Question

The brown residues that collect along the inner walls and bottoms of toliet tanks are made mostly of solid Fe(OH)3 that forms when FE^2+ ions in water are oxidized by dissolved O2. Write a balanced net ionic equation for this reaction in a slightly basic solution.

Sample Exercise 8.12 The brown residues that collect along the inner walls and bottoms of toilet tanks are made mostly of solid Fe(OH), that forms when Fe ions in water are oxidized by dissolved O2. Write a balanced net ionic equation for this reaction in a slightly basic solution. basto

Answer 1

Fe2+ and O2 are reacting in basic medium to form solid Fe(OH)3. Hence, the unbalanced equation we can write is

Fete + O2(aq) + Fe(OH)3(5)

Now, Fe is in +2 state in Fe2+ and +3 state in Fe(OH)3.

Hence, Fe2+ is being oxidised.

Hence, the oxidation half reaction is

Oridation : Fe2+ + Fe(OH)

O is in 0 oxidation state in O2. We can assume that O2 is being reduced to H2O (which is the reaction medium) in which O has an oxidation state of -2.

Hence, the reduction half reaction is

Reduction : 02 + H2O

Now, lets balance the oxidation half

Oridation : Fe2+ + Fe(OH)

Number of Fe atom is balanced.

To balance the number of O atoms, we add 3 H2O on the left.

Fe2+ + 3H2O → Fe(OH)

Now, there are 6 H on left and 3 H on right. Hence, we add 3 H+ on right to balance the number of H atom.

Fe2+ + 3H2O → Fe(OH)3 + 3H+

Since the reaction is being carried out in basic medium, we cannot have H+ in reaction, hence, we neutralise the H+ to H2O by adding equivalent amount of OH- on both side

Fe2+ + 3H2O +30H + Fe(OH)3 + 3H+ + 3OH → Fe2+ + 31,0 +30° + Fe(OH)3 + 350 Fe2+ + 3OH- + FelOH)2

Now, there are 2-3 = -1 charges on left and no charges on right. Hence, we add an electron on the right to balance the charges. Hence, the balanced oxidation half is

Fe2+ + 3OH + Fe(OH)3 te

Now, we will balance the reduction half

Reduction : 02 + H2O

To balance the O atoms, we multiply 2 on H2O

O2 + 2H2O

To balance the H atoms, we add 4H+ on left.

02 + 4H+ + 2H2O

Since medium is basic, we add 4OH- on both side

02 + 4H+ + 40H + 2H20 + 40H 702 + 4H20 +2H20 + 401 02 + 2H20 + 40H

Now, to balance the charges, we add 4 e on the left. Hence, the balanced reduction half is

O2 + 2H20 + 4e + 40H

Note that 4 e are involved in reduction half and 1 e involved in oxidation half, hence, to cancel the number of electrons when we add, we multiply 4 throughout the oxidation half reaction and add it with reduction half reaction.

Hence,

4 (Fe2+ + 3OH- + Fe(OH)3 + e-)

+

O2 + 2H20 + 4e + 40H

=

4Fe2+ + 12OH- +02 + 2H20 +46 +4Fe(OH)3 + 4 + 40H- +4Fe2+ + 120H+ O2 + 2H20 - 401 +4Fe(OH)3 + 401 - 40H → 4Fe2+ +80H +02+2H20 + 4Fe(OH)3

Hence, the balanced net ionic equation is

4F e +80H + O2(aq) + 2H200 + 4Fe(OH)3(5)