Electrode potential data are given below
`{:(Fe^(3+)(aq)+e^(-)rarrFe^(2+)(aq),E^(@)=+0.77V),(Al^(3+)(aq)+3e^(-)rarrAl(s),E^(@)=-1.66V),(Br_(2)(aq)+2e^(-)rarrBr^(-)(aq),E^(@)=+1.05V):}`
Based on the given data which statements is/are true?

Standard electrode potential data is given below : Fe^(3+) (aq) + e ^(-) rarr Fe^(2+) (aq) , E^(o) = + 0.77 V Al^(3+) (aq) + 3e^(-) rarr Al (s) , E^(o) = - 1.66 V Br_(2) (aq) + 2e^(-) rarr 2Br^(-) (aq) , E^(o) = +1.08 V Based on the data given above, reducing power of Fe^(2+) Al and Br^(-) will increase in the order :

Standard electrode potential data is given below : Fe^(3+) (aq) + e ^(-) rarr Fe^(2+) (aq) , E^(o) = + 0.77 V Al^(3+) (aq) + 3e^(-) rarr Al (s) , E^(o) = - 1.66 V Br_(2) (aq) + 2e^(-) rarr 2Br^(-) (aq) , E^(o) = +1.08 V Based on the data given above, reducing power of Fe^(2+) Al and Br^(-) will increase in the order :

Electrode potential data are given below: Fe^(3+)(aq)+e^(-)rarrFe^(2+)(aq):E^(@)=+0.77V Al^(3+)+3e^(-)rarrAl(s):E^(@)=-1.66V Br_(2)(aq)+2e^(-)rarr2Br^(-)(aq):E^(@)=+1.08V , Based on the data, the reducing power of Fe^(2+) Al and Br^(-) will increase in the order

Based on the data given below , the correct order of reducing power is : Fe^(3+)(aq) +erarrFe^(2+)(aq),E^@=+0.77V AI^(3+)(aq)+3erarrAI(s),E^@=-1.66V Br_2(aq) +2e rarr2Br^-(aq),E^@ =+1.08V

Standard reduction potentials of the half reactions are given below {:(F_(2(g))+2e^(-) rarr 2F^(-)""_((aq)),,E^(@)=+2.85V), (Cl_(2(g))+2e^(-) rarr 2Cl^(-)""_((aq)),, E^(@)=+1.36V), (Br_(2(l))+2e^(-) rarr 2Br^(-)""_((aq)),,E^(@)=+1.06V), (l_(2(s))+2e^(-) rarr 2l^(-)""_((aq)),, E^(@)=+0.53V):} The strongest oxidising and reducing agents respectively are