Edit: okay, here is the full explanation. I guess the character limit is longer if you edit an existing post?
A nice gen chem mnemonic for redox: OIL RIG = Oxidation Is Loss of electrons, Reduction Is Gain of electrons. (oxidation also occurs when a molecule gains oxygen or loses hydrogen, while reduction is the reverse--this definition is more handy in organic chemistry courses than in intro/high school chem).
14. The reducing agent is the compound doing the reducing, so it is oxidized. Let's separate the equation into half reactions so we only have to pay attention to one thing at a time, and remember that NaOH dissociates so it's really Na+ and OH-:
2 Na --> 2 Na+
2H2O --> 2 OH- + H2
Start from the top: Na goes from being neutral to a charge of +, so it's losing electrons, which means it's oxidized, which means it's the reducing agent. 14 is B.
15. Back to OIL RIG: oxidation is loss. Cu loses 2 e- here, so it's D, oxidation.
17. You don't have a reaction here, but your knowledge of the periodic table and the fact that atoms like to have a full octet should tell you that sodium likes to lose an electron and oxygen likes to gain them. Just like in 14, sodium acts as the reducing agent, because it is oxidized. A, sodium.
18. Electronegativity is a measure of how much an element attracts electrons; the more electronegative elements are on the right side of the periodic table and prefer to take on negative charges. Also note that compounds are written cation first; the conclusion you should reach is that the electron transfer is such that S gave up e- and the 2 Cls received them. Written as half reactions:
S --> S(+2) + 2e-
Cl2 +2e- --> 2Cl(-)
(You know where the electrons go because the charge must be balanced on both sides of the reaction, so if one side is neutral and the other has a positive charge, add the e- to the positive side, and if one side is neutral and the other has a negative charge, you add the e- to the neutral side.)
Going through the answers: Cl went from neutral to negatively charged, so it gained e-. Thus, it was reduced--A is wrong. D states that Cl was reduced, so D is the answer.
19. (see 17 as well) Ca forms the ion Ca(+2) and Cl forms the ion Cl-, so Ca is oxidized and Cl is reduced. Cl is thus the oxidizing agent; A.
20. Double replacement and acid-base reactions don't involve changing the charge on any individual atom, and no electron transfer=no redox. C, combustion.
22. See 17 again. Because oxygen gains a negative charge, it gains e- and it is reduced. B.
23. See 19 again. Ca loses e- to become a cation; it is oxidized. B.
24. Redox inherently involves changes in the oxidation states of elements (which is a number related to the charge of an element), so you know the answer is B or C, and you should know that changing the atomic number of an element changes the identity of that element so that's not an option in redox chemistry! Electrons are transferred, C.
25. See 17 again. H is in the same column of the periodic table as Na, and prefers to lose e-, so it is oxidized and thus the reducing agent. A.
26. Ionic reactions are the easiest to balance because everything separates nicely and you don't need to calculate oxidation numbers. B.
30. CH4 + 2 O2 --> 2 H2O + CO2. You should know that O usually has an oxidation state of -2 and H has +1. The sum of the oxidation states of all the atoms in a neutral molecule should be zero. Let's find C's oxidation states, then, using this information.
In the reactants: CH4: 0 = 4(+1) + __
In the products: CO2: 0 = 2(-2) + __
Solving for the blank, C's O.S. in CH4 is -4 and in CO2 is +4. Thus, C is oxidized because a higher oxidation state means more oxidized. Because the C in CH4 is oxidized, O2 must be the oxidizing agent. To prove this to yourself, remember that a neutral molecule has an O.S. of 0, so the Os in O2 each have an O.S. of 0, while the ones on the right have an O.S. of -2, so O is indeed being reduced. (D)
32. See 17 and 22... Na loses e- to become Na+, so it is oxidized and it is the reducing agent. D.
33. See 25. This is pretty much the same question because the reducing agent is what gets oxidized. A.
34. To balance charges, multiple the reactions so they have the same number of e-.
(Cl2 + 2 e- --> 2 Cl) x 3 = 3 Cl2 + 6 e- --> 6 Cl
(Cr --> Cr(+3) + 3 e-) x 2 = 2 Cr --> 2 Cr(+3) + 6 e-
Now, add the half reactions. All the reactants in both equations go into the new reactions' reactants, etc...
3 Cl2 + 6 e- + 2 Cr --> 6 Cl + 2 Cr(+3) + 6 e-
The 6 e- cancel, so remove them.
3 Cl2 + 2 Cr --> 6 Cl + 2 Cr(+3)
35. __Fe2O3 + __CO --> __Fe + __CO2
Fe is neutral so it's oxidation state is 0. Take O's O.S. to be -2 again and work the rest out like in problem 30: Fe in Fe2O3 is +3, C in CO is +2, and C in CO2 is +4.
Consider the O.S. of each element its "charge" when you do redox with molecules, even though it isn't precisely the same thing. Thus, for Fe to go from an O.S./"charge" of +3 in the reactants to one of 0 in the products, it must gain 3 e-. The e- come from the element being oxidized, which is C here (its O.S. goes from +2 to +4). Let's look at just the reactants:
__Fe2O3 + __CO
There are 2 Fe already, each of which needs 3 e- for a total of 6 e- needed. Each mol of CO contributes 2 e-, so we need 3 mol CO for every 1 mol FE2O3. Plug that in and consider the full reaction:
Fe2O3 + 3 CO --> __Fe + __CO2
Start with C: you have 3 C on the left, so you need 3 on the right.
Fe2O3 + 3 CO --> __Fe + 3 CO2
This gives you 6 O on both sides, so C and O are good. Simply balance Fe and you are done.
Fe2O3 + 3 CO --> 2 Fe + 3 CO2