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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} $

1 Answer

Answer : $[3(-0.36)-2(-0.439)]V$
$Fe^{3+}+3e^-\rightarrow Fe$
$Fe^{2+}+2e^-\rightarrow Fe$
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$Fe^{3+}+e^-\rightarrow Fe^{2+}$
$\Delta G_1^{\large\circ}=-(3)F(-0.36V)$
$\Delta G_2^{\large\circ}=-(2)F(-0.439V)$
$\Delta G=\Delta G_1^{\large\circ}-\Delta G_2^{\large\circ}=-F[3(-0.36-2(-0.439)]V$
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
 

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