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)
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
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