A) 1.968 V
B) 2.0968 V
C) 1.0968 V
D) 0.0968 V
Correct Answer: C
Solution :
| [c] 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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