Question 17.8: Balance the equation K2Cr2O7 + FeCl2 + HCl→CrCl3 + KCl + Fe...

Use the Problem-Solving Strategy: Using the Change in Oxidation Numbers to Balance Oxidation–Reduction Reactions.

1. Assign oxidation numbers ( Cr and Fe have changed):

2. Write the oxidation and reduction steps. Balance the number of atoms and then balance the electrical charge using electrons:

Fe^{2+}→Fe^{3+} + 1 e^-      (oxidation)

Fe^{2+} loses 1 electron

2 Cr^{6+} + 6 e^- → 2 Cr^{3+}            (reduction)

2 Cr^{6+}gain 6 electrons

3. Balance the loss and gain of electrons. Multiply the oxidation step by 6 and the reduction step by 1 to equalize the transfer of electrons.

6 Fe^{2+} → 6 Fe^{3+} + 6 e^-          (oxidation)

6 Fe^{2+} lose 6 electrons

2 Cr^{6+} + 6 e^- → 2 Cr^{3+}            (reduction)

2 Cr^{6+}gain 6 electrons

4. Transfer the coefficients from the balanced redox equations into the original equation. (Note that one formula unit of K_2Cr_2O_7 contains two Cr atoms.) We need to use 1 K_2Cr_2O_7 , 2 CrCl_3 , 6 FeCl_2 , and 6 FeCl_3:

K_2Cr_2O_7 + 6 FeCl_2 + HCl →

2  CrCl_3 + KCl + 6 FeCl_3 + H_2O      (unbalanced)

5. Balance the remaining elements in the order K, Cl, H , O :

K_2Cr_2O_7 + 6 FeCl_2 + 14 HCl →

2  CrCl_3 + 2KCl + 6 FeCl_3 + 7 H_2O      (balanced)

6. Check: The final balanced equation contains 2 K atoms, 2Cr  atoms, 7 O atoms, 6 Fe atoms, 26 Cl atoms, and 14 H atoms on each side.