Two half cell reactions of an electrochemical cell are given below :
`MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-),toMn^(2+)(aq)+4H_(2)O(l),E^(@)=+1.51V`
`Sn^(2+)(aq)toSn^(4+)(aq)+2e^(-),E^(@)=+0.51V` Construct the redox equation from the two half cell reactions and predict if the reaction favours formation of reactant or product shown in the equation.

Explanation

At anode (oxidation) : `Sn^(2+)(aq)toSn^(4+)(aq)+2e^(-),E^(@)=+0.15V`
At Cathod (reduction0 : `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l),E^(@)=+1.51V` Thus over all redox equation is
`5Sn^(2+)(aq)+2MnO_(4)^(-)(aq)+16H^(+)(aq)to5Sn^(4+)(aq)+2Mn^(2+)(aq)+8H_(2)O(l)`
`E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)=0.15=1.36V`