JEE Main 12 Jan 2019 Morning Question 12
Question: The standard electrode potential $ E{}^\circ $ and its temperature coefficient $ ( \frac{dE^{o}}{dT} ) $ for a cell are 2 V and $ -5\times {10^{-4}}V,{K^{-1}} $ at 300 K respectively.
The cell reaction is $ Z{n _{(s)}}+Cu _{(aq)}^{2+}\xrightarrow[{}]{{}}Zn _{(aq)}^{2+}+C{u _{(s)}} $ The standard reaction enthalpy $ ({\Delta_r}H^{o}) $ at 300 K in $ kJ,mo{l^{-1}} $ is, [Use $ R=8,J,{K^{-1}}mo{l^{-1}} $ and $ F=96,000,C,mo{l^{-1}} $ ] [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ -412.8 $
B) 192.0
C) $ -384.0 $
D) 206.4
Show Answer
Answer:
Correct Answer: A
Solution:
$ {\Delta_r}H^{o}=-nFE^{o}+nFT( \frac{\Delta E^{o}}{\Delta T} ) $
$ =-2\times 96000\times 2+2\times 96000\times 300(-5\times {10^{-4}}) $
$ =-384000-28800=-412.8kJ,mo{l^{-1}} $
