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  4. The temperature coefficient of a cell whose cell reaction is Pb(.s.)+HgCl2(.aq.) arrow PbCl2(.aq.)+Hg(.l.) ((d E/d T))P=1.5× 10- 4VK- 1 at 298 The change in entropy for the given cell reaction will be

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Q. The temperature coefficient of a cell whose cell reaction is
$Pb\left(\right.s\left.\right)+HgCl_{2}\left(\right.aq\left.\right) \rightarrow PbCl_{2}\left(\right.aq\left.\right)+Hg\left(\right.l\left.\right)$
$\left(\frac{d E}{d T}\right)_{P}=1.5\times 10^{- 4}VK^{- 1}$ at 298
The change in entropy for the given cell reaction will be

NTA AbhyasNTA Abhyas 2022

Solution:

$\left(\frac{d E}{d T}\right)_{P}=\frac{\Delta S}{n F}$
$1.5\times 10^{- 4}=\frac{\Delta S}{2 \times 96500}$
$\Delta S=28.95JK^{- 1}mol^{- 1}$ .