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
3 answers
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)
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Net Redox Rxn:
Cr2O72- + 6Fe^+2 + 14H^+ = 2Cr^3 + 6Fe^+3 + 7HOH
Cr2O7^2- + 2(3e^-) => 2Cr+3 + 7HOH (Reduction; Cr2O7^2- is the oxidizing agent)
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Net Redox Rxn:
Cr2O72- + 6Fe^+2 + 14H^+ = 2Cr^3 + 6Fe^+3 + 7HOH
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2 answers