# Redox (1 Viewer)

##### Member
Hey guys, I'm so lost on how to do redox reactions. We did em in year 11 and I completely forgot. Say for this titration q:
25 mL of hydrogen peroxide was made up to 250 mL with water, from this solution 25 mL was titrated against 23.45 mL of 0.0200 mol L-1 potassium permanganate solution.
Calculate the concentration of the original solution of hydrogen peroxide, in g L-1.

How do I figure out what oxides/reduces?

#### Luukas.2

##### Well-Known Member
You can look at the list if standard reduction potentials for typical reactions.

This is an unusual combination as both hydrogen peroxide and permanganate are oxidising agents, but permanganate is much more powerful, so you will have oxidation of hydrogen peroxide to oxygen gas caused by reduction of permanganate. In acidic conditions, the half-equations and overall net ionic equation will be:

H2O2 ------> O2 + 2H+ + 2e-

MnO4- + 8H+ + 5e- ------> Mn2+ + 4H2O

2MnO4- (aq) + 6H+ (aq) + 5H2O2 (aq) ------> 5O2 (g) + 2Mn2+ (aq) + 8H2O (l)
However, the question does not indicate that acid is present and in neutral or basic conditions, the chemistry changes. Most notably, the permanganate tends to reduce to manganese(IV) oxide rather than to manganese(II) ions:

H2O2 + 2OH- ------> O2 + 2H2O + 2e-

MnO4- + 2H2O + 3e- ------> MnO2 + 4OH-

2MnO4- (aq) + 3H2O2 (aq) ------> 3O2 (g) + 2MnO2 (s) + 2H2O (l) + 2OH- (aq)
The stoichiometry for the answer thus depends on the conditions of the titration.

#### lolcti

##### Well-Known Member
I got so scared like that we would be getting doxxed cause of the title, my dumbass cant

##### Member
You can look at the list if standard reduction potentials for typical reactions.

This is an unusual combination as both hydrogen peroxide and permanganate are oxidising agents, but permanganate is much more powerful, so you will have oxidation of hydrogen peroxide to oxygen gas caused by reduction of permanganate. In acidic conditions, the half-equations and overall net ionic equation will be:

H2O2 ------> O2 + 2H+ + 2e-

MnO4- + 8H+ + 5e- ------> Mn2+ + 4H2O

2MnO4- (aq) + 6H+ (aq) + 5H2O2 (aq) ------> 5O2 (g) + 2Mn2+ (aq) + 8H2O (l)
However, the question does not indicate that acid is present and in neutral or basic conditions, the chemistry changes. Most notably, the permanganate tends to reduce to manganese(IV) oxide rather than to manganese(II) ions:

H2O2 + 2OH- ------> O2 + 2H2O + 2e-

MnO4- + 2H2O + 3e- ------> MnO2 + 4OH-

2MnO4- (aq) + 3H2O2 (aq) ------> 3O2 (g) + 2MnO2 (s) + 2H2O (l) + 2OH- (aq)
The stoichiometry for the answer thus depends on the conditions of the titration.
So you base which one oxidises/reduces off their standard reduction potentials? How did you know this: H2O2 ------> O2 + 2H+ + 2e- I can't find it on the HSC chemistry datasheet. Also, idk if this is a dumb question but how do you know what part of the compound to choose, like you only took into account permanganate as an oxidising agent, and not K.

#### Luukas.2

##### Well-Known Member
So you base which one oxidises/reduces off their standard reduction potentials? How did you know this: H2O2 ------> O2 + 2H+ + 2e- I can't find it on the HSC chemistry datasheet. Also, idk if this is a dumb question but how do you know what part of the compound to choose, like you only took into account permanganate as an oxidising agent, and not K.
There's a lot more to redox than is addressed in the syllabus, and what is there isn't covered well.

Oxidation states give you a good idea of what is likely to oxidise or reduce, by looking at what is present in unusual oxidation states.

Potassium permanganate, KMnO4, has K+, Mn7+, and O2-... the manganese(VII) is readily reducible as it gains electrons and goes to a more typical oxidation state.

Hydrogen peroxide, H2O2, has H+ and O-. Typically, one would expect the oxygen to be reduced to its oxide ion form, O2-, but it can also be oxidised to return to oxygen gas with the oxygen atoms in oxidation state 0. The potential for oxygen to be oxidised and reduced from the form in the peroxide ion is seen in the decomposition of hydrogen peroxide:

2 H2O2 (aq) --------> 2 H2O (l) + O2 (g)​

In this disproportionation decomposition, one O atom in each hydrogen peroxide molecule is oxidised and end up in the oxygen gas, whilst the other O atom is reduced and ends up in its -2 state in a water molecule.

A more comprehensive list of standard reduction potentials like this one will include:
• permanganate to manganese(II) under acidic conditions at +1.51 V
• permanganate to manganese(IV) oxide under neutral / basic conditions at +0.60 V
• oxygen to hydrogen peroxide under acidic conditions at +0.695 V
• hydrogen peroxide to water under acidic conditions at +1.763 V
• potassium ions to potassium metal at -2.93 V - showing that reduction of potassium ions certainly won't happen, which is hardly surprising given potassium metal reacts violently with water.
These suggest the acidic conditions answer will be highly favourable, but don't give a clear answer on the neutral / alkaline conditions... but then, the HSC syllabus barely touches redox not under acidic conditions. Whether you need to be able to take processes like

MnO4- to Mn2+
or
CH2CH2OH to CH3COOH​

and convert them to balanced half-equations is not clear - though the skill is useful - but extending it to addressing neutral or alkaline conditions seems improbable.

##### Member
There's a lot more to redox than is addressed in the syllabus, and what is there isn't covered well.

Oxidation states give you a good idea of what is likely to oxidise or reduce, by looking at what is present in unusual oxidation states.

Potassium permanganate, KMnO4, has K+, Mn7+, and O2-... the manganese(VII) is readily reducible as it gains electrons and goes to a more typical oxidation state.

Hydrogen peroxide, H2O2, has H+ and O-. Typically, one would expect the oxygen to be reduced to its oxide ion form, O2-, but it can also be oxidised to return to oxygen gas with the oxygen atoms in oxidation state 0. The potential for oxygen to be oxidised and reduced from the form in the peroxide ion is seen in the decomposition of hydrogen peroxide:

2 H2O2 (aq) --------> 2 H2O (l) + O2 (g)​

In this disproportionation decomposition, one O atom in each hydrogen peroxide molecule is oxidised and end up in the oxygen gas, whilst the other O atom is reduced and ends up in its -2 state in a water molecule.

A more comprehensive list of standard reduction potentials will include:
• permanganate to manganese(II) under acidic conditions at +1.51 V
• permanganate to manganese(IV) oxide under neutral / basic conditions at +0.60 V
• oxygen to hydrogen peroxide under acidic conditions at +0.695 V
• hydrogen peroxide to water under acidic conditions at +1.763 V
• potassium ions to potassium metal at -2.93 V - showing that reduction of potassium ions certainly won't happen, which is hardly surprising given potassium metal reacts violently with water.
These suggest the acidic conditions answer will be highly favourable, but don't give a clear answer on the neutral / alkaline conditions... but then, the HSC syllabus barely touches redox not under acidic conditions. Whether you need to be able to take processes like

MnO4- to Mn2+
or
CH2CH2OH to CH3COOH​

and convert them to balanced half-equations is not clear - though the skill is useful - but extending it to addressing neutral or alkaline conditions seems improbable.
I'm sort of getting it, so we look at the oxidation states in a given compound, find the one furtherst away and then that part oxidises/reduces. So in this question, Iron (II) sulfate was made up in a 250 mL solution. 25 mL of this solution was titrated against 0.020 mol L-1 potassium(VII) manganate solution. The titre needed to oxidise the Iron (II) sulfate was 24.20 mL. Could you go through the process of finding the oxidation/reduction part for this, I find all the calculations easy but this stuff keeps messing me up. Greatly appreciated Luukas.2

#### Luukas.2

##### Well-Known Member
I'm sort of getting it, so we look at the oxidation states in a given compound, find the one furtherst away and then that part oxidises/reduces. So in this question, Iron (II) sulfate was made up in a 250 mL solution. 25 mL of this solution was titrated against 0.020 mol L-1 potassium(VII) manganate solution. The titre needed to oxidise the Iron (II) sulfate was 24.20 mL. Could you go through the process of finding the oxidation/reduction part for this, I find all the calculations easy but this stuff keeps messing me up. Greatly appreciated Luukas.2
It should be potassium manganate(VII), indicating that the manganese atom is present in its +7 oxidation state... it is more commonly called potassium permanganate, KMnO4.

The "per" prefix indicates a higher oxidation state than occurs with the simple -ate ending, typically by the presence of an extra oxide anion. We see this in, for example, the oxy-acids of chlorine:
• Perchloric acid, HClO4, which reacts to form salts containing the perchlorate anion, ClO4-... in both species, the chlorine is in its +7 oxidation state.
• Chloric acid, HClO3, which reacts to form salts containing the chlorate anion, ClO3-... in both species, the chlorine is in its +5 oxidation state.
• Chlorous acid, HClO2, which reacts to form salts containing the chlorite anion, ClO2-... in both species, the chlorine is in its +3 oxidation state.
• Hypochlorous acid, HOCl, which reacts to form salts containing the hypochlorite anion, ClO-... in both species, the chlorine is in its +1 oxidation state.
In your question, which should say that the solution of iron(II) sulfate is acidified, the permanganate is reduced to manganese(II) (typical for acidified conditions) and the iron(II) is oxidised to iron(III).

OXidation: . . . . . . . . . . . . . . . . . . . . Fe2+ ------> Fe3+ + e-

REDuction: . . . . . . . . . . . . . . . . . . . . MnO4- + 8H+ + 5e- ------> Mn2+ + 4H2O

Overall, RED + 5 x OX: . . . . . . . . . . . . . . . . . . . . 5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) ------> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)​

This reaction won't work in neutral or basic conditions because iron(II) hydroxide and iron(III) hydroxide both have extremely low solubility... at pH = 7 at 25 oC, iron(II) hydroxide precipitates once the concentration of Fe2+ reaches 4.87 x 10-3 mol L-1, whilst iron(III) hydroxide will precipitate once the concentration of Fe3+ reaches 2.79 x 10-18 mol L-1.