Question

Balance the following ionic equations.

\[\ce{Cr2O^{2-}7 + Fe^{2+} + H+ -> Cr^{3+} + Fe^{3+} + H2O}\]

Answer 1

Step 1: Separate the equation into two half-reactions.

The oxidation number of various atoms are shown below:

\[\ce{\overset{+6}{Cr2}\overset{-2}{O}^{2-}7 + \overset{+2}{Fe}^{2+} + \overset{+1}{H+} -> \overset{+3}{Cr}^{3+} + \overset{+3}{Fe}^{3+} + \overset{+1}{H2}\overset{-2}{O}}\]

In this case, chromium undergo reduction, oxidation number decreases from +6 (in \[\ce{Cr2O^{2-}7}\]) to +3 (in \[\ce{Cr^{3+}}\])

\[\ce{Fe^{2+}}\] (O.N. = +2) changes to \[\ce{Fe^{3+}}\] (O.N. = +3). The species undergoing oxidation and reduction are:

Oxidation: \[\ce{Fe^{2+} -> Fe^{3+}}\]

Reduction: \[\ce{Cr2O^{2-}7 -> Cr^{3+}}\] 

Step 2: Balance each half-reaction separately as:

(a) \[\ce{Fe^{2+} -> Fe^{3+}}\]

(i) Balance all atoms other than  \[\ce{H}\] and \[\ce{O}\]. This step is not needed, because, it is already balanced.

(ii) The oxidation number on left is +2 and on right is +3. To account for the difference, the electron is added to the right as: \[\ce{Fe^{2+} -> Fe^{3+} + e-}\]

(iii) Charge is already balanced.

(iv) No need to add \[\ce{H}\] or \[\ce{O}\].

The balanced half-equation is: 

\[\ce{Fe^{2+} -> Fe^{3+} + e-}\] ......(i)

Consider the second half-equation

(b) \[\ce{Cr2O^{2-}7 -> Cr^{3+}}\] 

(i) Balance the atoms other than \[\ce{H}\] and \[\ce{O}\].

\[\ce{Cr2O^{2-}7 -> 2Cr^{3+}}\] 

(ii) The oxidation number of chromium on the left is +6 and on the right is +3. Each chromium atom must gain three electrons. Since there ar two \[\ce{Cr}\] atoms, add \[\ce{6e-}\] on the left.

\[\ce{Cr2O^{2-}7 + 6e^{-} -> 2Cr^{3+}}\] 

(iii) Since the reaction takes place in acidic medium add \[\ce{14H+}\] on the left to equate the net charge on both sides.

\[\ce{Cr2O^{2-}7 + 6e^{-} + 14H^{+} -> 2Cr^{3+}}\] 

(iv) To balance \[\ce{FI}\] atoms, add \[\ce{7H2O}\] molecu;es on the right.

\[\ce{Cr2O^{2-}7 + 6e^{-} + 14H^{+} -> 2Cr^{3+} + 7H2O}\]   ......(ii)

This is the balanced half-equation.

Step 3: Now add up the two half-equations. Multiply equation (i) by 6 so that electrons are balanced.

                                      \[\ce{Fe^{2+} -> Fe^{3+} + e- × 6}\]

    \[\ce{Cr2O^{2-}7 + 6e- + 14H+ -> 2Cr^{3+} + 7H2O}\]
                                                                                                        

\[\ce{6Fe^{2+} + Cr2O^{2-}7 + 14H+ -> 6Fe^{3+} + 2Cr^{3+} + 7H2O}\]

The balanced equation is: \[\ce{6Fe^{2+} + Cr2O^{2-}7 + 14H+ -> 6Fe^{3+} + 2Cr^{3+} + 7H2O}\]