Normal aluminimum electrode cupled with normal hydrogen electrode gives an emf of ` 1.66 V`. So the standard electrode potential of aluminimu is ,

Options

A. `-1.66V`
B. `+1.66V`
C. `-0.83V`
D. `+0.83V`

Answer

A

Explanation

`E_(cell)^(o)=1.66=E_(H^(+)//H_(2))^(o)-E_(Al^(3+)//Al)^(o)`
`=O-E_(Al^(3+)//Al)^(o)` or `E_(Al^(3+)//Al)=-1.66V`