would the half reactions for Ag(s) |AgCl(s) |Cl−(aq) ||
Cl−(aq) |Hg2Cl2(s) |Hg(§¤)be
Ag+Cl+e==> AgCl
2Cl+3Hg==>Hg2Cl2+Hg+e
Wouldn't the first one be
Ag + Cl^- ==> AgCl + e
And the second one be
2Cl^- + 2Hg ==> Hg2Cl2 + 2e
Both written as oxidations.
To determine the half reactions for the given electrochemical cell, we need to follow a few steps:
1. Identify the oxidation half reaction:
- The species that is being oxidized is Ag(s) (silver metal) in the first electrochemical half cell.
- Ag(s) loses one electron (e-) to form Ag+ (silver ion) in solution.
- Therefore, the oxidation half reaction is: Ag(s) → Ag+(aq) + e-
2. Identify the reduction half reaction:
- The species that is being reduced is Cl−(aq) (chloride ion) in both electrochemical half cells.
- In the first electrochemical half cell, Cl−(aq) gains one electron (e-) to form AgCl(s) (silver chloride solid).
- Therefore, the reduction half reaction is: Cl−(aq) + e- → AgCl(s)
- In the second electrochemical half cell, Cl−(aq) gains two electrons (2e-) to form Hg2Cl2(s) (mercury(I) chloride solid), and additionally, one more electron to form liquid mercury (Hg):
- Therefore, the reduction half reaction is: 2Cl−(aq) + 3Hg(§¤) + e- → Hg2Cl2(s) + Hg(§¤)
The half reactions for the given electrochemical cell are:
- Oxidation half reaction: Ag(s) → Ag+(aq) + e-
- Reduction half reaction: Cl−(aq) + e- → AgCl(s) (first electrochemical half cell)
- Reduction half reaction: 2Cl−(aq) + 3Hg(§¤) + e- → Hg2Cl2(s) + Hg(§¤) (second electrochemical half cell)