Write the balanced net ionic equation for the following?

a) Copper metal and zinc ion
b) zinc and cupric ion
c) Copper metal and hydrogen ion
d) zinc metal and hydrogen ion
e) iron metal and hydrogen ion
f) Chlorine (Cl2) and bromide ion (Br-)
g) Chlorine (Cl2) and iodide ion (I-)
h) Bromine (Br2) and iodide ion (I-)
i) Bromine (Br2)and chloride ion (Cl-)
j) Iodine ion (I2) and chloride ion (Cl-)
k) Iodine ion (I2) and bromide ion (Br-)
l) Iron (III) (Fe^+) + iodide ion (I-)
m) Iron (III) (Fe^+) + bromide ion (Br-)
n) Bromine (Br2)+copper metal (Cu)
o) Iodine (I2)+copper metal (Cu)

To write the balanced net ionic equation, we need to determine the chemical formulas of the compounds involved and their charges. Then, we balance the equation by ensuring that the number of atoms on both sides of the equation is the same, and the charges are balanced. Here are the balanced net ionic equations for each given reaction:

a) Copper metal (Cu) and zinc ion (Zn²⁺):
Cu(s) + Zn²⁺(aq) → Cu²⁺(aq) + Zn(s)

b) Zinc metal (Zn) and cupric ion (Cu²⁺):
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

c) Copper metal (Cu) and hydrogen ion (H⁺):
Cu(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂(g)

d) Zinc metal (Zn) and hydrogen ion (H⁺):
Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

e) Iron metal (Fe) and hydrogen ion (H⁺):
Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g)

f) Chlorine (Cl₂) and bromide ion (Br⁻):
Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(l)

g) Chlorine (Cl₂) and iodide ion (I⁻):
Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(s)

h) Bromine (Br₂) and iodide ion (I⁻):
Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(s)

i) Bromine (Br₂) and chloride ion (Cl⁻):
Br₂(l) + 2Cl⁻(aq) → 2Br⁻(aq) + Cl₂(aq)

j) Iodine ion (I₂) and chloride ion (Cl⁻):
I₂(aq) + 2Cl⁻(aq) → I₂(aq) + 2Cl⁻(aq)

k) Iodine ion (I₂) and bromide ion (Br⁻):
I₂(aq) + 2Br⁻(aq) → I₂(aq) + 2Br⁻(aq)

l) Iron (III) (Fe³⁺) and iodide ion (I⁻):
Fe³⁺(aq) + 2I⁻(aq) → Fe²⁺(aq) + I₂(s)

m) Iron (III) (Fe³⁺) and bromide ion (Br⁻):
Fe³⁺(aq) + 2Br⁻(aq) → FeBr₂(aq)

n) Bromine (Br₂) and copper metal (Cu):
Br₂(aq) + Cu(s) → CuBr₂(aq)

o) Iodine (I₂) and copper metal (Cu):
I₂(aq) + 2Cu(s) → 2CuI(s)

Please note that the states of matter (s, l, g, aq) are included to indicate the physical state of the substances in the reaction: (s) - solid, (l) - liquid, (g) - gas, (aq) - aqueous (dissolved in water).

a) Cu(s) + Zn^2+(aq) -> Cu^2+(aq) + Zn(s)

b) Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)

c) Cu(s) + 2H^+(aq) -> Cu^2+(aq) + H2(g)

d) Zn(s) + 2H^+(aq) -> Zn^2+(aq) + H2(g)

e) Fe(s) + 2H^+(aq) -> Fe^2+(aq) + H2(g)

f) Cl2(g) + 2Br^-(aq) -> 2Cl^-(aq) + Br2(l)

g) Cl2(g) + 2I^-(aq) -> 2Cl^-(aq) + I2(s)

h) Br2(l) + 2I^-(aq) -> 2Br^-(aq) + I2(s)

i) Br2(l) + 2Cl^-(aq) -> 2Br^-(aq) + Cl2(g)

j) I2(s) + 2Cl^-(aq) -> 2I^-(aq) + Cl2(g)

k) I2(s) + 2Br^-(aq) -> 2I^-(aq) + Br2(l)

l) 2Fe^3+(aq) + 6I^-(aq) -> 2FeI3(s)

m) 2Fe^3+(aq) + 6Br^-(aq) -> 2FeBr3(s)

n) Br2(l) + 2Cu(s) -> 2CuBr(s)

o) I2(s) + 2Cu(s) -> 2CuI(s)

Actually all of these can be worked using the table of reduction potentials but I will give that process last and show shortcuts for metals vs metal ions and for halogens vs halogen ions.

The metals are done by looking at the activity series of metals. Here is a simplified chart.
https://www.google.com/search?q=activity+series&client=firefox-a&hs=ztE&rls=org.mozilla:en-US:official&channel=sb&tbm=isch&imgil=p0Q-9UyLhIFQuM%253A%253Bhttps%253A%252F%252Fencrypted-tbn2.gstatic.com%252Fimages%253Fq%253Dtbn%253AANd9GcQb__UIJVurrkm81eRd62Oz3E4KOdP6ef7KX4jVGT3nGWzSp-cZTA%253B1188%253B1206%253BtAi9SFMo8FKrOM%253Bhttp%25253A%25252F%25252Fcatalog.flatworldknowledge.com%25252Fbookhub%25252Freader%25252F1790%25253Fe%2525253Daverill_1.0-ch04_s08&source=iu&usg=__seYwXMF6R8H1r_NEBYXqiv7QoX8%3D&sa=X&ei=SP1OU923KO7S2wWb9YHoDA&ved=0CCwQ9QEwAQ#facrc=_&imgdii=_&imgrc=p0Q-9UyLhIFQuM%253A%3BtAi9SFMo8FKrOM%3Bhttp%253A%252F%252Fimages.flatworldknowledge.com%252Faverill%252Faverill-fig04_022.jpg%3Bhttp%253A%252F%252Fcatalog.flatworldknowledge.com%252Fbookhub%252Freader%252F1790%253Fe%253Daverill_1.0-ch04_s08%3B1188%3B1206

Looking at the chart, any METAL will displace any metal ION BELOW it in the activity series. For example,
a.
Cu(s) + Zn^2+(aq) ==> no reaction because Zn^2+ (zinc ion is ABOVE Cu metal).
b.
Zn(s) + Cu^2+(aq) ==> Cu(s) + Zn^2+(aq)
Zn metal is above the Cu ion, therefore, Cu ion is replaced.

For the halogens, any HALOGEN will displace any HALOGEN ION BELOW it in the periodic table. For example,
F2 + 2Cl^- ==> Cl2 + 2F^- because F2 is above chlorine in the periodic table. BUT
I2 + 2Br^- --> Br2 + 2I^- will not occur because I2 is BELOW Br2 in the periodic table.

For the others (those not between halogens/halogen ions or not between metals and metal ions, use the table of reduction potentials.

Here is how you can do all of them.
Look up the reduction potential for the two half equations. Add the oxidation value to the reduction value. If the sum of the voltages is + the reaction will occur. If it is negative the reaction will not occur. For example,
iron(III)ion + iodide ion -->
Divide into
2Fe^3+ + 6I^- ==> 2Fe(s) + 3I2
You will find in the reduction potential table Fe^3+ + 3e ==> Fe(s) Eo value of -0.04
For the I2 half reaction the one listed
I2 + 2e ==> 2I^- Eo reduction is +0.54. You want the reverse of that which is
2I^- ==> I2 + 2e and Eo will be -0.54
Now add -0.04 +(-0.54) = -0.58 v. It is a negative number and the reaction as written will not occur.
Post your work if you get stuck.

mentally we reverse each and Fe at 0.04 is the larger positive number than -0.54 so you add the reverse of Fe