- Joined
- Jun 18, 2013
- Messages
- 362
- Reaction score
- 52
I got the answer right, but I'm not super sure why Q = anode / cathode.
Is anode = products because that's where the concentration of copper increases?
quick electrochem question
- Thread starter echoyjeff222
- Start date
I got the answer right, but I'm not super sure why Q = anode / cathode.
Is anode = products because that's where the concentration of copper increases?
- Joined
- Jul 31, 2014
- Messages
- 245
- Reaction score
- 120
I'm not sure if this is right but here's how I consider it. Consider an acid HA. The equilibrium expression is (A-)/(HA). Here the base is oxidized and the acid is reduced (has a H). Therefore, for an acid you could consider Q to be (oxidized)/(reduced). In a galvanic cell, Q would also be (oxidized)/(reduced), where oxidation occurs in the anode and reduction occurs in the cathode. Therefore, Q = (anode)/(cathode).
But I could be way off base haha.
But I could be way off base haha.
