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Given that standard electrode potentials...

Given that standard electrode potentials`(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"`
`Sn^(2+)(aq) to Sn^(4+)(aq)+2e^(-),E^(@)=-0.15" V"`
Construct the redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured.

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Please note that less the `E^(@)` value of an electrode, more is its reducing power. The increasing order of reducing power is :
`Ag^(+)//Ag lt Cu^(2+)//Cu lt Fe^(2+)//Fe lt Mg^(2+)//Mg lt K^(+) lt K`.
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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.

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