Why does electrolysis of aqueous concentrated sodium bromide produce bromine at the anode?
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It is often stated that electrolysis of any aqueous concentrated halogen compound will produce the halogen at the anode, despite it having a more postitive standard electrode potential than the oxidation of hydroxide, while when it is dilute oxygen will be produced, but what is the reason for this?
Why does the principle of concentrated solutions during electrolysis only apply for halogens and not any other element. For example, why isn’t sodium produced at the cathode when concentrated sodium chloride is electrolysed?
electrolysis
asked Apr 20 at 6:46
Anthony PAnthony P
572
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add a comment |
$begingroup$
It is often stated that electrolysis of any aqueous concentrated halogen compound will produce the halogen at the anode, despite it having a more postitive standard electrode potential than the oxidation of hydroxide, while when it is dilute oxygen will be produced, but what is the reason for this?
Why does the principle of concentrated solutions during electrolysis only apply for halogens and not any other element. For example, why isn’t sodium produced at the cathode when concentrated sodium chloride is electrolysed?
electrolysis
asked Apr 20 at 6:46
Anthony PAnthony P
572
$endgroup$
add a comment |
$begingroup$
It is often stated that electrolysis of any aqueous concentrated halogen compound will produce the halogen at the anode, despite it having a more postitive standard electrode potential than the oxidation of hydroxide, while when it is dilute oxygen will be produced, but what is the reason for this?
Why does the principle of concentrated solutions during electrolysis only apply for halogens and not any other element. For example, why isn’t sodium produced at the cathode when concentrated sodium chloride is electrolysed?
electrolysis
asked Apr 20 at 6:46
Anthony PAnthony P
572
$endgroup$
It is often stated that electrolysis of any aqueous concentrated halogen compound will produce the halogen at the anode, despite it having a more postitive standard electrode potential than the oxidation of hydroxide, while when it is dilute oxygen will be produced, but what is the reason for this?
Why does the principle of concentrated solutions during electrolysis only apply for halogens and not any other element. For example, why isn’t sodium produced at the cathode when concentrated sodium chloride is electrolysed?
electrolysis
electrolysis
asked Apr 20 at 6:46
Anthony PAnthony P
572
asked Apr 20 at 6:46
Anthony PAnthony P
572
asked Apr 20 at 6:46
Anthony PAnthony P
572
asked Apr 20 at 6:46
Anthony PAnthony P
572
asked Apr 20 at 6:46
Anthony PAnthony P
572
572
add a comment |
add a comment |
1 Answer
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There is important to know the actual potential depends on actual activities of reagents ( approximately concentrations for dilute solutions).
Additionally, the electrode potential is the thermodynamic quantity. If some net reaction kinetic is involved, other potentials affects the actual potential at the solution border, like the diffusion potential and dielectric layer potential.
We consider reactions
$$begin{align}
ce{Br2 + 2e- &<=> 2 Br-}\
ce{O2 + 2 H2O + 4e- &<=> 4 OH-}\
end{align}$$
The initial bromine activity is very low.
Activity/concentration ratio of ions typically decreases with concentration, as seen in the Debye-Huckel equation,
but for high salt concentrations it often jumps up the roof.
Therefore, activity of bromides is very high and the actual potential gets low.
$$E_{c_{ce{Br2}}/c_{ce{Br-}}}=E^{circ}_{c_{ce{Br2}}/c_{ce{Br-}}} + 0.059 log left( frac{a_{ce{Br2}}}{a_{ce{Br-}}}right)$$
OTOH,
The potential for hydroxide oxidation is pushed up for neutral solutions to the middle between standard redox potentials for acidic and alkaline solutions $$(+0.401 + +1.229)/2=+0.815$$
$$E_{c_{ce{O2}}/c_{ce{OH-}}}=E^{circ}_{c_{ce{O2}}/c_{ce{OH-}}} + 0.059log left( frac{a_{ce{O2}}}{a_{ce{OH-}}}right)$$
For $ce{NaCl}$ solutions, difference of standard potentials is too high to produce metallic sodium, unless one uses mercury as the cathode, where happens the hydrogen, related to its reaction kinetics. This is used for production of sodium hydroxide, where sodium amalgam reacts with water.
answered Apr 20 at 8:11
PoutnikPoutnik
1,534311
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
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active
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$begingroup$
There is important to know the actual potential depends on actual activities of reagents ( approximately concentrations for dilute solutions).
Additionally, the electrode potential is the thermodynamic quantity. If some net reaction kinetic is involved, other potentials affects the actual potential at the solution border, like the diffusion potential and dielectric layer potential.
We consider reactions
$$begin{align}
ce{Br2 + 2e- &<=> 2 Br-}\
ce{O2 + 2 H2O + 4e- &<=> 4 OH-}\
end{align}$$
The initial bromine activity is very low.
Activity/concentration ratio of ions typically decreases with concentration, as seen in the Debye-Huckel equation,
but for high salt concentrations it often jumps up the roof.
Therefore, activity of bromides is very high and the actual potential gets low.
$$E_{c_{ce{Br2}}/c_{ce{Br-}}}=E^{circ}_{c_{ce{Br2}}/c_{ce{Br-}}} + 0.059 log left( frac{a_{ce{Br2}}}{a_{ce{Br-}}}right)$$
OTOH,
The potential for hydroxide oxidation is pushed up for neutral solutions to the middle between standard redox potentials for acidic and alkaline solutions $$(+0.401 + +1.229)/2=+0.815$$
$$E_{c_{ce{O2}}/c_{ce{OH-}}}=E^{circ}_{c_{ce{O2}}/c_{ce{OH-}}} + 0.059log left( frac{a_{ce{O2}}}{a_{ce{OH-}}}right)$$
For $ce{NaCl}$ solutions, difference of standard potentials is too high to produce metallic sodium, unless one uses mercury as the cathode, where happens the hydrogen, related to its reaction kinetics. This is used for production of sodium hydroxide, where sodium amalgam reacts with water.
answered Apr 20 at 8:11
PoutnikPoutnik
1,534311
$endgroup$
add a comment |
$begingroup$
There is important to know the actual potential depends on actual activities of reagents ( approximately concentrations for dilute solutions).
Additionally, the electrode potential is the thermodynamic quantity. If some net reaction kinetic is involved, other potentials affects the actual potential at the solution border, like the diffusion potential and dielectric layer potential.
We consider reactions
$$begin{align}
ce{Br2 + 2e- &<=> 2 Br-}\
ce{O2 + 2 H2O + 4e- &<=> 4 OH-}\
end{align}$$
The initial bromine activity is very low.
Activity/concentration ratio of ions typically decreases with concentration, as seen in the Debye-Huckel equation,
but for high salt concentrations it often jumps up the roof.
Therefore, activity of bromides is very high and the actual potential gets low.
$$E_{c_{ce{Br2}}/c_{ce{Br-}}}=E^{circ}_{c_{ce{Br2}}/c_{ce{Br-}}} + 0.059 log left( frac{a_{ce{Br2}}}{a_{ce{Br-}}}right)$$
OTOH,
The potential for hydroxide oxidation is pushed up for neutral solutions to the middle between standard redox potentials for acidic and alkaline solutions $$(+0.401 + +1.229)/2=+0.815$$
$$E_{c_{ce{O2}}/c_{ce{OH-}}}=E^{circ}_{c_{ce{O2}}/c_{ce{OH-}}} + 0.059log left( frac{a_{ce{O2}}}{a_{ce{OH-}}}right)$$
For $ce{NaCl}$ solutions, difference of standard potentials is too high to produce metallic sodium, unless one uses mercury as the cathode, where happens the hydrogen, related to its reaction kinetics. This is used for production of sodium hydroxide, where sodium amalgam reacts with water.
answered Apr 20 at 8:11
PoutnikPoutnik
1,534311
$endgroup$
add a comment |
$begingroup$
There is important to know the actual potential depends on actual activities of reagents ( approximately concentrations for dilute solutions).
Additionally, the electrode potential is the thermodynamic quantity. If some net reaction kinetic is involved, other potentials affects the actual potential at the solution border, like the diffusion potential and dielectric layer potential.
We consider reactions
$$begin{align}
ce{Br2 + 2e- &<=> 2 Br-}\
ce{O2 + 2 H2O + 4e- &<=> 4 OH-}\
end{align}$$
The initial bromine activity is very low.
Activity/concentration ratio of ions typically decreases with concentration, as seen in the Debye-Huckel equation,
but for high salt concentrations it often jumps up the roof.
Therefore, activity of bromides is very high and the actual potential gets low.
$$E_{c_{ce{Br2}}/c_{ce{Br-}}}=E^{circ}_{c_{ce{Br2}}/c_{ce{Br-}}} + 0.059 log left( frac{a_{ce{Br2}}}{a_{ce{Br-}}}right)$$
OTOH,
The potential for hydroxide oxidation is pushed up for neutral solutions to the middle between standard redox potentials for acidic and alkaline solutions $$(+0.401 + +1.229)/2=+0.815$$
$$E_{c_{ce{O2}}/c_{ce{OH-}}}=E^{circ}_{c_{ce{O2}}/c_{ce{OH-}}} + 0.059log left( frac{a_{ce{O2}}}{a_{ce{OH-}}}right)$$
For $ce{NaCl}$ solutions, difference of standard potentials is too high to produce metallic sodium, unless one uses mercury as the cathode, where happens the hydrogen, related to its reaction kinetics. This is used for production of sodium hydroxide, where sodium amalgam reacts with water.
answered Apr 20 at 8:11
PoutnikPoutnik
1,534311
$endgroup$
There is important to know the actual potential depends on actual activities of reagents ( approximately concentrations for dilute solutions).
Additionally, the electrode potential is the thermodynamic quantity. If some net reaction kinetic is involved, other potentials affects the actual potential at the solution border, like the diffusion potential and dielectric layer potential.
We consider reactions
$$begin{align}
ce{Br2 + 2e- &<=> 2 Br-}\
ce{O2 + 2 H2O + 4e- &<=> 4 OH-}\
end{align}$$
The initial bromine activity is very low.
Activity/concentration ratio of ions typically decreases with concentration, as seen in the Debye-Huckel equation,
but for high salt concentrations it often jumps up the roof.
Therefore, activity of bromides is very high and the actual potential gets low.
$$E_{c_{ce{Br2}}/c_{ce{Br-}}}=E^{circ}_{c_{ce{Br2}}/c_{ce{Br-}}} + 0.059 log left( frac{a_{ce{Br2}}}{a_{ce{Br-}}}right)$$
OTOH,
The potential for hydroxide oxidation is pushed up for neutral solutions to the middle between standard redox potentials for acidic and alkaline solutions $$(+0.401 + +1.229)/2=+0.815$$
$$E_{c_{ce{O2}}/c_{ce{OH-}}}=E^{circ}_{c_{ce{O2}}/c_{ce{OH-}}} + 0.059log left( frac{a_{ce{O2}}}{a_{ce{OH-}}}right)$$
For $ce{NaCl}$ solutions, difference of standard potentials is too high to produce metallic sodium, unless one uses mercury as the cathode, where happens the hydrogen, related to its reaction kinetics. This is used for production of sodium hydroxide, where sodium amalgam reacts with water.
answered Apr 20 at 8:11
PoutnikPoutnik
1,534311
edited Apr 20 at 12:38
answered Apr 20 at 8:11
PoutnikPoutnik
1,534311
answered Apr 20 at 8:11
PoutnikPoutnik
1,534311
answered Apr 20 at 8:11
PoutnikPoutnik
1,534311
1,534311
add a comment |
add a comment |
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