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8.1: Balancing redox reactions

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    164370
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    In studying redox chemistry, it is important to begin by learning to balance electrochemical reactions. Simple redox reactions (for example, H2 + I2 → 2 HI) can be balanced by inspection, but for more complex reactions it is helpful to have a foolproof, systematic method. The electron-ion method allows one to balance redox reactions regardless of their complexity. We illustrate this method with two examples.

    Example 1 - Acidic Solution

    I- is oxidized to IO3- by MnO4-, which is reduced to Mn2+. This reaction takes place in acidic solution.

    How can this reaction be balanced? In the electron-ion method we follow a series of ten steps:

    Step 1: Write out the (unbalanced) reaction and identify the elements that are undergoing redox.

    MnO4- + I- → IO3- + Mn2+     (acid)

    Step 2: Assign the oxidation states to each element on both sides of the equation.

    • MnO4-       Mn = +7        O = -2
    • I-            I = -1
    • IO3-        I = +5            O = -2
    • Mn2+       Mn = +2

    Step 3: Separate the reaction into two half reactions, balancing the element undergoing redox in each.

    red.    MnO4-Mn2+            Manganese is being reduced

    ox.      I-IO3-                    Iodine is being oxidized

     

    Step 4: Add the appropriate number of electrons to each half-reaction. Electrons are always reactants in the reduction half-reaction and products in the oxidation half reaction.

    red.    MnO4-  +  5 e-    →    Mn2+        Mn is reduced from +7 to +2. This adds 5 electrons.

    ox.     I-    →    IO3-  +  6 e-                I is oxidized from -1 to +5. This removes 6 electrons.

     

    Step 5: Balance the charge of each half-reaction by adding hydrogen ions, H+. Since this reaction occurs in acidic solution, there are ample H+ ions for charge balance.

    red.    MnO4-  +  5 e-  +  8H+    →    Mn2+        Both sides of the equation have a total 2+ charge.

    ox.     I-    →    IO3-  +  6 e-  +  6 H+               Both sides of the equation have a total 1- charge.

     

    Step 6: Balance the oxygen and hydrogen atoms by adding H2O to the appropriate side of each half-reaction.

    red.    MnO4-  +  5 e-  +  8 H+    →    Mn2+  +  4H2O

    ox.     I-  +  3 H2O    →    IO3- + 6 e-  +  6 H+

     

    Step 7: Multiply each half-reaction by the appropriate coefficient so that there are equal numbers of electrons as reactants and products.

    red.    (MnO4-  +  5 e-  +  8 H+    →    Mn2+  +  4H2O) x 6

    ox.     (I-  +  3 H2O    →    IO3- + 6 e-  +  6 H+x 5

     

    Step 8:  Combine the half-reactions into an overall equation.

    6 MnO4-  +  5 I-  +  30 e-  +  48 H+  +  15 H2O    →    6 Mn2+  +  5 IO3-  +  30 e-  +  30 H+  +  24 H2O

     

    Step 9:  Cancel the H+, electrons, and H2O:

    6 MnO4-  +  5 I-  30 e-  +  48 H +   15 H2O    →     6 Mn2+  + 5 IO3-  +  30 e-  +  30 H+24 H2O

    The overall balanced reaction is therefore:

    6 MnO4-(aq) +  5 I-(aq)  +  18 H+(aq)    →    6 Mn2+(aq)  +  5 IO3-(aq) + 9 H2O()

     

    Step 10:  Check your work by making sure that all elements and charges are balanced.

     

    Example 2 - Basic Solution

    The chromate ion (CrO42-) oxidizes Fe(OH)2 to Fe(OH)3. The CrO42- ion is reduced to Cr(OH)3. This reaction takes place in basic solution.

    How can this reaction be balanced? In the electron-ion method we follow a series of ten steps:

    Step 1: Write out the (unbalanced) reaction and identify the elements that are undergoing redox.

    CrO42-  +  Fe(OH)2    →    Cr(OH)3  +  Fe(OH)3        (basic)

    Step 2: Assign the oxidation states to each element on both sides of the equation.

    • CrO42-       Cr = +6        O = -2
    • Fe(OH)2   Fe = +2       O = -2            H = +1
    • Cr(OH)3   Cr = +3       O = -2            H = +1
    • Fe(OH)​3​​​   Fe = +3       O = -2            H = +1

    Step 3: Separate the reaction into two half reactions, balancing the element undergoing redox in each.

    red.     Cr O4-Cr(OH)3            Chromium is being reduced

    ox.     Fe(OH)2 → Fe(OH)3       Iron is being oxidized

     

    Step 4: Add the appropriate number of electrons to each half-reaction. Electrons are always reactants in the reduction half-reaction and products in the oxidation half reaction.

    red.    CrO42-  +  3 e-    →    Cr(OH)3        Cr is reduced from +6 to +3. This adds 3 electrons.

    ox.    Fe(OH)2    →    Fe(OH)3  +  e-          Fe is oxidized from +2 to +3. This removes 1 electron.

     

    Step 5: Balance the charge of each half-reaction by adding hydroxide ions, OH-. Since this reaction occurs in basic solution, there are ample OH- ions for charge balance.

    red.    CrO42-  +  3 e-    →    Cr(OH)3  +  5 OH-        Both sides of the equation have a total 5- charge.

    ox.     Fe(OH)2  +  OH-    →     Fe(OH)3  +  e-         Both sides of the equation have a total 1- charge.

     

    Step 6: Balance the oxygen and hydrogen atoms by adding H2O to the appropriate side of each half-reaction.

    red.    CrO42-  +  3 e-  +  4 H2O    →    Cr(OH)3  +  5 OH-

    ox.     Fe(OH)2  +  OH-    →     Fe(OH)3  +  e-             This half-reaction is already balanced.

     

    Step 7: Multiply each half-reaction by the appropriate coefficient so that there are equal numbers of electrons as reactants and products.

    red.    CrO42-  +  3 e-  +  4 H2O    →    Cr(OH)3  +  5 OH-

    ox.     (Fe(OH)2  +  OH-    →     Fe(OH)3  +  e-x 3

     

    Step 8:  Combine the half-reactions into an overall equation.

    CrO42-  +  3 Fe(OH)2  +  3 e-  +  3 OH-  +  4 H2O    →    Cr(OH)3  +  3 Fe(OH)3  +  3 e-  +  5 OH-

     

    Step 9:  Cancel the OH-, electrons, and H2O:

    CrO42-  +  3 Fe(OH)2  +  3 e-  +  3 OH-  +  4 H2O    →    Cr(OH)3  +  3 Fe(OH)3  +  3 e-  +  5 OH-

    The overall balanced reaction is therefore:

    CrO42-(aq)  +  3 Fe(OH)2(s)  +  4 H2O(ℓ)    →    Cr(OH)3(s)  +  3 Fe(OH)3(s)  +  2 OH-(aq)

     

    Step 10:  Check your work by making sure that all elements and charges are balanced.

    Contributors


    8.1: Balancing redox reactions is shared under a CC BY-SA license and was authored, remixed, and/or curated by LibreTexts.

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