Electron Transfer Theory
Write an label the oxidation and reduction half-reaction equations.
a) Ni(s) + Cu(NO3)2(aq) -> Cu(s)+ Ni(No3)2(aq)
oxidation - Ni(s) -> Ni2+(aq) + 2e-
reduction - Cu2+(aq) + 2e- -> Cu(s)
b) Pb(s) + Cu(NO3)2(aq) -> Cu(s) + Pb(NO3)2(aq)
oxidation - Pb(s) -> Pb2+(aq) + 2e-
reduction - Cu2+(aq) +2e -> Cu(s)
c) Ca(s) + 2HNO3(aq) -> H2(g) + Ca(NO3)2(aq)
oxidation - Ca(s) -> Ca2+(aq) + 2e-
reduction - 2H+(aq) + 2e- -> H2(g)
d) 2Al(g)+ Fe2O3(s) -> 2Fe(l) + Al2O3(s)
oxidation - 2Al ->
I have no idea what to do with the last one. If I made any mistakes or did anything wrong can someone correct them.
Oxidation Numbers
Assign oxidation numbers to chloride in each of the follow chemicals.
HCl = +1?
Cl2 = 0?
NaClO = ?
Cl- = -1?
HClO3 = either +1 or -3?
ClO3- = -3?
KClO2 - +2?
ClO2 = +2?
HClO4 = +4?
I have no idea if I'm doing these right. Can someone explain oxidation numbers and correct those?
Also can someone do an example in this and explain how they got it. I don't understand the concept.
Assign oxidation numbers to manganese in each of the following chemicals.
MnO4^2-(aq)
Use oxidation numbers to identify the oxidation and reduction atoms
MnO4-(aq) + H2Se(g) + H+(aq) -> Se(s) + Mn2+(aq) + H2O(l)
Can someone please do those examples and explain how they got the answers because I don't understand oxidation numbers.
The first three equations are ok.
d) 2Al(g)+ Fe2O3(s) -> 2Fe(l) + Al2O3(s)
oxidation - 2Al ->
Al goes from 0 oxidation state on the left to zero on the right so it is oxidized. Fe goes from +3 on the left to 0 on the right so it is reduced. I assume you can write the half equations.
have no idea what to do with the last one. If I made any mistakes or did anything wrong can someone correct them.
Oxidation Numbers
Assign oxidation numbers to chloride in each of the follow chemicals.
HCl = +1? No, H is +1 so Cl is -1
Cl2 = 0? right
NaClO = ? Na is +1, O is -2; therefore, Cl is what to make NaClO zero. Cl must be +1.
Cl- = -1? right
HClO3 = either +1 or -3? H is +1, O is -2, there are 3 of them to make total on O of -6; therefore, what must Cl be to have HClO3 zero. Cl must be +5
ClO3- = -3? +5. O is -6 and Cl must be +5 to leave a -1 charge on the ion.
KClO2 - +2? K is +1, O is -4, therefore, Cl must be +3
ClO2 = +2? Cl must be +4 to zero out the -4 from O2
HClO4 = +4? H is +1, O is -8; therefore, Cl must be +7
I have no idea if I'm doing these right. Can someone explain oxidation numbers and correct those?
Also can someone do an example in this and explain how they got it. I don't understand the concept.
Assign oxidation numbers to manganese in each of the following chemicals.
MnO4^2-(aq) O is -8 so Mn must be +7 to leave a -1 charge on the ion.
Use oxidation numbers to identify the oxidation and reduction atoms
MnO4-(aq) + H2Se(g) + H+(aq) -> Se(s) + Mn2+(aq) + H2O(l)
Mn goes from +7 on the left to +2 on the right. Se goes from -2 on the left to 0 on the right.
Can someone please do those examples and explain how they got the answers because I don't understand oxidation numbers.
I think you have guessed at some of the answers and we don't want to do that, do we?. I think I have a good site with minimal rules that can clear this up for you if you want it. I hope I turned off the bold for your work and used bold for mine. If I didn't I assume you will know what you typed and what I typed, bold or not
Minimal but simple rules to live by for redox. Oxidation numbers are nothing more than a book keeping method.
http://dbhs.wvusd.k12.ca.us/webdocs/Redox/Meaning-of-Redox.html
(Broken Link Removed)
To label the oxidation and reduction half-reaction equations, you need to identify the species that lose and gain electrons in the chemical equation. Here are the solutions for the given reactions:
a) Ni(s) + Cu(NO3)2(aq) -> Cu(s) + Ni(No3)2(aq)
Oxidation half-reaction: Ni(s) -> Ni2+(aq) + 2e-
Reduction half-reaction: Cu2+(aq) + 2e- -> Cu(s)
b) Pb(s) + Cu(NO3)2(aq) -> Cu(s) + Pb(NO3)2(aq)
Oxidation half-reaction: Pb(s) -> Pb2+(aq) + 2e-
Reduction half-reaction: Cu2+(aq) + 2e- -> Cu(s)
c) Ca(s) + 2HNO3(aq) -> H2(g) + Ca(NO3)2(aq)
Oxidation half-reaction: Ca(s) -> Ca2+(aq) + 2e-
Reduction half-reaction: 2H+(aq) + 2e- -> H2(g)
d) 2Al(g)+ Fe2O3(s) -> 2Fe(l) + Al2O3(s)
Oxidation half-reaction: 2Al -> 2Al3+(aq) + 6e-
Reduction half-reaction: Fe2O3(s) + 6e- -> 2Fe(l)
For the oxidation numbers:
HCl: Since hydrogen usually has an oxidation number of +1, and chloride is a halogen with a typical oxidation number of -1, the oxidation number of Cl in HCl is -1.
Cl2: Chlorine molecules are uncharged, so each Cl atom has an oxidation number of 0.
NaClO: Since the oxidation number of Na is typically +1 and oxygen is usually -2, we can solve for Cl's oxidation number using the equation: (+1) + Cl + (-2) = 0. Therefore, Cl's oxidation number in NaClO is +1.
Cl-: In ionic compounds, as in this case where Cl has gained an electron, the oxidation number of Cl is -1.
HClO3: Let's solve for Cl's oxidation number by using the equation: (+1) + Cl + 3(-2) = 0. Therefore, Cl's oxidation number in HClO3 is +5.
ClO3-: Let's solve for Cl's oxidation number by using the equation: Cl + 3(-2) = -1. Therefore, Cl's oxidation number in ClO3- is +5.
KClO2: Let's solve for Cl's oxidation number by using the equation: (+1) + Cl + 2(-2) = 0. Therefore, Cl's oxidation number in KClO2 is +3.
ClO2: Let's solve for Cl's oxidation number by using the equation: Cl + 2(-2) = 0. Therefore, Cl's oxidation number in ClO2 is +4.
HClO4: Let's solve for Cl's oxidation number by using the equation: (+1) + Cl + 4(-2) = 0. Therefore, Cl's oxidation number in HClO4 is +7.
For the example with manganese:
MnO4^2-(aq)
Since the overall charge of the ion is 2-, and oxygen is usually -2, we have (4 * O) + 2 + (x) = -2. Solving for x, we get x = +7. So, manganese's oxidation number in MnO4^2-(aq) is +7.
For the example with the oxidation and reduction atoms:
MnO4-(aq) + H2Se(g) + H+(aq) -> Se(s) + Mn2+(aq) + H2O(l)
In MnO4-, the oxidation number of oxygen is -2. Since there are four oxygen atoms, the overall oxidation number from oxygen is -8. In order to balance the equation, the sum of the oxidation numbers must equal the overall charge. Therefore, the oxidation number of manganese (Mn) is +7.
In H2Se, hydrogen typically has an oxidation number of +1. Since selenium (Se) is less electronegative than hydrogen, selenium has an oxidation number of -2. Therefore, the oxidation number of hydrogen changes from +1 to 0, indicating reduction.
In H+, hydrogen typically has an oxidation number of +1. In this case, it remains +1.
In Se, selenium typically has an oxidation number of -2. In this case, it remains -2.
In Mn2+, the overall charge of the ion is 2+. Therefore, the oxidation number of manganese (Mn) is +2.
In H2O, hydrogen typically has an oxidation number of +1 and oxygen typically has an oxidation number of -2. In this case, hydrogen remains +1 and oxygen remains -2.
So, in the given reaction, Mn and H2Se undergo oxidation, while Se and H2O undergo reduction.