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Question: Why electrolysis of NaBr and NaI gives \( B{r_2} \) ​ and \( {I_2} \) respectively while that of NaF...

Why electrolysis of NaBr and NaI gives Br2B{r_2} ​ and I2{I_2} respectively while that of NaF gives O2{O_2} ​ instead of F2{F_2} ​ ?

Explanation

Solution

The given question can be solved by taking into account their redox potential. Remember that the more positive the redox potential the better the Oxidizing agent. Fluorine has the highest +ve Redox potential.

Complete answer:
The compounds given to us are NaBr, NaI and NaF. The products that form at the cathodes and anodes during the electrolysis process depend on the redox potential of the reactions that occur at the respective electrodes. Let us first consider the electrolysis of NaF. The aqueous solution will contain H2O(l),Na+,F−{H_2}{O_{(l)}},N{a^ + },{F^ - } ions. The reaction occurring at the anode is:
2H2O→4H(aq)++O2(g)+4e−(E0=−1.23V)2{H_2}O \to 4{H_{(aq)}}^ + + {O_{2(g)}} + 4{e^ - }({E^0} = - 1.23V) -- (1)
2F−(aq)→F2(g)+2e−(E0=−2.87V)2{F^ - }(aq) \to {F_2}(g) + 2{e^ - }({E^0} = - 2.87V) --- (2)
Anode is where oxidation occurs and oxidation is the process of loss of electrons. The -ve sign indicates oxidation process happening. As the reaction will be written in reverse the E0{E^0} will have an opposite sign. The more the positive the value of E0{E^0} the more its tendency to get oxidized. From the above values we can conclude that (1) has more positive value than (2) hence reaction (1) will occur. The product at the anode will be O2{O_2} and not F2{F_2}
Let now us take into account the E0{E^0} values of reaction (1) with the below given:
2Br−→Br2+2e−(E0=−1.09V)2B{r^ - } \to B{r_2} + 2{e^ - }({E^0} = - 1.09V)
2I−→I2+2e−(E0=−0.54V)2{I^ - } \to {I_2} + 2{e^ - }({E^0} = - 0.54V)
The E0{E^0} values of both the reactions are more positive than that of (1), therefore, the products obtained at the anode will be Br2B{r_2} ​ and I2{I_2} .

Note:
The reactions occurring at the cathode can also be taken into account for determining the products. The reactions at cathode are:
2H2O(l)→H2(g)+2OH−(aq)(E0=−0.83V)2{H_2}{O_{(l)}} \to {H_{2(g)}} + 2O{H^ - }_{(aq)}({E^0} = - 0.83V) -- (3)
Na(aq)++e−→Na(s)(E0=−2.71V)Na_{(aq)}^ + + {e^ - } \to N{a_{(s)}}({E^0} = - 2.71V) -- (4)
The reduction potential of Na is lower than that of water, hence the tendency of Na to get reduced will be less. The reaction with higher reduction potential reaction (3) will take place at cathode forming water.