How do you balance the chemical reactions?

Answer 1

SO42- + 4H+ + 2e- SO2(g) + 2H2O(l)

Answer 2

To illustrate I will use the example of the oxidation of Cu^+ to Cu^2+ with permanganate.

Write each half equation out (if they have not been provided for you). Oxygen atoms require the addition of 2 H^+ ions each, to produce water. You need to know what each species is being oxidised or reduced to in order to do this.

In this example permanganate is being reduced to Mn^2+

Cu^+ -> Cu^2+ + e^-

MnO4^- + 8H^+ +5e^- -> Mn^2+ + 4H2O

Then work out the ratios of the reactants based on the ratios of electrons (1:5 in this case) and add the equations together. The electrons will cancel out.

MnO4^- +5Cu^+ 8H^+ -> 5Cu^2+ + Mn^2+ + 4H2O

Answer 3

I'm not quite sure what you mean by "balance a reaction in acidic solution." The only thing I can think of that you could mean is how do balance a redox reaction in acidic solution.

For specific instructions on how to balance a redox reaction in both acidic and basic solutions, see the Related Questions links to the left.

Answer 4

Getting the proportions right between the 2 sides. You have to have the same number of each atom on both sides (law of conversation of matter). 2O2+C4 = 4CO2 You have 4 oxygen atoms, and 4 carbon atoms. When you combine them you get 8 atoms all together. They are the same on both sides. Say you had 2 of the carbon molecules though. 2O2+2C4 = 4CO2 Then the sides wouldn't be equal and you would need to have 4 dioxide molecules on the reactants side and 8 carbon dioxide molecules on the products. 4O2+2C4 = 8CO2 So that both sides would have 16 atoms.

Answer 5

you should draw a table: left right Na+H2O NaOJ+H2 its just a matter of trial and error till both sides elements are equal.

Answer 6

by calculation of difference in oxidation numbers for each atom on both sides of reaction