Question

According to the Nernst equation, the potential of an electrode changes by $59.2mV$ whenever the ratio of the oxidized and the reduced species changes by a factor of $10$ at ${25}^{o}C$. What would be the corresponding change in the electrode potential if the experiment is carried out at ${30}^{o}C$?

• A. $59.2mV$
• B. $71.0mV$
• C. $60.2mV$
• D. $Noneofabove$

Answer 1

Answer: (A)

$59.2mV$

$M^{n+}+n{e}^{-}⟶M$
$At298K$,
$E_1=E^0-\frac{0.0592}{n}log\frac{\left[Reduced\right]}{\left[Oxidised\right]}$
On changing the ratio by factor 10
$E_2=E^0-\frac{0.0592}{n}log10\frac{\left[Reduced\right]}{\left[Oxidised\right]}$
$E_2=E^0-\frac{0.0592}{n}log\frac{\left[Reduced\right]}{\left[Oxidised\right]}-\frac{0.0592}{n}log10$
$E_2=E_1-\frac{0.0592}{n}or{E}_2-E_1=\frac{59.2}{n}mV$
The case taken in the problem has n=1
$E_2-E_1=59.2mV$