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# discharge at anode a/c to E values

• electrolysis of MgBr2(aq) gives bromine at anode. why not O2? the standard electrode potential for 2Br......Br2  + 2e is -1.07 V , for OH- to discharge it is -0.40 V so why not Oxygen should be produced at anode as the value of OH is more positive than Br??

• Originally posted by hoay:

electrolysis of MgBr2(aq) gives bromine at anode. why not O2? the standard electrode potential for 2Br......Br2  + 2e is -1.07 V , for OH- to discharge it is -0.40 V so why not Oxygen should be produced at anode as the value of OH is more positive than Br??

This isn't an alkali solution, so you have to consider H2O instead of OH-. The standard oxidation potential of H2O to O2, is less positive (or more negative) than the standard oxidation potential of Br- to Br2.

But in practice, the molarity of H2O isn't standard either. So in practice, you will obtain a mixture of both O2 and Br2 generated at the anode.

For A level purposes, tell your students that the I2 and Br2 will be obtained readily, Cl2 will be obtained only if concentrated Cl- aqueous solution is used, and F2 will not be obtained unless molten F- is used.

• -deleted-
• okay. got it.

One more thing...

for an ion to discharge at either cathode or anode how much positive the value of standard elctrode potential ?

• Originally posted by hoay:

okay. got it.

One more thing...

for an ion to discharge at either cathode or anode how much positive the value of standard elctrode potential ?

Can you rephrase your question, preferably by using an actual A level exam question as an example?
• If we have a solution of S2O8-2 (+2.01V) ion and Ag+ (+0.80V) ion then which wil be discharged at cathode? Assuming both have 1 M solution.

• Originally posted by hoay:

If we have a solution of S2O8-2 (+2.01V) ion and Ag+ (+0.80V) ion then which wil be discharged at cathode? Assuming both have 1 M solution.

An excellent trick question you can use to test your students!

Although the standard reduction potential of peroxodisulfate(VI) aka peroxydisulfate(VI) to sulfate(VI) is more positive than the standard reduction potential of silver(I) to silver(0), and hence peroxodisulfate(VI) aka peroxydisulfate(VI) will seem to be more thermodynamically favored to be reduced to sulfate(VI) at the cathode, but that does NOT occur at all.

(Note that the term "discharged" can only be used specifically if the species loses a charge, ie. Ag+ discharged to Ag, but when S2O8 2- is reduced to SO4 2-, the negative charge remains, so advise your students to write "oxidized at anode" and "reduced at cathode" instead of loosely (and often erroneously) misusing the term "discharge").

The reason for this, is because the peroxodisulfate(VI) aka peroxydisulfate(VI) ion is ANIONIC, and hence electrostatically migrates to the ANODE (instead of cathode), but cannot be further oxidized (so some other anion present, or H2O, will have to be oxidized instead, exactly which, depends on their oxidation potentials).

The Ag+ CATION will migrate to the CATHODE and be reduced to Ag(s) instead. So in this setup, to answer your question, Ag+ is reduced instead of the peroxodisulfate(VI) aka peroxydisulfate(VI).

• although not related to this topic but i have a query about sigma bond.

A sigma bond is the bond formed by heah-to-head overlap or end-on overlap. Is this the reason why it is stronger than a pi bond , which is formed by side way overlapping?

• Originally posted by hoay:

although not related to this topic but i have a query about sigma bond.

A sigma bond is the bond formed by heah-to-head overlap or end-on overlap. Is this the reason why it is stronger than a pi bond , which is formed by side way overlapping?

Yes, correct.
• What about the maximun roration in sigma-bond formation?

• Originally posted by hoay:

What about the maximun roration in sigma-bond formation?

• Originally posted by UltimaOnline:

An excellent trick question you can use to test your students!

Although the standard reduction potential of peroxodisulfate(VI) aka peroxydisulfate(VI) to sulfate(VI) is more positive than the standard reduction potential of silver(I) to silver(0), and hence peroxodisulfate(VI) aka peroxydisulfate(VI) will seem to be more thermodynamically favored to be reduced to sulfate(VI) at the cathode, but that does NOT occur at all.

(Note that the term "discharged" can only be used specifically if the species loses a charge, ie. Ag+ discharged to Ag, but when S2O8 2- is reduced to SO4 2-, the negative charge remains, so advise your students to write "oxidized at anode" and "reduced at cathode" instead of loosely (and often erroneously) misusing the term "discharge").

The reason for this, is because the peroxodisulfate(VI) aka peroxydisulfate(VI) ion is ANIONIC, and hence electrostatically migrates to the ANODE (instead of cathode), but cannot be further oxidized (so some other anion present, or H2O, will have to be oxidized instead, exactly which, depends on their oxidation potentials).

The Ag+ CATION will migrate to the CATHODE and be reduced to Ag(s) instead. So in this setup, to answer your question, Ag+ is reduced instead of the peroxodisulfate(VI) aka peroxydisulfate(VI).

yes this is for Electrolytic cell

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