Two half reactions of an electro-chemical cell are given below:
`MnO_4(aq)+8H^+ (aq)+5e^(-) to Mn^(2+) (aq)+4H_2O(l), 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 are predict if the reaction is reactant or product favoured.

Explanation

Reduction potential of the first half reaction is more than that of the second half reaction,

`E^@=0.15`
`5Sn^(2+)+2MnO_4^(-) to Sn^(4+) +2Mn^(2+)`
`E_(cell)^@=E_(cathode)-E_(anode)`
`=1.51-0.15=1.36`
As the cell potential has positive value, the reaction is product favoured and is spontaneous.