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

4

$begingroup$

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 7 hours ago

andselisk

19.7k665128

asked 9 hours ago

w_ww_w

232

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 | 

4

$begingroup$

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 7 hours ago

andselisk

19.7k665128

asked 9 hours ago

w_ww_w

232

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 7 hours ago

andselisk

19.7k665128

asked 9 hours ago

w_ww_w

232

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 7 hours ago

andselisk

19.7k665128

asked 9 hours ago

w_ww_w

232

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 7 hours ago

andselisk

19.7k665128

asked 9 hours ago

w_ww_w

232

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 7 hours ago

andselisk

19.7k665128

edited 7 hours ago

andselisk

19.7k665128

asked 9 hours ago

w_ww_w

232

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 9 hours ago

w_ww_w

232

1 Answer
1

active

oldest

votes

5

$begingroup$

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 6 hours ago

answered 8 hours ago

PoutnikPoutnik

1,284310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Apply the same principle of availability and stability.
  $endgroup$
  – Poutnik
  7 hours 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @w_w I mean ability to survive. $ce{OH-}$ or anions of weak acids like $ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides. Generally, choice duch a reaction form, where reactants on both sides may exist in abundance.
  $endgroup$
  – Poutnik
  6 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I forgot to ask: If we have a neutral solution can we act and choose the equation as if we want to keep it neutral or is this not necessarily the case?
  $endgroup$
  – w_w
  2 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

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 6 hours ago

answered 8 hours ago

PoutnikPoutnik

1,284310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Apply the same principle of availability and stability.
  $endgroup$
  – Poutnik
  7 hours 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @w_w I mean ability to survive. $ce{OH-}$ or anions of weak acids like $ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides. Generally, choice duch a reaction form, where reactants on both sides may exist in abundance.
  $endgroup$
  – Poutnik
  6 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I forgot to ask: If we have a neutral solution can we act and choose the equation as if we want to keep it neutral or is this not necessarily the case?
  $endgroup$
  – w_w
  2 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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 6 hours ago

answered 8 hours ago

PoutnikPoutnik

1,284310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Apply the same principle of availability and stability.
  $endgroup$
  – Poutnik
  7 hours 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @w_w I mean ability to survive. $ce{OH-}$ or anions of weak acids like $ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides. Generally, choice duch a reaction form, where reactants on both sides may exist in abundance.
  $endgroup$
  – Poutnik
  6 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I forgot to ask: If we have a neutral solution can we act and choose the equation as if we want to keep it neutral or is this not necessarily the case?
  $endgroup$
  – w_w
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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 6 hours ago

answered 8 hours ago

PoutnikPoutnik

1,284310

$endgroup$

share|improve this answer

edited 6 hours ago

answered 8 hours ago

PoutnikPoutnik

1,284310

answered 8 hours ago

PoutnikPoutnik

1,284310

answered 8 hours ago

PoutnikPoutnik

1,284310