A) + 0.4V
B) \[-1.75V\]
C) + 1.25V
D) + 1.75V
Correct Answer: A
Solution :
\[Ni/N{{i}^{2+}}[1.0M]||A{{u}^{3+}}[1.0M]/Au\] \[E_{cell}^{o}(A{{u}^{3+}}/Au)=0.15V\] \[E_{cell}^{o}(N{{i}^{2+}}/Ni)=-0.25V\] \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] \[=0.150-(-0.25)\] \[=+0.4V\]You need to login to perform this action.
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