A) 1.56 V
B) -1.56V
C) -0.036 V
D) + 0.036V
Correct Answer: A
Solution :
Half-cell reactions are (i) \[\operatorname{Zn} \xrightarrow{{}}\,\,Z{{n}^{2\,+}} + 2{{e}^{-}} E{}^\circ \,\,=\,\,+ 0.763\,V\] (ii) \[{{\operatorname{Ag}}^{+}}\,\,+\,\,e\,\xrightarrow{{}} Ag, \,E{}^\circ = \,0.799 V\,\,\times \,\,2\] From Eqs. (i) and (ii), we get \[\operatorname{Zn}\,\,+\,\,Ag\,\,+\,\,\xrightarrow{{}}\,\,Z{{n}^{2\,+}} + Ag\] \[\operatorname{E}{}^\circ =\, 0.763\,\,+\,\,0.799\,\,=\, 1.56V\]
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