Question

balance the following redox reaction in basic solution Ag(s)+CN-(aq)+ O2(g) -->Ag (CN)2- (aq)

cyanide is not participating in the electrochemical reaction but does associate with the silver ion formed in a second association reaction. Balance the electrochemistry first and then add in the cyanide in by adding an associative reaction to the redox reaction.

Answer 1

Step1- Given unbalanced redox equation is

Ag(s) + CN⁻(aq) + O₂(g) → Ag(CN)₂(aq)

Step-2- The unbalanced equation with its ionic compounds dissociated.

Ag(s) + CN⁻(aq) + O₂(g) → Ag⁺(aq) + 2 CN⁻(aq)

As per the equation Ag is in solid form cyanide (CN) in aqueous form means ionic state and O₂ is in dissolved gas form.

In the first Half reaction Ag undergoes oxidation by giving an electron  

Step 2- The unbalanced net ionic equation, without the spectator ion, CN⁻₍aq)

Ag(s) + O₂(g) → Ag⁺(aq)

Step 3- The unbalanced equations of two half-reactions are (first oxidation and then reduction)

The oxidation half-reaction is:

Ag(s) → Ag⁺(aq)

The reduction half-reaction involves oxygen reduces to water.

O₂(g) → 2 H₂O(l)

Step 4- Balance each half-reaction equation. The oxidation half-reaction is balanced.

For the reduction half-reaction, balance hydrogen by adding hydrogen ions.

O₂(g) + 4 H⁺(aq) → 2 H₂O

Since the reaction takes place in a basic solution, eliminate H⁺(aq) by adding an equal number of OH⁻(aq) to both sides of the equation.

O₂(g) + 4 H⁺(aq) + 4 OH⁻(aq) → 2 H₂O(l) + 4 OH⁻(aq)

Subtract 2 water molecules from each side to eliminate redundant water molecules.

O₂(g) + 2 H₂O(l) → 4 OH⁻(aq)

Step 5- Use electrons to balance the charge in each half-reaction equation.

Ag(s) → Ag+(aq) + e⁻

O₂(g) + 2 H₂O + 4 e− → 4 OH⁻(aq)

Step 6- Equalize the electron transfer in the two half-reaction equations.

Multiply the oxidation half-reaction equation by 4

4 Ag(s) → 4 Ag⁺(aq) + 4 e⁻

O₂(g) + 2 H₂O(l) + 4 e⁻ → 4 OH⁻(aq)

Step 7- Add the two half-reaction equations. Include the spectator ion, CN^–(aq).

4 Ag(s) + O₂(g) + 2 H₂O → 4 Ag⁺(aq) + 4 OH⁻(aq)

4 Ag(s) + 8 CN⁻(aq) + O₂(g) + 2 H₂O → 4 Ag(CN)₂⁻(aq) + 4 OH⁻(aq)

The balanced equation is

4 Ag(s) + 8 CN⁻(aq) + O₂(g) + 2 H₂O → 4 Ag(CN)₂⁻(aq) + 4 OH⁻(aq)