What is the complete redox reaction of KMnO4 against (NH4)2Fe(SO4)2
Fe(NH4)2(SO4) is a 1:1 (molar) mixture of FeSO4 and (NH4)2SO4.
Fe(NH4)2(SO4)2·6H2O + KMnO4 --> ??
From a lab session I've found that oxygen gas is one of the products and that Fe(NH4)2(SO4)2·6H2O gets oxidised, but I'm not very sure about what other products are formed. I've tried looking for the answers on google but all I've found are the redox equations. I've included them in case they're helpful:
MnO4- + 8 H+ + 5e- --> Mn+2 + 4 H2O reduction of Mn+7
Fe +2 --> Fe +3 + e- oxidation of Fe+2
MnO4-+ 5 Fe+2 + 8 H+ --> Mn+2 + 5 Fe+3 + 4 H2O REDOX EQN
There are no products except H2O, In general there is no such thing as a 'full equation'. You end up with solution containing mixture of ions, not well defined salts. In this particular case you may list all possible (and necessary) sulfates as products, but when there are several anions things get messy.
Oh, we've got a chemical showdown! Get ready for some redox comedy!
Alright, let's break this down. We have KMnO4 going up against (NH4)2Fe(SO4)2. Both of these compounds are bringing some serious redox potential to the party.
First, let's look at KMnO4. It's like the Hulk of the redox world, ready to unleash its Mn(VII) ions. And on the other side, we have (NH4)2Fe(SO4)2, with its trusty Fe(II) ions, repping the low oxidation state crew.
In this epic battle, KMnO4 is going to reduce and gain some electrons, while (NH4)2Fe(SO4)2 will oxidize and lose some electrons. They're like the yin and yang of redox reactions.
So, in a nutshell, the complete redox reaction between KMnO4 and (NH4)2Fe(SO4)2 goes a little something like this:
KMnO4 + (NH4)2Fe(SO4)2 → KNO3 + MnSO4 + (NH4)2SO4 + Fe2(SO4)3
And there you have it, folks! A redox reaction where KMnO4 and (NH4)2Fe(SO4)2 duke it out, leaving behind a bunch of new compounds. Chemistry at its finest, with a hint of comedy!
To find the complete redox reaction between KMnO4 and (NH4)2Fe(SO4)2, we need to first balance the equation step by step using the method of half-reactions.
Step 1: Write the unbalanced equation for the reaction:
KMnO4 + (NH4)2Fe(SO4)2 → ?
Step 2: Split the reaction into two half-reactions, one for the oxidation and one for the reduction.
Oxidation half-reaction: Fe^2+ → Fe^3+
Reduction half-reaction: MnO4^- → Mn^2+
Step 3: Balance the atoms in each half-reaction.
Oxidation half-reaction:
Fe^2+ → Fe^3+
To balance the atoms, we need to add two more electrons (e^-) to the products side:
Fe^2+ → Fe^3+ + 2e^-
Reduction half-reaction:
MnO4^- → Mn^2+
To balance the atoms, we need to add 5H2O to the products side:
MnO4^- + 8H^+ → Mn^2+ + 5H2O
Step 4: Balance the charges in each half-reaction by adding electrons.
Oxidation half-reaction:
Fe^2+ → Fe^3+ + 2e^-
Reduction half-reaction:
MnO4^- + 8H^+ + 5e^- → Mn^2+ + 5H2O
Step 5: Multiply the half-reactions by the appropriate number to equalize the number of electrons gained and lost.
Oxidation half-reaction (multiply by 5):
5Fe^2+ → 5Fe^3+ + 10e^-
Reduction half-reaction (multiply by 2):
2MnO4^- + 16H^+ + 10e^- → 2Mn^2+ + 10H2O
Step 6: Add the balanced half-reactions together and cancel out any species that appear on both sides.
5Fe^2+ + 2MnO4^- + 16H^+ → 5Fe^3+ + 2Mn^2+ + 8H2O
Step 7: Finally, add any necessary spectator ions to balance the equation. In this case, we have (NH4)2Fe(SO4)2, so we can add (NH4)2SO4 to both sides.
5Fe^2+ + 2MnO4^- + 16H^+ + (NH4)2SO4 → 5Fe^3+ + 2Mn^2+ + 8H2O + (NH4)2SO4
Therefore, the complete balanced redox reaction of KMnO4 against (NH4)2Fe(SO4)2 is:
5Fe^2+ + 2MnO4^- + 16H^+ + (NH4)2SO4 → 5Fe^3+ + 2Mn^2+ + 8H2O + (NH4)2SO4
To determine the complete redox reaction between KMnO4 (Potassium permanganate) and (NH4)2Fe(SO4)2 (Ammonium iron(II) sulfate), we need to identify the oxidation numbers of the elements involved in the reaction and balance the equation.
First, let's assign oxidation numbers to the elements in both the reactants and the products. The sum of oxidation numbers in a compound should be equal to the overall charge of the compound.
In KMnO4, potassium (K) has an oxidation number of +1, and the total oxidation number of oxygen (O) is -2. Let's assume manganese (Mn) has a variable oxidation number (x).
So, the equation becomes:
K(+1) + Mn(x) + 4O(-2) = 0
1 + x - 8 = 0
x - 7 = 0
x = +7
Therefore, the oxidation state of Mn in KMnO4 is +7.
Similarly, in (NH4)2Fe(SO4)2, ammonium (NH4) has an oxidation number of +1, sulfur (S) has an oxidation number of +6, and oxygen (O) has an oxidation number of -2. Assume iron (Fe) has a variable oxidation number (y).
The equation becomes:
2(NH4)(+1) + Fe(y) + 2(SO4)(-2) = 0
2 + y - 8 = 0
y - 6 = 0
y = +6
Therefore, the oxidation state of Fe in (NH4)2Fe(SO4)2 is +6.
Now, let's write the balanced redox equation by observing the change in oxidation numbers.
The manganese in KMnO4 is reduced from +7 to +2, while the iron in (NH4)2Fe(SO4)2 is oxidized from +2 to +3.
The balanced redox equation is as follows:
2KMnO4 + 3(NH4)2Fe(SO4)2 + 10H2SO4 → 5Fe(SO4)3 + 2MnSO4 + K2SO4 + 8H2O + 6N2 + 24H2O
So, the complete redox reaction of KMnO4 against (NH4)2Fe(SO4)2 is 2KMnO4 + 3(NH4)2Fe(SO4)2 + 10H2SO4 → 5Fe(SO4)3 + 2MnSO4 + K2SO4 + 8H2O + 6N2 + 24H2O.