The standard reduction potentials of `Ca^(2+)|Ca, Zn^(2+)|Zn, H^(+)|(1)/(2)H_(2), (1)/(2)Br_(2)|Br^(-), (1)/(2)Cl_(2)|Cl^(-)` electrodes are `-2.87, -0.76, 0.00, +1.06 and +1.36V`, respectively. Among `Ca, Zn, H_(2), Br_(2) and Cl_(2)`, which one is the strongest reductant, and which one is the strongest oxidant?

Explanation

Electrode having higher standard reduction potential will have tendency to get reduced easily. Thus the increasing order of the standard potential of the given electrodes is :
`E_(Ca^(2+)|Ca)^(@)ltE_(Zn^(2+)|Zn)^(@)ltE_(H^(+)|(1)/(2)H_(2))^(@)ltE_((1)/(2)Br_(2)|Br^(-))^(@)ltE_((1)/(2)Cl_(2)|Cl^(-))^(@)`
As `(1)/(2)Cl_(2)|Cl^(-) and Ca^(2+)|Ca` having maximum and minimum electrode potential, respectively, `Cl_(2)` has the highest tendency to get reduced whereas `Ca^(2+)` has the highest tendency to get oxidised. Thus `Cl_(2)` will be the strongest oxidant and `Ca^(2+)` will be the strongest reductant among the given elements.