Balance the chemical equation for the following redox reaction under acidic aqueous conditions with the smallest whole-number coefficients possible using the half-reaction method.What is the coefficient of C₂H₅OH? Cr₂O₇²^-(aq)+ C₂H₅OH(aq)$\to$ Cr³⁺(aq)+ CO₂(g)

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The half-reaction for the reduction of Cr₂O₇^2- is:Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂OThe half-reaction for the oxidation of C₂H₅OH is:C₂H₅OH + 2H₂O → 2CO₂ + 6H⁺ + 6e^-To balance the number of electrons transferred, we multiply the first half-reaction by 3 and the second half-reaction by 1:3Cr₂O₇^2- + 42H⁺ + 18e^- → 6Cr³⁺ + 21H₂OC₂H₅OH + 2H₂O → 2CO₂ + 6H⁺ + 6e^-We can now add the two half-reactions to get the overall balanced equation:3Cr₂O₇^2- + C₂H₅OH + 44H⁺ → 6Cr³⁺ + 2CO₂ + 23H₂OThe coefficient of C₂H₅OH is 1.

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