This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Here are the answers to your questions: 1a) Step 1: Determine the oxidation states of Cr and C in the reactants and products. In Cr_2O_7^2-: 2(Cr) + 7(-2) = -2 2(Cr) = +12 Cr = +6. In Cr^3+: Cr = +3. Chromium's oxidation state decreases from +6 to +3, so Cr_2O_7^2- is reduced. In C_2O_4^2-: 2(C) + 4(-2) = -2 2(C) = +6 C = +3. In CO_2: C + 2(-2) = 0 C = +4. Carbon's oxidation state increases from +3 to +4, so C_2O_4^2- is oxidized. Step 2: Identify the oxidizing and reducing agents. The species that is reduced is the oxidizing agent. The species that is oxidized is the reducing agent. Oxidizing agent: Cr_2O_7^2- Reducing agent: C_2O_4^2- 1b) The given reaction is: Cr_2O_7^2- + C_2O_4^2- + H^+ → Cr^3+ + CO_2 + H_2O Step 1: Write the unbalanced half-reactions. Oxidation: C_2O_4^2- → CO_2 Reduction: Cr_2O_7^2- → Cr^3+ Step 2: Balance atoms other than O and H. Oxidation: C_2O_4^2- → 2CO_2 Reduction: Cr_2O_7^2- → 2Cr^3+ Step 3: Balance O atoms by adding H_2O. Oxidation: C_2O_4^2- → 2CO_2 (O atoms are already balanced) Reduction: Cr_2O_7^2- → 2Cr^3+ + 7H_2O Step 4: Balance H atoms by adding H^+ (since it's an acidic medium). Oxidation: C_2O_4^2- → 2CO_2 (H atoms are already balanced) Reduction: 14H^+ + Cr_2O_7^2- → 2Cr^3+ + 7H_2O Step 5: Balance charge by adding electrons (e^-). Oxidation: C_2O_4^2- → 2CO_2 + 2e^- (Charge: -2