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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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EMF of the cell $Sn,SnCl_2(0.5M)/AgCl,Ag$ is $0.430V$ at $25^{\large\circ}$C and $0.448$ at $0^{\large\circ}$C. The free energy change$(\Delta G)$, enthalpy change($\Delta H)$ and entropy change $(\Delta S)$ of the cell $rx_n$ at $25^{\large\circ}C$ are

$\begin{array}{1 1}(A)\;-80KJ,-124.30KJ,-138.6KJ\\(B)\;-82.99KJ,-124.30KJ,-138.6KJ\\(C)\;-80KJ,-24.30KJ,-130.6KJ\\(D)\;82.99KJ,124.30KJ,138.6KJ\end{array}$

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1 Answer

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$\Delta G=-mFE_{cell}$
$\quad\;\;=-2(96500)(0.430)$
$\quad\;\;=-82.99KJ$
$Rx_n$ are :
anode$\rightarrow Sn\rightarrow Sn^{+2}+2e^-$
cathode$\rightarrow 2Ag^++2e^-\rightarrow 2Ag$
$\Delta S=nF\big(\large\frac{\Delta E}{\Delta T}\big)$
$\quad\;\;=\large\frac{2\times 96500\times 0.018}{-25}$
$\quad\;=-138.6J$
$\Delta H=\Delta G+T\Delta S$
$\quad\;\;=-82990+(298)(-138.6J)=-124.3KJ$
Hence (B) is the correct answer.
answered Dec 10, 2013 by sreemathi.v
 

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