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Electrolysis of water: Which equations to use? (IB Chem)

1

$begingroup$

There is a list of standard electrode potentials at 298 K from the p. 23 of IB Data Booklet 2016. Which of the following equations (forward/backward reactions), from the two possible ones involving the discharge of hydrogen gas and the other two with oxygen gas discharge, should I use for the oxidation and reduction of water in electrolytic cells?

$$
begin{array}{cc}
hline
ce{text{Oxidized species} <=> text{Reduced species}} & E^⦵(pu{V}) \
hline
begin{align}
ce{H2O(l) + e- &<=> 0.5 H2(g) + OH-(aq)} \
ce{H+(aq) + e- &<=> 0.5 H2(g)} \
ce{0.5 O2(g) + H2O(l) + 2 e- &<=> 2 OH-(aq)} \
ce{0.5 O2(g) + 2 H+(aq) + 2 e- &<=> H2O(l)}
end{align}
&
begin{array}{r}
-0.83 \
0.00 \
+0.40 \
+1.23
end{array}
\
hline
end{array}
$$

(Unless the use of any of these equations cannot be generalized — for a concise explanation of why this is so and what to do then I would be equally grateful.)

physical-chemistry electrochemistry redox water reduction-potential

share|improve this question

edited 32 mins ago

andselisk

19.7k665128

asked 2 hours ago

w_ww_w

82

New contributor

w_w is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

$begingroup$

There is a list of standard electrode potentials at 298 K from the p. 23 of IB Data Booklet 2016. Which of the following equations (forward/backward reactions), from the two possible ones involving the discharge of hydrogen gas and the other two with oxygen gas discharge, should I use for the oxidation and reduction of water in electrolytic cells?

$$
begin{array}{cc}
hline
ce{text{Oxidized species} <=> text{Reduced species}} & E^⦵(pu{V}) \
hline
begin{align}
ce{H2O(l) + e- &<=> 0.5 H2(g) + OH-(aq)} \
ce{H+(aq) + e- &<=> 0.5 H2(g)} \
ce{0.5 O2(g) + H2O(l) + 2 e- &<=> 2 OH-(aq)} \
ce{0.5 O2(g) + 2 H+(aq) + 2 e- &<=> H2O(l)}
end{align}
&
begin{array}{r}
-0.83 \
0.00 \
+0.40 \
+1.23
end{array}
\
hline
end{array}
$$

(Unless the use of any of these equations cannot be generalized — for a concise explanation of why this is so and what to do then I would be equally grateful.)

physical-chemistry electrochemistry redox water reduction-potential

share|improve this question

edited 32 mins ago

andselisk

19.7k665128

asked 2 hours ago

w_ww_w

82

New contributor

w_w is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

There is a list of standard electrode potentials at 298 K from the p. 23 of IB Data Booklet 2016. Which of the following equations (forward/backward reactions), from the two possible ones involving the discharge of hydrogen gas and the other two with oxygen gas discharge, should I use for the oxidation and reduction of water in electrolytic cells?

$$
begin{array}{cc}
hline
ce{text{Oxidized species} <=> text{Reduced species}} & E^⦵(pu{V}) \
hline
begin{align}
ce{H2O(l) + e- &<=> 0.5 H2(g) + OH-(aq)} \
ce{H+(aq) + e- &<=> 0.5 H2(g)} \
ce{0.5 O2(g) + H2O(l) + 2 e- &<=> 2 OH-(aq)} \
ce{0.5 O2(g) + 2 H+(aq) + 2 e- &<=> H2O(l)}
end{align}
&
begin{array}{r}
-0.83 \
0.00 \
+0.40 \
+1.23
end{array}
\
hline
end{array}
$$

(Unless the use of any of these equations cannot be generalized — for a concise explanation of why this is so and what to do then I would be equally grateful.)

physical-chemistry electrochemistry redox water reduction-potential

share|improve this question

edited 32 mins ago

andselisk

19.7k665128

asked 2 hours ago

w_ww_w

82

New contributor

w_w is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 32 mins ago

andselisk

19.7k665128

asked 2 hours ago

w_ww_w

82

New contributor

w_w is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 32 mins ago

andselisk

19.7k665128

asked 2 hours ago

w_ww_w

82

New contributor

w_w is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 32 mins ago

andselisk

19.7k665128

edited 32 mins ago

andselisk

19.7k665128

asked 2 hours ago

w_ww_w

82

New contributor

w_w is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

w_ww_w

82

1 Answer
1

active

oldest

votes

2

$begingroup$

For the acidic electrolysis, use the reactions where $ce{H+}$ occurs.

As $ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product.

Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance.
$ce{OH-}$ or anions of weak acids like $ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides.

But note that using reactions with half of a molecule is not necessery.

$$begin{align}
ce{O2(g) + 4H+(aq) + 4e- &<=> 2 H2O(l)}\
ce{2H+(aq) + 2e- &<=> H2(g)}
end{align}$$

For the alkaline electrolysis, similarly, use the reactions where $ce{OH}$- occurs.

$$begin{align}
ce{2 H2O(l) + 2e- &<=> H2(g) + 2 OH^-(aq)}\
ce{O2(g) + 2 H2O(l) + 4e- &<=> 4 OH^-(aq)}
end{align}$$

share|improve this answer

edited 4 mins ago

answered 2 hours ago

PoutnikPoutnik

1,254310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, @Poutnik. Could you also explain to me how I could apply this knowledge to, for example, the electrolysis of aqueous sodium chloride? A textbook example uses equations from both categories you mentioned.
  $endgroup$
  – w_w
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Apply the same principle of availability and stability.
  $endgroup$
  – Poutnik
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @w_w If you are happy with the answer, feel free to upvote (might not be available due to low rep) and accept it (grey tick mark below).
  $endgroup$
  – andselisk
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Poutnik, I suppose that by "availability" you mean concentration of the salt, but I am not sure how to tackle "stability"... Does "stability" have to do with electrode potential values? I am too verdant to figure this out by myself.
  $endgroup$
  – w_w
  16 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @andselisk, Thank you for the edits and the reminder!
  $endgroup$
  – w_w
  16 mins ago
  

  
  
  
  

 | 
show 2 more comments

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2

$begingroup$

For the acidic electrolysis, use the reactions where $ce{H+}$ occurs.

As $ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product.

Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance.
$ce{OH-}$ or anions of weak acids like $ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides.

But note that using reactions with half of a molecule is not necessery.

$$begin{align}
ce{O2(g) + 4H+(aq) + 4e- &<=> 2 H2O(l)}\
ce{2H+(aq) + 2e- &<=> H2(g)}
end{align}$$

For the alkaline electrolysis, similarly, use the reactions where $ce{OH}$- occurs.

$$begin{align}
ce{2 H2O(l) + 2e- &<=> H2(g) + 2 OH^-(aq)}\
ce{O2(g) + 2 H2O(l) + 4e- &<=> 4 OH^-(aq)}
end{align}$$

share|improve this answer

edited 4 mins ago

answered 2 hours ago

PoutnikPoutnik

1,254310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, @Poutnik. Could you also explain to me how I could apply this knowledge to, for example, the electrolysis of aqueous sodium chloride? A textbook example uses equations from both categories you mentioned.
  $endgroup$
  – w_w
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Apply the same principle of availability and stability.
  $endgroup$
  – Poutnik
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @w_w If you are happy with the answer, feel free to upvote (might not be available due to low rep) and accept it (grey tick mark below).
  $endgroup$
  – andselisk
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Poutnik, I suppose that by "availability" you mean concentration of the salt, but I am not sure how to tackle "stability"... Does "stability" have to do with electrode potential values? I am too verdant to figure this out by myself.
  $endgroup$
  – w_w
  16 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @andselisk, Thank you for the edits and the reminder!
  $endgroup$
  – w_w
  16 mins ago
  

  
  
  
  

 | 
show 2 more comments

2

$begingroup$

For the acidic electrolysis, use the reactions where $ce{H+}$ occurs.

As $ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product.

Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance.
$ce{OH-}$ or anions of weak acids like $ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides.

But note that using reactions with half of a molecule is not necessery.

$$begin{align}
ce{O2(g) + 4H+(aq) + 4e- &<=> 2 H2O(l)}\
ce{2H+(aq) + 2e- &<=> H2(g)}
end{align}$$

For the alkaline electrolysis, similarly, use the reactions where $ce{OH}$- occurs.

$$begin{align}
ce{2 H2O(l) + 2e- &<=> H2(g) + 2 OH^-(aq)}\
ce{O2(g) + 2 H2O(l) + 4e- &<=> 4 OH^-(aq)}
end{align}$$

share|improve this answer

edited 4 mins ago

answered 2 hours ago

PoutnikPoutnik

1,254310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, @Poutnik. Could you also explain to me how I could apply this knowledge to, for example, the electrolysis of aqueous sodium chloride? A textbook example uses equations from both categories you mentioned.
  $endgroup$
  – w_w
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Apply the same principle of availability and stability.
  $endgroup$
  – Poutnik
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @w_w If you are happy with the answer, feel free to upvote (might not be available due to low rep) and accept it (grey tick mark below).
  $endgroup$
  – andselisk
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Poutnik, I suppose that by "availability" you mean concentration of the salt, but I am not sure how to tackle "stability"... Does "stability" have to do with electrode potential values? I am too verdant to figure this out by myself.
  $endgroup$
  – w_w
  16 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @andselisk, Thank you for the edits and the reminder!
  $endgroup$
  – w_w
  16 mins ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

For the acidic electrolysis, use the reactions where $ce{H+}$ occurs.

As $ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product.

Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance.
$ce{OH-}$ or anions of weak acids like $ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides.

But note that using reactions with half of a molecule is not necessery.

$$begin{align}
ce{O2(g) + 4H+(aq) + 4e- &<=> 2 H2O(l)}\
ce{2H+(aq) + 2e- &<=> H2(g)}
end{align}$$

For the alkaline electrolysis, similarly, use the reactions where $ce{OH}$- occurs.

$$begin{align}
ce{2 H2O(l) + 2e- &<=> H2(g) + 2 OH^-(aq)}\
ce{O2(g) + 2 H2O(l) + 4e- &<=> 4 OH^-(aq)}
end{align}$$

share|improve this answer

edited 4 mins ago

answered 2 hours ago

PoutnikPoutnik

1,254310

$endgroup$

share|improve this answer

edited 4 mins ago

answered 2 hours ago

PoutnikPoutnik

1,254310

answered 2 hours ago

PoutnikPoutnik

1,254310

answered 2 hours ago

PoutnikPoutnik

1,254310