Do this by either the oxidation method or half reactionmethod. Cr 2 O 7 2- +CH 3

Question

Do this by either the oxidation method or half reactionmethod. Cr2O72- +CH3CH2OH + H+--->Cr3+ + CH3COOH + H2O Do this by either the oxidation method or half reactionmethod. Cr2O72- +CH3CH2OH + H+--->Cr3+ + CH3COOH + H2O

Explanation / Answer

Cr2O72- +CH3CH2OH ---> Cr3+ +CH3COOH Separate into oxidation and reduction halfreactions Oxidation: CH3CH2OH --> CH3COOH Reduction: Cr2O72---> Cr3- Balance allelements besides oxygen and hydrogen. Then balance oxygen by addingwater to the side that isdeficient Oxidation:CH3CH2OH + H2O --> CH3COOH Reduction: Cr2O72--->2Cr3-+ 7H2O Balance the H by adding H+ to the sidethat is deficient Oxidation: CH3CH2OH + H2O--> CH3COOH + 4H+ Reduction: Cr2O72- +14H+ -->2Cr3- + 7H2O Balance charge by adding electrons to themore positive side of theequations Oxidation:CH3CH2OH + H2O -->CH3COOH + 4H++4e- Reduction: Cr2O72- +14H+ + 18e- -->2Cr3- + 7H2O Multiply a half-reaction by an integervalue so that the number of electrons is thesame Oxidation:18[CH3CH2OH +H2O --> CH3COOH +4H++ 4e-]= 18CH3CH2OH +18H2O -->18CH3COOH +72H++72e- Reduction: 4[Cr2O72-+ 14H+ + 18e- --> 2Cr3-+ 7H2O] = 4Cr2O72- +56H+ +72e- -->8Cr3-+ 28H2O Combine reactions into one. Cancel whatoccurs simultaneously on the left andright 4Cr2O72-+ 18CH3CH2OH--> 4Cr3- +18CH3COOH + 10H2O + 16H+

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