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
I don't agree.
6Fe^+2 => 6Fe^+3 + 6e^- (Oxidation; Fe^+2 is the Reducing Agent)
Cr2O7^2- + 2(3e^-) => 2Cr+3 + 7HOH (Reduction; Cr2O7^2- is the oxidizing agent)
----------------------------------
Net Redox Rxn:
Cr2O72- + 6Fe^+2 + 14H^+ = 2Cr^3 + 6Fe^+3 + 7HOH
for more DrReb048(at)g m a i l (dot) c o m
To determine the oxidation half-reaction for the given equation:
1. Identify the elements that change their oxidation state from one side of the equation to the other. In this case, the elements that change are chromium (Cr) and iron (Fe).
2. Write the half-reaction for each element, showing the changes in oxidation states. The half-reaction for chromium is:
Cr^+3 → Cr2O7^-2
The half-reaction for iron is:
Fe^+2 → Fe^+3
3. Determine which half-reaction represents oxidation. In oxidation reactions, the oxidation state of an element increases.
In this case, the oxidation state of chromium increases from +3 to +6, indicating that it undergoes oxidation.
4. Therefore, the oxidation half-reaction for the given equation is:
Cr^+3 → Cr2O7^-2
The correct answer is: Cr^+3 → Cr2O7^-2