Arrange the following metals in increasing order of their reducing power.
`E_(K^(+)//K)^(@)=-2.93V, E_(Ag^(+)//Ag)^(@)=+0.80V, E_(Al^(3)//Al)^(@)=-1.66V E_(Au^(3+)//Au)^(@)=+1.40 V,
E_(Li^(+)//Li)^(@)=-3.05 V`

Options

A. `Li lt K lt Al lt Ag lt Au`
B. `Au lt Ag lt Al lt K lt Li`
C. `K lt Al lt Au lt Ag lt Li`
D. `Al lt Ag lt Au lt Li lt K`

Answer

B

Explanation

Lower the electrode potential, better is the reducing power. The electrode potential increases in the order of `Li^(+)//Li (-3.05 V), K^(+)//K (-2.93 V), Al^(3+)//Al, (-1.66 V), Ag^(+)//Ag (+0.80 V) ` and `Au^(3+)//Au (+1.40 V)`. Hence, reducing power of metals will be `Au lt Ag lt Al lt K lt Li`.