A) \[-2.13\times {{10}^{5}}J\]
B) \[+2.13\times {{10}^{5}}J\]
C) \[-4.67\times {{10}^{7}}J\]
D) \[+4.67\times {{10}^{7}}J\]
Correct Answer: A
Solution :
Zinc is oxidised and \[A{{g}_{2}}O\] is reduced. \[E_{cell}^{o}=E_{A{{g}_{2}}O/Ag}^{o}(red)+E_{Zn/Z{{n}^{2+}}}^{o}(OX)\] \[=0.344+0.76=1.104V\] \[\Delta H=-nFE_{cell}^{o}=-2\times 96500\times 1.104J\] \[=-2.13\times {{10}^{5}}J\]You need to login to perform this action.
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