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 theis reaction favours formation of reaction or product shown in the equation.

Explanation

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.