Oxidation Reduction Potentials: Nernst's Law

Exercise 5

            

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We consider the following redox system: $Cr_2O_7^{2-}$ // $Cr^{3+}$     $E^o$= $1.36 \;V$ What is his →           potential at $0^oC$ and pH=1 for $[Cr_2O_7^{2-}]$ $=$ $0.1 \frac{mol}{L}$ and $[Cr^{3+}]$ $=$ $0.4 \frac{mol}{L}$?

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

$Cr_2O_7^{2-}$ + 6 $e^- $ +$14H^+$ $\leftrightarrows$ $2Cr^{3+}+7H_2O$

$E$ $ =$ $ E^o+\frac{8.3145\cdot (273.15)}{5\cdot96 485}\;\cdot 2.3 \cdot log\frac{[Cr_2O_7^{2-}]\cdot[H^+]^{14}}{[Cr^{3+}]^2}$ $ = $ $1.36 + 0.011\cdot log(\frac{0.1\cdot 10^{-14}}{0.4^2})$ $ =$ $1.20\;V$