Question

Obtain the balanced equation, using the lowest possible whole numbers, for the redox reaction in acidic...

Obtain the balanced equation, using the lowest possible whole numbers, for the redox reaction in acidic solution: CH3OH(aq) + Cr2O72−(aq) → CH2O(aq) + Cr3+(aq) How many H+ appear in the final balanced equation, and which side are they on?

A) 8, reactant side

B) 8, product side

C) 14, reactant side

D) 6, product side

E) 12, reactant side

Answer is A. Please explain step by step.

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Homework Answers

Answer #1

Cr2O72- + CH3OH ---------------> Cr3+ + CH2O

oxidation : CH3OH   -------------> CH2O

reduction : Cr2O72-   -------------> Cr+3

1) balance oxidation half reaction

Cr2O72- ------------> 2Cr3+

Cr2O72- -------------> 2Cr3+ + 7H2O

14H+ + Cr2O72- -------------> 2Cr3+ + 7H2O

6e– + 14H+ + Cr2O72- ---------------> 2Cr3+ + 7H2O

balance reduction half reaction :

CH3OH ------------> CH2O

CH3OH -----------> CH2O + 2H+

(CH3OH ------------> CH2O + 2H+ + 2e– ) x 3

3CH3OH --------------> 3CH2O + 6H+ + 6e-

Add the 2 equations, cancelling out the electrons and any other common terms

3CH3OH + 8H+ + Cr2O72- → 2Cr3+ + 7H2O + 3CH2O

so in reactant side 8 H+ appear in the balanced equation.

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