I'm having a little trouble determining oxidation numbers for elements in a redox reaction
KMnO4+HCl--> MnCl2+Cl2+H2O+KCl
So far what I have for the reactants side is: K= +1 Mn= +7 O4= -2 H= +1 Cl= -1
And for the products side I have:
Mn= +2 Cl2= -1 Cl2(diatomic)= 0 H2=+2 O= -2 K= +1 Cl= -1
Please tell me if these are right, and I wasn't sure about H2, but I'm guessing it could either be +2 or +1, since it was H20 I think it would be +1, but I am not entirely sure :(
You are right. It is always better to determine the oxidation state of EACH atom; therefore, H in H2O is +1 EACH and +2 total. O is -2 for the one. In that vein, Cl in MnCl2 is -1 each and -2 total.
Ah, ok, so another question I have is how would you determine which elements are oxidized and reduced?
You remember the definitions. Oxidation is the loss of electrons; reduction is the gain of electrons.
So Mn changes from +7 to +2 which is a gain of electrons. That means MnO4^- is reduced to Mn^2+.
Cl changes from -1 in HCl to zero in Cl2; that is a loss of electrons so Cl^- is oxidized (that's for Cl in HCl going to Cl2. The other Cl ions in HCl that go to MnCl2 are spectator ions since they don't change oxidation state.
Thanks for your help! Makes lots of sense :)
To determine the oxidation numbers of elements in a redox reaction, you need to apply some rules:
1. For elements in their elemental form (e.g., Cl2, O2, H2), the oxidation number is always zero. This means that your oxidation numbers for Cl2 and O2 are correct with 0.
2. The oxidation number of an alkali metal ion (Group 1A) is +1, so your oxidation number for K is correct with +1.
3. In compounds, oxygen usually has an oxidation number of -2. This applies to the O4 in KMnO4. There are four oxygen atoms, so their combined oxidation number is -8. Since the compound is neutral, the sum of the oxidation numbers must be zero. Therefore, the oxidation number for Mn can be calculated by the equation: (+7) + (-8*4) = -1. Thus, your oxidation number for Mn in KMnO4 is correct with +7.
4. Hydrogen generally has an oxidation number of +1 when bonded to nonmetals and -1 when bonded to metals. In HCl, Cl has an oxidation number of -1 (as you correctly noted) and H must therefore have an oxidation number of +1.
Recapping the reactants side of the equation:
K = +1
Mn = +7
O4 = -2
H = +1
Cl = -1
Now, let's analyze the products side:
1. MnCl2: Since Cl has an oxidation number of -1 in HCl on the reactants side, it retains that oxidation number in the product MnCl2. Therefore, the oxidation number for Mn in MnCl2 can be calculated: (+1) + (-1*2) = 0. Thus, your oxidation number for Mn in MnCl2 is correct with +2.
2. Cl2: As mentioned earlier, elemental forms have an oxidation number of zero. So, your oxidation number for Cl2 is correct with 0.
3. H2O: In H2O, O has an oxidation number of -2, as usual. Since the compound is neutral, the sum of the oxidation numbers must be zero. Therefore, the oxidation number for H in H2O can be calculated by the equation: (+2) + (-2) = 0. Thus, your oxidation number for H in H2O is correct with +2.
4. KCl: K has an oxidation number of +1 (as it had in the reactant side) and Cl retains its oxidation number of -1. So, your oxidation numbers for K and Cl in KCl are correct with +1 and -1, respectively.
Recapping the products side of the equation:
Mn = +2
Cl2 = 0
H2 = +2
O = -2
K = +1
Cl = -1
Regarding your question about the oxidation number for H2, in H2O, hydrogen has an oxidation number of +1, not +2. This is because oxygen is more electronegative than hydrogen, so hydrogen donates its electron to oxygen, resulting in an oxidation number of +1. Therefore, your assumption that H2 is +1 is correct.
Overall, your oxidation numbers for the reactants and products in the given redox reaction are correctly deduced.