| \[Z{{n}^{2}}+2{{e}^{-}}\to Zn\left( s \right);\text{ }E{}^\circ =-\,0.76V\] |
| \[C{{a}^{2+}}+2{{e}^{-}}\to Ca\left( s \right);\text{ }E{}^\circ =-\,2.87\,V\] |
| \[M{{g}^{2+}}+2{{e}^{-}}\to Mg\left( s \right);\text{ }E{}^\circ =-\,2.36\,V\] |
| \[N{{i}^{2}}+2{{e}^{-}}\to Ni\left( s \right);\text{ }E{}^\circ =-\,0.25\,V\] |
A) \[Ca<Mg<Zn<Ni\]
B) \[Ni<Zn<Mg<Ca\]
C) \[Zn<Mg<Ni<Ca\]
D) \[Ca<Zn<Mg<Ni\]
Correct Answer: B
Solution :
Higher is the oxidation potential better is the tendency to get oxidised and better is the reducing power \[\therefore \,\,\,\text{ }Ni<Zn<Mg<Ca\] is the correct reducing power of the metal
You need to login to perform this action.
You will be redirected in
3 sec
