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

Answer

A

Explanation

`E_("cel")^(@)=E_(2H^(+)//H_(2))^(@)-E_(Al^(3+)//Al)^(@)`
`1.66 =0-E_(Al^(3+)//Al)^(@)`
`therefore E_(Al^(3+)//Al)^(@)= -1.66V`