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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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Given the $rx_n, \;2Fe^{3+}(aq)+Sn^{2+}(aq)\rightarrow 2Fe^{2+}(aq)+Sn^{4+}(aq)$ what is the equ.constant of given $rx_n$

Given:

$Fe^{3+}(aq)+2e^-\rightarrow Fe^{2+}(aq)\;E^{\large\circ}=0.77V$

$Sn^{4+}+2e^-\rightarrow Sn^{2+}(aq)\;E^{\large\circ}=0.75V$

$\begin{array}{1 1}(A)\;9\times 10^{20}V&(B)\;9.55\times 10^{20}V\\(C)\;8.23\times 10^{10}V&(D)\;9.55\times 10^{15}V\end{array}$

Can you answer this question?
 
 

1 Answer

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The electrochemical cell corresponding to given $rx_n$ is,
$Sn^{2+},Sn^{4+}/Fe^{3+},Fe^{2+}$
$E^{\large\circ}_{cell}=e^{\large\circ}_{Fe^{3+}/Fe^{2+}}-E^{\large\circ}_{Sn^{4+}/Sn^{2+}}=0.77V-0.15V=0.62V$
So,$\log K=\large\frac{2E^{\large\circ}}{0.0591}$
$\Rightarrow\large\frac{ 2\times 0.62}{0.0591}$
$\Rightarrow 20.98$
$K=10^{20.98}=9.55\times 10^{20}$
Hence (B) is the correct answer.
answered Dec 10, 2013 by sreemathi.v
 

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