If the specified ions are both in solution in water, the half reaction is:
SO4-2 + H2O + 2 e-1 => SO3-2 + 2 OH-1.
These are the ions and their charges: Cu+2 SO3-2The charges have to add up to zero, so one +2 copper ion cancels out one -2 sulfite ion. Therefore, the formula is CuSO3.
The brown precipitate is MnO2, it is the product of the following redox in neutral environment:For each half reaction:MnO4- + 4H++ 3e- → MnO2 + 2H2O REDUCTIONHSO3- + H2O → SO42- + 3H+ + 2e- OXIDATIONOverall: Permanaganate half reaction x2 + Bisulfite half reactionx32MnO4- + 3HSO3- + 8H+ + 3H2O + 6e- → 2MnO2 + 3SO42- + 4H2O + 9H+ + 6e-Cancelling like terms:2MnO4- + 3HSO3- → 2MnO2 + 3SO42- + H3O+Note that all species are balanced, all charges are balanced and the brown precipitate MnO2 is accounted for. The reaction also predicts that acid is formed in the process.
Equations that separate the oxidation from the reduction parts of the reaction
it will kill everybody and you will die
Br2 + (2e)- --> 2 Br- 2I- --> I2 + (2e)-
No! Looks like half a combustion reaction and I do not mean the half reaction method. C2H6O + 3O2 -> 2CO2 + 3H2O
MgO + 2HCl --> MgCl2 + H2O
Cl2 --> 2Cl-
An oxidation half-reaction
They show the oxidation an reduction half's of a reaction seperately
A redox half reaction is a reduction or an oxidation reaction. He half reaction does not occur by itself it much be coupled so that he electron released for another to be accepted.
These are the ions and their charges: Cu+2 SO3-2The charges have to add up to zero, so one +2 copper ion cancels out one -2 sulfite ion. Therefore, the formula is CuSO3.
half reaction
The yin and yang needs to keep balanced because in the circle half of it is black and half of it is white so if its not balanced it wont be even.
Oxidant half reaction: 2H+ + 2e- --> H2 Reductant half reaction: Fe --> Fe2+ + 2e- Tribune ions: 2Cl- --> 2Cl-
One-half of Avogadro's Number, or about 3.011 X 1023.
They show the oxidation an reduction halves of a reaction