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The EMF of electrode -concentration cell $Zn-Hg(C_1)/Zn^{+2}(aq)/Zn-Hg(C_2)$ at $25^{\large\circ}C$ if the concentration of zinc amalgam are $C_1=2g$ per 100g of mercury and $C_2=1g$ per 100g of mercury.

$\begin{array}{1 1}(a)\;8.8\times 10^{-4}V\\(b)\;8.8\times 10^{-5}V\\(c)\;8.8\times 10^{-6}V\\(d)\;8.8\times 10^{-3}V\end{array}$

1 Answer

The cell reaction in this case are
At cathode :$Zn^{+2}+2e^-\rightarrow Zn(C_2)$
At anode :$Zn(C_1)\rightarrow Zn^{+2}+2e^-$
Net reaction : $Zn(C_1)\rightarrow Zn(C_2)$
$E=\large\frac{0.059}{2}$$\log \large\frac{C_2}{C_1}$
$\Rightarrow -0.0295\log (\large\frac{1}{2})$
$\Rightarrow 8.8\times 10^{-3}V$
Hence (d) is the correct answer.
answered Feb 18, 2014 by sreemathi.v

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