A) 308.08 kJ
B) 154.4 KJ
C) 77.2 kJ
D) -154.4 kJ
Correct Answer: B
Solution :
[b] \[E_{cell}^{{}^\circ }=E_{{{H}_{2}}}^{{}^\circ }-E_{Ag}^{{}^\circ }\] \[=0-0.80=-0.80\] \[\Delta G{}^\circ =-nFE_{cell}^{{}^\circ }\] \[=-2\times 96500\times \left( -0.8 \right)\] \[=+154.4\text{ kJ}\]
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