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Calculate EMF of concentration cell consisting of zinc electrodes,one immersed in a solution of 0.01M $Zn^{+2}$ ions and the other in a solution of 0.1M $Zn^{+2}$ ions at $25^{\large\circ}C$.The two solution are separated by a salt bridge

$\begin{array}{1 1}(a)\;0.03V\\(b)\;0.09V\\(c)\;1.2V\\(d)\;-0.03V\end{array}$

1 Answer

The cell may be represented as $Zn/Zn^{+2}(0.01M)\parallel Zn^{+2}(0.1M)/Zn$
The EMF of cell is given by
$E_{cell}=\large\frac{RT}{nF}$$ln \large\frac{C_2}{C_1}$
$\Rightarrow \large\frac{0.0591}{2}$$\log \large\frac{0.1}{0.01}$
$\Rightarrow 0.03$V
Hence (a) is the correct answer.
answered Feb 18, 2014 by sreemathi.v

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