0313 $

Taking into account the conversion ln → log: $E$ $ =$ $ 0.75+\frac{2.3\;0.0313}{1}log(0.1)$ $ =$ $0.68\;V$