(Where \[{{E}^{o}}\] for \[N{{i}^{2+}}/Ni\] is \[-0.25\text{ }V,\] |
\[{{E}^{o}}\] for \[A{{u}^{3+}}/Au\]is\[0.150\text{ }V\]) |
A) \[+1.25\,V\]
B) \[-1.75\,V\] nmn
C) \[+1.75\text{ }V\]
D) \[+0.4\text{ }V\]
Correct Answer: D
Solution :
\[Ni/N{{i}^{2+}}[1.0M]\,\,||A{{u}^{3+}}[1.0\,M]/Au\] |
\[E_{cell}^{o}\left( A{{u}^{3+}}/Au \right)=0.15\,V\] |
\[E_{cell}^{o}\left( N{{i}^{2+}}/Ni \right)=-0.25\,V\] |
\[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] |
\[=0.150-(-0.25)\] |
\[=+0.4\,V\] |
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