Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Given that $E^{\large\circ}_{Fe^{3+}|Fe}$ and $E^{\large\circ}_{Fe^{2+}|Fe}$ are -0.36V and -0.439V,respectively.The value of $E^{\large\circ}_{Fe^{3+},Fe^{2+}|Pt}$ would be

$\begin{array}{1 1}(-0.36-0.439)V\\(-0.36+0.439)V\\(3(-0.36)+2(-0.439))V\\(3(-0.36)-2(-0.439))V\end{array} $

Can you answer this question?

1 Answer

0 votes
Answer : $[3(-0.36)-2(-0.439)]V$
$Fe^{3+}+3e^-\rightarrow Fe$
$Fe^{2+}+2e^-\rightarrow Fe$
$Fe^{3+}+e^-\rightarrow Fe^{2+}$
$\Delta G_1^{\large\circ}=-(3)F(-0.36V)$
$\Delta G_2^{\large\circ}=-(2)F(-0.439V)$
$\Delta G=\Delta G_1^{\large\circ}-\Delta G_2^{\large\circ}=-F[3(-0.36-2(-0.439)]V$
Hence ,$E^{\large\circ}=[3(-0.36)-2(-0.439)]V$
answered Jul 21, 2014 by sreemathi.v
edited Jul 21, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App