1.

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

Answer»

Solution :`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.


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