Two half-reactions of an electrochemical cell are given below : MnO_(4)^(-)(aq) + 8H^(+) (aq) + 5e^(-) toMn^(2+) (aq) + 4H2_O (I), E° = 1.51 V Sn^(2+) (aq) toSn^(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.

Two half-reactions of an electrochemical cell are given below : MnO_(4

1.	Two half-reactions of an electrochemical cell are given below : MnO_(4)^(-)(aq) + 8H^(+) (aq) + 5e^(-) toMn^(2+) (aq) + 4H2_O (I), E° = 1.51 V Sn^(2+) (aq) toSn^(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.
Answer» Solution :`MnO_(4)^(-) (aq) + 8H^(+) (aq)+ 5e^(-) to Mn^(2+)(aq) + 4H_(2)O (l), E^(@) = +1.51 V`………..(i) `Sn^(2+)(aq) to Sn^(4-) (aq) + 2E^(-), E^(@) = +0.15 V`……(ii) Multiply EQUATION (i) by 2 and equation (ii) by 5, we have `2MnO_(4)^(-) (aq) + 16H^(+) (aq) to 2Mn^(2+)(aq) + 8H_(2)O (l), E^(@) =+1.51 V` `5Sn^(2+)(aq) to 5Sn^(4+) (aq) + 10E^(-), E^(@) = +0.15 V` Adding equation (i) and (ii), we GET `5Sn^(2+) (aq)to 5Sn^(4+) (aq) to 2Mn^(2+) (aq) + 5Sn^(4+)(aq) + 8H_(2)O (l) E^(@) = +1.66 V` As the value is +ve, the reaction is PRODUCT favoured.