Final answer:
The balanced redox reaction is obtained by balancing the electrons transferred in the oxidation and reduction half-reactions, which for this equation requires multiplying the oxidation half-reaction by 2 and the reduction half-reaction by 3, resulting in a balanced equation of 3Ni²⁺ (aq) + 2Co(s) → 2Co³⁺ (aq) + 3Ni(s).
Step-by-step explanation:
To balance the given redox reaction 3Ni²⁺ (aq) + 2Co(s) → 2Co³⁺ (aq) + 3Ni(s), we need to identify and balance the half-reactions for both oxidation and reduction processes.
The oxidation half-reaction involves the cobalt metal (Co) being oxidized to cobalt(III) ions (Co³⁺), which can be represented as:
Oxidation (x2):
Co(s) → Co³⁺ (aq) + 3e¯
The reduction half-reaction involves the nickel(II) ions (Ni²⁺) being reduced to nickel metal (Ni), which is:
Reduction (x3):
Ni²⁺ (aq) + 2e¯ → Ni(s)
Following step #6 in the balancing process, we need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. Thus, we multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3:
- 2 Co(s) → 2 Co³⁺ (aq) + 6e¯
- 3 Ni²⁺ (aq) + 6e¯ → 3 Ni(s)
Adding these two half-reactions together, we get the balanced redox equation:
3Ni²⁺ (aq) + 2Co(s) → 2Co³⁺ (aq) + 3Ni(s)
This equation is now balanced in terms of both mass and charge.