Answer:
H₂(g) + 2Ag⁺(aq) → 2Ag(s) + 2H⁺ (aq)
Explanation:
In an electrochemical cell , the cell notation is written in a way , that first is the oxidation half is written followed by the reduction , which are separated by the salt bridge ,
hence , from the question , the cell notation is given as -
Pt(s) | H₂(g) | H⁺(aq) || Ag⁺(aq) | Ag(s)
Hence , from the above reaction , the oxidation half cell and reduction half cell as given as -
Oxidation -
H₂(g) → H⁺ (aq)
Reduction -
Ag⁺(aq) → Ag(s)
Now , balancing the reaction , by adding electrons on the required side -
Oxidation -
H₂(g) → 2H⁺ (aq) + 2e⁻
Reduction -
Ag⁺(aq) + e⁻ → Ag(s)
The above reaction can also be written as -
2Ag⁺(aq) + 2e⁻ → 2Ag(s)
Now , to overall cell reaction can be written as adding the respective half cells and eliminating the electrons , i.e. ,
H₂(g) + 2Ag⁺(aq) → 2Ag(s) + 2H⁺ (aq)
