can anyone teach me how to do redox (equations) step by step? i dont get any of it. 10th grade chemistry?
- 1 month ago
guys please help me on my questions please
- lenpol7Lv 72 months ago
The Classic Example is iron(Fe) and potassium dichromate. in acidified solution.
Taking the dichromate half equation first
Cr2O7^2- Remember oxygen is always '-2' oxidation state.
Hence"cr + 7(-2) = -2
2Cr - 14 = -2
2Cr = 12
Cr = +6 (oxidation state) this is reduced to (+3)
Hence Cr2O7^2- + 14H^+ 6e^- = 2Cr^3+ 7H2O
Notice the Cr is now 2Cr^3+
The 7 oxygens become 7 water
We 'back' calculate the hydrogens from the water as '14'
Then we balance the charges
-2 + 14 -(x)e = 2(+3)
-2 + 14 - xe^- = +6
xe^- = +6 - 14 + 2 = -6
So the 'x' of electrons is '6'
So the half equation is
Cr2O7^2- + 14H^+ 6e^- = 7H2O + 2Cr^3+ is the reductant side.
This will oxidise Iron(II) to Iron)III)
The half equation for iron is
Fe^2+ = Fe^3+ + e^-
We notice that there are 6 electrons in the dichromate half equation, so we need 6 e^- in the iron half equation.
Hence 6Fe^2+ = 6Fe^3+ + 6e^-
We now have two half equations viz.
Cr2O7^2- + 14H^+ 6e^- = 7H2O + 2Cr^3+
6Fe^2+ = 6Fe^3+ + 6e^-
We now add the two half equations, this will eliminate the electrons (e^-)
Cr2O7^2- + 14H^+ + 6Fe^2+ = 2Cr^3+ + 7H2O + 6Fe^3+
Notice that the elements/atoms balance, and more particularly the charges balance, at '+26' on both sides.
Hope that helps!!!!
- davidLv 72 months ago
If you want step-by-step .. Try Kahn Academy .. an online teaching site. It could be a very detailed set of instructions ... Give a google search for Kahn Academy and the lesson you need .. Redox Reactions
- How do you think about the answers? You can sign in to vote the answer.
- Anonymous2 months ago
Since your Google must be broken I Googled it for you. Here’s a 4 minute video. Pay attention.Source(s): I hope you fix your Google.