No equation above, but this is an example.
2H+ + 2e- + 1/2O2 --> H2O
This is the equation, of the electron transport chain, that shows the final electron acceptor, oxygen, taking two protons and two electrons and forming water.
This,
2e-
represents the two electrons and the negative charge they carry.
The balanced equation for the corrosion of copper is: 2Cu(s) + O2(g) + H2O(l) + 2e- -> 2Cu2+(aq) + 2OH-(aq)
The balanced chemical equation for the conversion of H2(g) to H(aq) and H-(aq) is: 2H2(g) - 2H(aq) 2e-
Balanced chemical equation: Zn (s) + 2HCl (aq) --> ZnCl2 (aq) + H2 (g) Oxidation half-reaction: Zn (s) --> Zn2+ (aq) + 2e- Reduction half-reaction: 2H+ (aq) + 2e- --> H2 (g)
The formation of an iron(II) ion involves the loss of two electrons from a neutral iron atom. The equation for this process can be written as Fe → Fe^2+ + 2e-.
The balanced half reaction for the oxidation of hydrogen peroxide is: 2H2O2 - O2 2H 2e-
Br2 + (2e)- --> 2 Br- 2I- --> I2 + (2e)-
The balanced equation for the corrosion of copper is: 2Cu(s) + O2(g) + H2O(l) + 2e- -> 2Cu2+(aq) + 2OH-(aq)
The balanced chemical equation for the conversion of H2(g) to H(aq) and H-(aq) is: 2H2(g) - 2H(aq) 2e-
oxidation
Balanced chemical equation: Zn (s) + 2HCl (aq) --> ZnCl2 (aq) + H2 (g) Oxidation half-reaction: Zn (s) --> Zn2+ (aq) + 2e- Reduction half-reaction: 2H+ (aq) + 2e- --> H2 (g)
No
So42- + 4H+ + 2e- goes to SO2(g) + 2H20(I)
So42- + 4H+ + 2e- goes to SO2(g) + 2H20(I)
CIO H2O 2e- CI- 2OH
The formation of an iron(II) ion involves the loss of two electrons from a neutral iron atom. The equation for this process can be written as Fe → Fe^2+ + 2e-.
The balanced half reaction for the oxidation of hydrogen peroxide is: 2H2O2 - O2 2H 2e-
The half-reaction equation representing the reduction of iron (II) ion (Fe^2+) is: Fe^2+ + 2e^- -> Fe