Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺[tex]_{aq}[/tex] || Ag⁺[tex]_{aq}[/tex] | Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu[tex]_{s}[/tex] ⇄ Cu²⁺[tex]_{aq}[/tex] + 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺[tex]_{aq}[/tex] + 2e⁻ ⇄ Ag[tex]_{s}[/tex]
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu[tex]_{s}[/tex] ⇄ Cu²⁺[tex]_{aq}[/tex] + 2e⁻
Ag⁺[tex]_{aq}[/tex] + e⁻ ⇄ Ag[tex]_{s}[/tex]
we multiply the second reaction by 2 to balance up:
2Ag⁺[tex]_{aq}[/tex] + 2e⁻ ⇄ 2Ag[tex]_{s}[/tex]
The net reaction equation:
Cu[tex]_{s}[/tex] + 2Ag⁺[tex]_{aq}[/tex] + 2e⁻⇄ Cu²⁺[tex]_{aq}[/tex] + 2e⁻ + 2Ag[tex]_{s}[/tex]
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu[tex]_{s}[/tex] + 2Ag⁺[tex]_{aq}[/tex] ⇄ Cu²⁺[tex]_{aq}[/tex] + 2Ag[tex]_{s}[/tex]
