Given that the standard electrode` (E^(@))` of metals are :
`K^(+)//K=- 2.93 V, Ag^(+)//Ag = 0.80 V, Cu^(2+)//Cu = 0. 34 V, `
` Mg^(2+)//Mg =- 2.37 V, Cr^(3+)// Cr =- 0.74 V, Fe^(2+)//Fe =- 0.44 V`.
Arrange these metals in an increasing order of their reducing power.
Or
Two half -reactions of an electrochemical cell are given below :
`MnO_4^(-)`(aq) + 8 `H^(+)` (aq) + 5 `e^(-)` `rarr``Mn^(2+)` (aq) + 4 `H_(2) O` (l) , `E^(@)` = + 1.51 V,
` Sn^(2+) (aq) to Sn^(4+) (aq) + 2e^(-) , E^(@) = + 0.15^(V)` construct redox equation and predict if the reaction is reactant or product favoured.

Explanation

Metals in an increasing order of their reducing power .
`Ag^(+)//Ag, Cu^2//Cu, Fe^(2+)//Fe,Cr^(3+)//Cr, Mg^(2+)//Mg, K^(+) //K`
` i.e." " Ag lt Cu lt Fe lt Cr lt Mg lt K`
Or
Cathode : `[MnObar(4) (aq) + 8 H^(=) (aq) + 5 e^(-) to Mn^(2+) (aq) + 4 H_(2) O (l) ]xx 2`
Anode : `[Sn^(2+) (aq) to Sn^(4+) (aq) + 2 e^(-) ] xx 5`
Cell reaction : `2MnO^(-)._(4)(aq)+ 16H^(+)(aq)+5 Sn^(2+)(aq) to 2Mn^(2)+5 Sn^(4+)(aq)+8 H_(2)O`
`E^(@)._("cell") = E_("Cathode-")E _ ("anode") = 1.51 - (-0.15) = 1.66 V`
`:. ` EMF of the cell is positive. Hence the reaction favoured product.