What is the oxidation half reaction for the following equation: Cr2O7^-2 +Fe+2 - - > Cr^+3 + Fe^+3
Cr^+3 - - > Cr2O7^-2
Fe^+2 - - > Fe^+3
Fe^+3 - - > Fe^+2
Cr2O7^-2 - -> Cr^+3***
Did you leave off the electrons? What you have check is the reduction reaction but that isn't right because the electrons have been omitted.
Oxidation is the loss of eletrons.
Fe^2+ ==> Fe^3+ + e is the oxidation half reaction.
Also note that neither A nor D are balanced. They don't have the right number of Cr atoms, nor O atoms, and the charges don't balance.
Something is wrong with the Zn/Cu question above. You have answered it correctly but the computer won't show my answer when I try to post it.
The oxidation half-reaction in the given equation is:
Cr2O7^-2 → Cr^+3
To determine the oxidation half-reaction in the given equation, we need to identify which species is being oxidized and which species is being reduced.
In this case, we can see that the chromium ion (Cr^+3) is being reduced to the dichromate ion (Cr2O7^-2). Therefore, the half-reaction for the reduction of chromium can be written as:
Cr2O7^-2 - -> Cr^+3
This half-reaction shows the conversion of Cr2O7^-2 to Cr^+3 by gaining 3 electrons.
By contrast, the iron ion (Fe^+2) is being oxidized to the iron(III) ion (Fe^+3). Therefore, the half-reaction for the oxidation of iron can be written as:
Fe^+2 - - > Fe^+3
This half-reaction shows the conversion of Fe^+2 to Fe^+3 by losing 1 electron.
It's important to note that the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction in order to balance the overall equation.