electrochem, please help!?

Question

Zn(s) / ZnCl2(aq) // Cl-(aq)/ Cl2(g)/ C(s)

In the electrochemical cell described by the cell diagram above, what reaction occurs at the anode?
a) Zn-> Zn2+ +2e-
b) Zn2+ + 2e- --> Zn
c) Cl2 + 2e -> 2Cl-
d) 2Cl- --> Cl2+ 2e
e) Zn+ Cl2 --> ZnCl2

answer is A but i dont get it
i dont get what the "diagram" is trying to tell me either

by the way, the "/" slanted lines in the diagram are supposed to be straight lines but i made them slanted to make it look less confusing

Answer 1

Okay well there is a little saying to help remember this stuff. LEO goes GER. This means Losing Electrons Oxidiation and Gaining Electrons Reduction. An anode will always be oxidized. Thus, this is oxidation. This only leaves you with the answer A or D. You should have a table of the energy potential, this should help you know which way the reaction runs and which one is the anode and cathode.

Answer 2

Anode by definition is where oxidation occurs, so the only right answer is A. The diagram shows you what the constituents of the galvanic cell are - anode chamber is Zn and ZnCl2 (with the reaction Zn = Zn2+ + 2e), the cathode chamber it's cloride ions and chlorine (which would be a chlorine electrode) - Cl2 + 2e = 2Cl-. C does not participate in the electrochemical reaction, it's just there to conduct electrons (although some of it will react with chlorine)

Answer 3

the cell diagram is always laid out a certain way. these are just conventions so there's no reason for them really but it's important to remember them nonetheless. i guess it's a bit like what side of the road you drive on, the choice doesn't matter but it's important that everyone agrees on the answer :)

the reduction reaction occurs on the right hand side of the cell diagram, and that electrode is called the cathode. I remember this by noting that reduction and right both begin with R, although there are other memory tricks. on the left hand side you have oxidation at the anode, which i remember by noting that both words begin with vowels (Oxidation - Anode).

so the reaction occuring at the anode is oxidation, on the left hand side. clearly it's not c) or d) since these don't involve zinc. e) is not a half reaction (no electrons in the reaction, and reactions occuring at each electrode must be half reactions or it would not be an electrochemical cell). b) is a reduction, and we know that the reaction occuring on the left must be an oxidation. that leaves a).

Answer 4

In an electrochemical cell, it is always good to have a copy of the standard electrode potential (See this page for full list: http://en.wikipedia.org/wiki/Table_of_standard_electrode_potentials). Don't worry about the cell diagram, good to know the basic behind the equation.

From here, we can see that
1. Zn2+(aq) + 2e− → Zn(s) −0.76 V
2. C(s) + 4H+ + 4e− → CH4(g) +0.13 V
3. Cl2(g) + 2e− → 2Cl−(aq) +1.36 V

Since the values are given in their ability to be reduced, the bigger the standard reduction potentials, the easier they are to be reduced, in other words, they are simply better oxidizing agents. For example, Cl2 reduces easily and is therefore a good oxidizing agent. In contrast, Zn2+ would rather undergo oxidation (hence a good reducing agent). Thus Zn2+/Zn whose standard reduction potential is -0.76V can be oxidized by any other electrode whose standard reduction potential is greater than -0.76V.

Once we establish this, we will know know that Zn prefers to be oxidised rather than reduced while Cl2 and C act as oxidising agent. Oxidising Zn gives 2 electrons. We know that the flow of electrons is always from anode–to–cathode outside of the cell. I always remember cathode as being shorter (being reduced, if you are familiar with a malay term "ketot" mean shorter). Since electrons are coming from Zn, therefore Zn is the anode and thus Zn is oxidised to Zn2+. Hence the answer is A.

The cell diagram then is just a way from showing the anode–to–cathode outside of the cell. How to read the cell diagram, : First, the reduced form of the metal to be oxidized at the anode (Here is Zn) is written . This is separated from its oxidized form by a vertical line (oxidation changes). The double vertical lines represent the bridge on the cell. Finally, the oxidized form of the metal/elements to be reduced at the cathode, is written, separated from its reduced form by the vertical line.

Simply, it's Oxidation // Reduction. For the order, always remember from "what oxidised to what// what reduced to give what"