K2Cr2О7, + KI+ HSO4 = Cr2(SO4)3 + I2 + K2S04 + H20

Let's analyze and balance the chemical equation: **Unbalanced equation:** \[ \text{K}_2\text{Cr}_2\text{O}_7 + \text{KI} + \text{HSO}_4^- \rightarrow \text{Cr}_2(\text{SO}_4)_3 + \text{I}_2 + \text{K}_2\text{SO}_4 + \text{H}_2\text{O} \] --- ### Step 1: Assign oxidation states - K₂Cr₂O₇: K is +1, O is -2, Cr is +6 - KI: I is -1 - HSO₄⁻: H +1, S +6, O -2 - Cr₂(SO₄)₃: Cr +3, S +6 - I₂: I is 0 - K₂SO₄: K +1, S +6, O -2 - H₂O: H +1, O -2 ### Step 2: Identify what's oxidized and reduced - Iodide (I⁻) in KI is oxidized to I₂ (0) - Chromium (Cr⁶+ in dichromate) is reduced to Cr³+ in Cr₂(SO₄)₃ ### Step 3: Write half-reactions **Oxidation:** \[ 2 I^- \rightarrow I_2 + 2 e^- \] **Reduction:** \[ \text{Cr}_2\text{O}_7^{2-} + 14 H^+ + 6 e^- \rightarrow 2 Cr^{3+} + 7 H_2O \] *(Note: In basic or neutral medium, H⁺ can be balanced with H₂O or OH⁻)* **Since HSO₄⁻ is present, the medium is slightly acidic, so we'd balance accordingly.** ### Step 4: Balance the overall reaction **Assuming acidic medium:** - The chromate reduction: \[ \text{Cr}_2\text{O}_7^{2-} + 14 H^+ + 6 e^- \rightarrow 2 Cr^{3+} + 7 H_2O \] - Iodide oxidation: \[ 2 I^- \rightarrow I_2 + 2 e^- \] Multiply the oxidation half-reaction by 3 to balance electrons with reduction: \[ 6 I^- \rightarrow 3 I_2 + 6 e^- \] Now, combine: \[ \text{Cr}_2\text{O}_7^{2-} + 14 H^+ + 6 e^- + 6 I^- \rightarrow 2 Cr^{3+} + 7 H_2O + 3 I_2 \] --- ### Step 5: Incorporate the reactants and products - Recognize that K₂Cr₂O₇ provides Cr₂O₇²⁻ - K and other ions are spectator ions balancing charge and atoms Given the initial reactants and final products, the balanced overall reaction in acidic solution is: \[ K_2Cr_2O_7 + 6 KI + 8 H_2SO_4 \rightarrow 2 Cr_2(SO_4)_3 + 6 I_2 + K_2SO_4 + 4 H_2O \] --- ### **Final balanced equation:** \[ \boxed{ K_2Cr_2O_7 + 6 KI + 8 H_2SO_4 \rightarrow 2 Cr_2(SO_4)_3 + 6 I_2 + K_2SO_4 + 4 H_2O } \] --- ### **Summary:** - 1 molecule of potassium dichromate reacts with 6 molecules of potassium iodide. - 8 molecules of sulfuric acid provide the acidic medium. - Products are chromium sulfate, iodine, potassium sulfate, and water. Let me know if you'd like a detailed step-by-step explanation or help with a different problem!