The standard reduction potential for the reaction `Sn^(4+)+2e^(-) rarr Sn^(2+)` is + 0.15V. Calculate the free energy change of the reaction.

Explanation

`Sn^(4+)+2e^(-) rarr Sn^(2+) " "E^(@)=0.15V`
Given : n = 2 electrons
F = 96495 coulombs
Formula : `DeltaG=-nFE^(@)`
Solution : `Delta G=-2 times 96495 times 0.15`
`" "=28.948`
Free energy = -28.948 kJ.