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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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Calculate the stability constant of the complex $[Zn(NH_3)_4]^{2+}$ formed in the $rx_n$:$Zn^{2+}+4NH_3\rightleftharpoons[Zn(NH_3)_4]^{2+}$. The stability constant 'K' is

Given :

$E^{\large\circ}_{Zn^{2+}/Zn}=-0.76V$

$E^{\large\circ}_{Zn(NH_{34}^{2+}/Zn^{\large\circ}}=-1.03V$

$\begin{array}{1 1}(A)\;3\times 10^{10}&(B)\;1.38\times 10^9\\(C)\;2.28\times 10^9&(D)\;9.1\times 10^6\end{array}$

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1 Answer

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$E^{\large\circ}_{cell}=E^{\Large\circ}_{Zn^{2+}/Zn}-E^{\Large\circ}_{[Zn(NH_3)_4]^{2+}/Zn}=-0.76-(-1.03V)$
$\Rightarrow 0.27V$
As,
$\log K=\large\frac{2E^{\large\circ}}{0.0591}$
$\qquad\;=\large\frac{2(0.27)}{0.0591}$
$\qquad\;=9.14$
$\therefore K=1.38\times 10^9$
Hence (B) is the correct answer.
answered Dec 10, 2013 by sreemathi.v
 

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