Two half cell reaction 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^(@)=+0.15V Construct the redox reaction from the two half cell reaction and predict if this reaction favours formation of reactants or products shown in the equation.

Two half cell reaction of an electrochemical cell are given below: Mn

1.	Two half cell reaction 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^(@)=+0.15V Construct the redox reaction from the two half cell reaction and predict if this reaction favours formation of reactants or products shown in the equation.
Answer» Solution :Multiplying first EQN. by 2 and second eqn. by 5 to cancel out the electrons, and adding, the redox reaction will be `2MnO_(4)^(-)+16H^(+)+5Sn^(2+)to2Mn^(2+)+5Sn^(4+)+8H_(2)O` OXIDATION potential of 2nd reaction`=-0.15` V (as the given value is reduction potential). Hence, if oxidation occurs on `Sn^(2+)//Sn^(4+)` electrode, `E_(CELL)^(@)=1.51+(-0.15)V=1.36V`, i.e., it will be +ve, hence, reaction will FAVOUR FORMATION of products as represented above.