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The emf of cell $Zn/ZnCl_2(0.05mol\;dm^{-3})/Ag^+/AgCl(s)/Ag$ is 1.015V at 298K,the silver electrode being positive,while the temperature coefficient of its emf is $-0.00492VK^{-1}$.Write down the equation for reaction occuring when the cell is allowed to discharge and calculate changes in entropy (s) accompanying this reaction,at 298K?

$\begin{array}{1 1}(a)\;95JK^{-1}mol^{-1}&(b)\;100JK^{-1}mol^{-1}\\(c)\;-95JK^{-1}mol^{-1}&(d)\;-110JK^{-1}mol^{-1}\end{array}$

1 Answer

$\Delta G=\Delta H-T\Delta S$
$\Delta S=\large\frac{\Delta H-\Delta G}{T}$
$\Delta S_{298}=nf(\large\frac{\partial E}{\partial T})_P$
$\Rightarrow-2\times 96500\times 0.0049KJmol^{-1}$
$\Rightarrow -95JK^{-1}mol^{-1}$
Hence (c) is the correct answer.
answered Feb 18, 2014 by sreemathi.v
 

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