A) 2.00 V
B) 1.25 V
C) -1.25 V
D) 1.75 V
Correct Answer: D
Solution :
For the cell, \[\text{Ni }\!\!|\!\!\text{ N}{{\text{i}}^{2+}}\text{ }\!\!|\!\!\text{ }\!\!|\!\!\text{ A}{{\text{u}}^{3+}}|Au\] Given, \[E_{N{{i}^{2+}}/Ni}^{o}=-0.25\,V\] \[E_{A{{u}^{3+}}/Au}^{o}=+\,1.5\,V\] Here \[Ni\]I anode and Au is cathode. \[\therefore \] \[{{E}_{cell}}={{E}_{C}}-{{E}_{A}}\] \[=1.5-(-0.25)\] \[=1.5+0.25\] \[=1.75\,V\]
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