To balance the redox equation for the reaction between potassium dichromate K2Cr2O7 and ethanol C2H5OH in acidic medium, we first need to identify the half-reactions for oxidation and reduction.Oxidation half-reaction ethanol to acetic acid :C2H5OH CH3COOHReduction half-reaction dichromate to chromium III ion :Cr2O7^2- 2Cr^3+Now, we balance the half-reactions:1. Balance the atoms other than O and H:Oxidation: C2H5OH CH3COOH already balanced Reduction: Cr2O7^2- 2Cr^3+ already balanced 2. Balance the O atoms by adding H2O:Reduction: Cr2O7^2- 2Cr^3+ + 7H2O3. Balance the H atoms by adding H+:Oxidation: C2H5OH CH3COOH + 2H+Reduction: 14H+ + Cr2O7^2- 2Cr^3+ + 7H2O4. Balance the charges by adding electrons e- :Oxidation: C2H5OH + 2H+ CH3COOH + 2e-Reduction: 14H+ + Cr2O7^2- + 6e- 2Cr^3+ + 7H2O5. Make the number of electrons equal in both half-reactions and add them together:3 C2H5OH + 2H+ CH3COOH + 2e- 2 14H+ + Cr2O7^2- + 6e- 2Cr^3+ + 7H2O 6. Combine the half-reactions:3C2H5OH + 6H+ + 28H+ + 2Cr2O7^2- 3CH3COOH + 6e- + 4Cr^3+ + 14H2O + 12e-7. Simplify and cancel out common species:3C2H5OH + 34H+ + 2Cr2O7^2- 3CH3COOH + 4Cr^3+ + 14H2OThis is the balanced redox equation for the reaction between potassium dichromate and ethanol in acidic medium.To calculate the oxidation number of chromium in the dichromate ion before and after the reaction:Before the reaction:In Cr2O7^2-, the oxidation number of oxygen is -2. Let the oxidation number of chromium be x.2x + 7 -2 = -22x - 14 = -22x = 12x = +6So, the oxidation number of chromium in the dichromate ion before the reaction is +6.After the reaction:In Cr^3+, the oxidation number of chromium is +3.The chromium in the dichromate ion is reduced from an oxidation state of +6 to +3 during the reaction.