Given the standard electrode potentials, `K^(+) | K = -2.93 V`,`Ag^(+) | Ag = 0.80 V` , `Hg_(2)^(2+) | Hg = 0.79 V` , `Mg_(2)^(2+) | Mg = -2.73 V` , `Cr_(2)^(3+) | Cr = -0.74 V`. Arrange these metals in increasing order of their reducing power.

Given the standard electrode potentials, K^(+) // K = - 2.93 V, Ag^(2+) //Ag = 0.80 V Hg^(2+) //Hg = 0.79 V , Mg^(2+) //Mg = - 2.37 V, Cr^(3+) //Cr = - 0.74 V Arrange these metals in their increasing order of reducing power.

Given, the standard electrode potentials, K^+//K = -2.93 V, Ag^+//Ag = 0.80 V, Hg^(2+)//Hg =0.79 V, Mg^(2+)//Mg = - 2.37 V, Cr^(3+)//Cr = - 0.74 V Arrange these metals in their increasing order of reducing power.

Given the standard electrode potential value of some metals. K^(+)//K = -2.93V , Ag^(+)//Ag = 0.80V , Hg^(+)//Hg = 0.79V , Mg^(2+)//Mg + -2.37V , Cr^(3+)//Cr = -0.74V . Arrange these metals in their increasing order of reducing power.

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.

Given the standard electrode potential value of some metals K^+//K = -2.93V,Ag^+//Ag=0.80V , Hg^(2+)//Hg= 0.79V , Mg^(2+)/MG=-2.37V , Cr^(3+)//Cr=-0.74V Arrange these metals in their increasing order of reducing power.