Write the oxidation and reduction half-reactions for a silver-chromium voltaic cell. Identify the anode, cathode, and the direction of electron flow.
Ag+ + e- Ag
Cr Cr2+ + 2e- or Cr Cr3+ + 3e- ????? Which is the correct equation for the chromium side of the reaction?
Anode: chromium
Cathode: silver
Electron flow: chromium to silver
What you have looks ok to me. Cr will go to Cr^3+.
The correct equation for the oxidation half-reaction on the chromium side is:
Cr → Cr3+ + 3e-
Anode: Chromium (Cr) - this is where the oxidation half-reaction occurs
Cathode: Silver (Ag) - this is where the reduction half-reaction occurs
Direction of electron flow: From the anode (chromium) to the cathode (silver)
To identify the oxidation and reduction half-reactions for a silver-chromium voltaic cell, you need to understand the process of oxidation-reduction (redox) reactions.
Oxidation involves the loss of electrons, while reduction involves the gain of electrons. In the given reactions, silver is being reduced since it gains an electron, and chromium is being oxidized since it loses electrons.
The correct equation for the chromium side of the reaction should be: Cr → Cr3+ + 3e-.
Now, let's identify the anode, cathode, and the direction of electron flow.
The anode is where oxidation occurs, which means it's the site where electrons are being released. In this case, the chromium side is where oxidation occurs, so the anode is chromium.
The cathode is where reduction occurs, meaning it's where electrons are being gained. In this case, the silver side is where reduction occurs, so the cathode is silver.
The direction of electron flow is from the anode to the cathode. In other words, electrons flow from the chromium side (anode) to the silver side (cathode).
So, to summarize:
Oxidation half-reaction (anode): Cr → Cr3+ + 3e-
Reduction half-reaction (cathode): Ag+ + e- → Ag
Anode: Chromium
Cathode: Silver
Electron flow: Chromium to Silver
I hope this explanation helps you understand the process! Let me know if you have any further questions.