Two half cell reactions of an electrons of an cell are given below :
`MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-) to Mn^(2+)(aq)+4H_(2)O(l) , E^(@)=+1.51 V`
`Sn^(2+)(aq)toSn^(4+)(aq)+2e^(-),E^(@)=+0.15 V.`
Construct a redox equation from the two half cell reactions and predict if this reaction favours the formation of reactants or products as shown in the equation.

Explanation

The cell is represented as :
`Sn^(2+)(aq)|Sn^(4+)(aq)||MnO_(4)^(-)(aq)|Mn^(2+)(aq)`
`"At cathode " : MnO_(4)^(-) (aq) +8H^(+)(aq)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l)]xx2`
`{:("At anode ":,[Sn^(2+)(aq)to Sn^(4+)(aq)+2e^(-)]xx5),("overall reaction":,bar (2MnO_(4)^(-)(aq)+16H^(+)(aq)+5Sn^(2+)(aq)rarr2Mn^(2+)(aq)+5.Sn^(4+)(aq)+8H_(2)O(l))):}`
`E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=^(@)MnO_(4)^(-)|Mn^(2+)-E_(Sn^(4+))^(@)|Sn^(2+)=1.51-(0.15)=1.36 V`
Since `E^(@)` cell is positive the reaction favours the formation of products.