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Oxidation-Reduction Potentials: Nernst's Law

Exercise 1

      

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We consider the following redox system: $Zn^{2+}$ // $Zn$     $E^o$= $-0.76 \;V$ What is his →           potential under the standard conditions for $[Zn^{2+}]$ $=$ $0.1 \frac{mol}{L}$ ?

$E$ $ =$ $ E^o+\frac{0.059}{n}log[Zn^{2+}]$

$n\; =\; 2$ because $Zn^{2+}$ + 2 $e^- $ $\leftrightarrows$ $Zn$

$E$ $ =$ $ -0.76+\frac{0.059}{2}log\; 0.1$ $ =$ $-0.79\;V$