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

Explanation

`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.