Oxidation is loss of electrons and reduction is gain of electrons. In this case the Cl2 molecule is changed to two Cl- ions by gaining electrons, so the Cl2 is reduced. The Br in the NaBr salt is in the form of Br- ions. The Br- ions are changed to neutral Br atoms (a Br2 molecule) by losing electrons and so are oxidised.
Reduced means 'gain of electrons' so Cl2 was reduced since it went from having 17 e- each to 18. (Cl-)
3 mole FeCl2 will react with 6 mole NaOH (stoechiometric mole ratio: 1 FeCl2 to 2 NaOH), so 3 mole FeCl2 (= 6 added -3 used) will be left over.
Plus twelve volts if you're in physics.
Sodium carbonate plus hydrochloric acid gives sodium chloride plus water plus carbon dioxide.
No reaction
it = 6153
my reactant is Cl2 plus 2NaBr, my product is 2NaBr2. Which one is oxidized, what is reduced. How can I tell which one is which?
I think this is right... Cl2 + 2NaBr = 2NaCl + Br2
NAD+ is the oxidized and NADH is the reduced form.
2NaBr (s) + Cl2 (g) --------> 2NaCl (s) + Br2 (g)
Cu is being reduced and Zn is being oxidized, hence Zn + CuSO4 --> ZnSO4 + Cu
Cu is oxidized. The oxidation number goes from 0 in Cu to +2 in CuSO4. S is reduced. The oxidation number goes from +6 in H2SO4 to +4 in SO2. The oxidizing agent is H2SO4 since it causes Cu to be oxidized. The reducing agent is Cu since it causes S in H2SO4 to be reduced.
NAD+ isn't oxidised, it can be reduced by H to form NADH
Tin and iron are being oxidized because to be oxidized is to lose electrons. Chlorine is being reduced, because it gains electrons.
2NaCl + H2SO4 = Na2SO4 + 2HCl
Br2 + 2NaI -> 2NaBr + I2
The type of reaction represented by Zn plus 2NaCl yields 2Na plus ZnCl2 is an impossible one. It would be the other way around. 2Na + ZnCl2 --------> 2NaCl + Zn
2NaCl + Ni (for A+)