A) 1.968 V
B) 2.0968 V
C) 1.0968 V
D) 0.0968 V
Correct Answer: C
Solution :
Given, \[\Delta _{{{H}_{2}}O(l)}^{o}=-237.2\] \[\Delta _{C{{O}_{2}}(g)}^{o}=-394.4\] \[\Delta G_{{{C}_{5}}{{H}_{12}}(g)}^{o}=-8.2\] \[{{C}_{5}}{{H}_{12}}+16{{O}_{2}}\xrightarrow[{}]{{}}5C{{O}_{2}}+6{{H}_{2}}O\] \[\Delta {{G}^{o}}=5\times \Delta G_{C{{O}_{2}}}^{o}+6\times \Delta G_{{{H}_{2}}O}^{o}\] \[-\left( \Delta G_{{{C}_{5}}{{H}_{12}}}^{o}+\Delta G_{{{O}_{2}}}^{o} \right)\] \[=5\times (-394.4)+6\times (-237.2)-(-8.2)+0\] \[=-3387\,\text{kJ/mol}\] In pentane-oxygen fuel cell 32 electrons are involved. \[\Delta {{G}^{o}}=nFE_{cell}^{o}\] \[-3387\times {{10}^{3}}=32\times 96500\times E_{cell}^{o}\] \[E_{cell}^{o}=\frac{-3.387\times {{10}^{3}}}{32\times 96500}\] \[=\frac{3387}{3088}\,=1.0968\,\text{V}\]
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