A) \[-176kJ\]
B) \[-234.7kJ\]
C) \[+123.5\text{ }kJ\]
D) \[-167.26\text{ }kJ\]
Correct Answer: D
Solution :
The cell reaction is \[C{{d}_{(s)}}+2AgC{{l}_{(s)}}\xrightarrow[{}]{{}}2A{{g}_{(s)}}+Cd_{(aq)}^{2+}\] \[+2Cl_{(aq)}^{-}\] \[{{E}_{1}}=0.6915\text{ }V\]at\[0{}^\circ C\] \[{{E}_{2}}=0.6753V\] at\[25{}^\circ C\] NOW \[\frac{\partial {{E}_{cell}}}{\partial T}=\frac{{{E}_{2}}-{{E}_{1}}}{{{T}_{2}}-{{T}_{1}}}\] \[=\frac{0.6753-0.6915}{298-273}\] \[=-6.48\times {{10}^{-4}}\] \[\Delta S=nF\left[ \frac{\partial {{E}_{cell}}}{\partial T} \right]\] Now we put the value \[\Delta S=2\times 96500(-6.48\times {{10}^{-4}})=-125.064\] We know that \[\Delta G=-nF{{E}_{cell}}\] \[=-2\times 96500\times 0.6753\] \[=-1.303\times {{10}^{5}}\] As, \[\Delta G=\Delta H-T\Delta S\] For calculating \[\underset{(25{}^\circ C)}{\mathop{\Delta H}}\,=\Delta G+T\Delta S\] \[=-1.303\times {{10}^{5}}+298(-125.064\text{ }kJ)\] \[\Delta H=-1.6726\times {{10}^{5}}joules\] \[=-167.26\text{ }kJ\]
You need to login to perform this action.
You will be redirected in
3 sec
