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