Fluorine gas can't be prepared Sodium fluoride and MnO2 with conc. H2SO4 . Why ?
Answers
The principle needed here is that if there are two possible redox reactions, the one with the more positive voltage is the one that will happen. This is because positive voltage reactions are spontaneous (release energy) and negative voltage reactions are non-spontaneous (require energy to happen). Thus, if there are two negative voltage reactions the one with the least negative voltage will happen because it is the most spontaneous.
These are the possible reactions we have to analyze because the species we would start with would be Na+, F-, and H2O because it is a sodium fluoride solution:
4F- ==> 2F2 + 4e- (-2.889 V)
2H2O ==> O2 + 4H+ + 4e- (-1.229 V)
4Na+ + 4e- ==> 4Na (-2.714 V)
The 4Na+ + 4e- ==> 4Na reduction reaction will happen because it is the only reduction reaction.
The reactions left are the oxidation of F- and the oxidation of water. Of these the water one has the least negative voltage so it will happen instead of the F- one.
Our two equations then are
4Na+ + 4e- ==> 4Na (-2.714 V)
2H2O ==> O2 + 4H+ + 4e- (-1.229 V)
So the net reaction is 4Na+ + 2H2O ==> 4Na + O2 + 4H+ (-3.943 V)
Thus the reaction produces oxygen gas instead of fluorine gas.
Note the net reaction has a negative voltage which means it is non spontaneous and we need a 3.943 V battery to force it to happen. But with that 3.943 V battery the H2O oxidation will happen instead of the F- oxidation because its reaction voltage is less negative than the F- oxidation reaction voltage.
Explanation:
it undergoes redox reactions with them