Question 18.PS.4: Balancing Redox Equations for Reactions in Basic Solutions I......

Cd(s) + Ni_2O_3 (s) + 3 H_2O(\ell) → Cd(OH)_2 (s) + 2  Ni(OH)_2 (s)

Strategy and Explanation   Use the systematic approach given below.

Step 1:    Recognize whether the reaction is an oxidation-reduction process.Then determine what is reduced and what is oxidized. This is a redox reaction because the oxidation number of Cd changes from 0 in Cd metal to +2 in Cd(OH)_2, so Cd metal is oxidized. The oxidation number of each Ni changes from +3 in Ni_2O_3 to +2 in Ni(OH)_2, so the Ni is reduced.

Step 2:    Break the overall unbalanced equation into half-reactions.

Cd( s ) → Cd(OH)_2(s)              (oxidation half-reaction)

Ni_2O_3(s) → Ni(OH)_2(s)                (reduction half-reaction)

Step 3:    Balance the atoms in each half-reaction. First, balance all atoms except the O and H atoms; do them last. Balance each half-reaction as if it were in an acidic solution to start. Balance O by adding H_2O and balance H by adding H^+. We will revert to using OH^+ ions characteristic of basic solutions later in the process.

In the Cd half-reaction, the Cd atoms are balanced. Adding two water molecules on the left balances the two O atoms on the right, but this leaves four H atoms on the left and only two on the right. Adding two H^+ ions on the right balances H atoms in this half-reaction.

2  H_2O(\ell) + Cd(s) → Cd(OH)_2(s) + 2  H^+(aq)

For the Ni_2O_3 half-reaction, a coefficient of 2 is needed for Ni(OH)_2 because there are two Ni atoms on the left. To balance the four O atoms and four H atoms now on the right with the three O atoms on the left requires one water molecule and two H^+ ions on the left.

2  H^+(aq) + H_2O(\ell) + Ni_2O_3(s) → 2  Ni(OH)_2(s)

Step 4:    Balance the half-reactions for charge using electrons. The Cd half-reaction produces 2  e^- as a product.

2  H_2O(\ell) + Cd(s) → Cd(OH)_2(s) + 2  H^+(aq) + 2  e^-            (balanced)

The Ni_2O_3 half-reaction requires 2  e^- as a reactant.

2  H^+(aq) + H_2O(\ell) + Ni_2O_3(s) + 2  e^- → 2  Ni(OH)_2(s)              (balanced)

Step 5:    Multiply the half-reactions by appropriate factors so that the reducing agent produces as many electrons as the oxidizing agent accepts. The Cd half-reaction produces two electrons, and the Ni_2O_3 half-reaction accepts two, so the electrons are balanced.

Step 6:    Because H^+ does not exist at any appreciable concentration in a basic solution, remove H^+ by adding an appropriate amount of OH^- to both sides of the equation. H^+ and OH^- react to form H_2O. In the Cd half-reaction, add two OH^- ions to each side to get

2  OH^-(aq) + 2  H_2O(\ell) + Cd(s) → Cd(OH)_2(s) + 2  H_2O(\ell) + 2 e^-

On the product side, two OH^- ions plus two H^+ ions form two H_2O molecules. For the Ni_2O_3 half-reaction, add two OH^- ions to each side to get

3  H_2O(\ell) + Ni_2O_3(s) + 2  e^- → 2  Ni(OH)_2(s) + 2  OH^-(aq)

Step 7:    Add the half-reactions to give the net reaction, and cancel reactants and products that appear on both sides of the reaction arrow.

\cancel{2  OH^-(aq)} + \cancel{2  H_2O(\ell)} + Cd(s) → Cd(OH)_2(s) + \cancel{2  H_2O(\ell)} + 2 e^-

\underline{3  H_2O(\ell) + Ni_2O_3(s) + 2  e^- → 2  Ni(OH)_2(s) + \cancel{2  OH^-(aq)}}

Cd( s ) + Ni_2O_3(s) + 3  H_2O((\ell) → Cd(OH)_2(s) + 2  Ni(OH)_2(s)

Step 8:    Check the final results to make sure both atoms and charge are balanced. The net equation is balanced. In the final equation, there are no net charges on either side of the reaction arrow, and the numbers of atoms of each kind on each side of the reaction arrow are equal.