Pre-Exam 3 email questions...

A few questions have come in by email, here they are:

----Question-----

I have a question regarding problem 9 on exam 3a from spring 2013. I understand that the calculated x value doesn't fit under the assumptions. I see where the first two values come from. I was wondering where the -1.238 x 10^-3 came from?

{(x)(x)} / (0.516 – x) = 2.40 x 10^-3

0.516 is the initial concentration and 2.40 x 10^-3 is Ka.

 x^2 + (2.40 x 10^-3)x + (-1.238 x 10^-3 ) = 0

----Answer-----

In order to use the quadratic formula, we have to solve the equation to the form:

ax² + bx + c = 0

The (-1.238x10^-3) term comes from (2.40x10^-3)(0.516).

----Question-----
I was going through some of the old exams and the problems that use the Henderson - Hasselbalch. In those examples when it asked for the concentration of the conjugate base over the concentration of conjugate acid, but in the key only the moles of both are put in those places. Why? 
----Answer-----

This is a little mathematical shortcut. Since both the conjugate acid and conjugate base are in the same total volume of solution, the volumes mathematically cancel so I left them out. For example, if we had a buffer made from 0.65mols of HA and 0.55mols of A^-1 in 800.0mL of buffer solution, that last part of the Henderson-Hasselbalch would look like:

You can always keep the volume in there and calculate the actual concentrations of each component, you should get exactly the same answer either way.

----Question-----
I dont understand how you can derive pH values from pka's given in the question. I also dont understand how pH can be calculated at each eq point in the titration curve as in number 11 on spring 2013 where it asks what indicator to use.. it says use the 2 pkas...how does this help?
----Answer-----

There are a couple ways that pK_a (or pK_b) can lead to a pH. One possibility is in a question like "What is the expected pH of a 0.618M solution of ammonium nitrate solution?" In this question, you can set up a K_a-type equilibrium for ammonium ions and use the K_a of ammonium to calculate [H₃O⁺] and pH. This is similar to the problem I posted yesterday (http://chemistryingeneral.blogspot.com/2014/03/neutral-salts.html). This method can be used to calculate the initial pH for a titration. It would also work to approximate the pH of an equivalence point. Let's think about that...

For the titration of phosphite ions with hydrochloric acid, we can calculate the initial pH by setting up a K_b-type equililbrium and using the K_b of phosphite ion to calculate [OH^-1] and pOH and pH. At the first equivalence point in this titration, we have a solution that we can think of as HPO₃^-2(aq) because we have added just enough acid to complete the following equation exactly once:

PO₃^-3(aq) + H⁺(aq)  <=> HPO₃^-2(aq)

Between equivalence points, we have buffering regions of the titration curve... at the mid-point of this buffering region, the pH is equal to the pK_a of the weak acid of the mixture. If we know the pK_a (and therefore the pH) on either side of the equivalence point we're interested in, we can get a pretty reliable estimate of the pH of that equivalence point. On the titration curve below, if the pK_as that bound the equivalence point are 6 and 9, the pH of the equivalence point should be right between them at pH = 7.5.

Keep an eye on the weather... Unless MSUM officially closes campus tomorrow, we will have class and the exam as planned. I'll be there at 7:30.

Neutral salts

The conjugate of a strong acid or strong base is neutral. That's just something we tend to accept, memorize, and move on. But why? Those two words are a big part of the reason I'm a chemist.

Let's take a look at a strong acid and see if we can make sense of this. Of the typical strong acids, nitric is usually the weakest, and nitric is also the only one that might have a K_a listed in standard tables. The stronger strong acids have really useful K_a values listed in the tables like "large" or "strong"... I don't have a "large" button on my calculator, so let's just use nitric acid and we'll hopefully see why the other strong acids would follow the same trend if we had a value for their K_a.

The K_a for nitric acid is usually listed at around 25. That means the K_b for nitrate ions is:

That's a REALLY weak K_b, but we can go ahead and calculate the pH of a solution just like in any other situation. How about the problem: What is the expected pH of a 0.500M solution of sodium nitrate? {By the way, we could make similar arguments to show that the sodium ions don't affect the pH, but we'll save those for another day...} As with all good equilibrium problems, it's probably not a bad idea to start with a table:

Now we can set up the K_b expression and plug in the numbers we have:

We should be able to simplify that with some assumptions... Let's assume that "x" is much smaller than 0.500 and much larger than 10^-7. That gets us the simplified expression:

Solving this expression, we get x = 1.41x10^-8, which gives us a pOH = -log(1.41x10^-8) = 7.85, and pH = 6.15. Hmm, that's not neutral, that's acidic. The whole point of this was to prove that nitrate was a neutral ion. This is a disaster.

BUT WAIT!

We made some assumptions. We didn't check our assumption after we solved for "x". This is why I always tell you to check assumptions... We assumed that "x" would be much smaller than 0.500, which it is, but we also assumed that "x" was much larger than 10^-7, which it absolutely is not! So the assumptions we made were an oversimplification of the problem and that's where we entered the danger zone. Looking back at our K_b expression, we can only simplify it to:

That's still going to require the quadratic formula to solve. I'll let you work out the details, but the result should be that x = 7.96x10^-9. That means:

[OH^-1]eq = 10^-7 + (7.96x10^-9) = 1.08x10^-7 M

pOH = -log(1.08x10^-7) = 6.9667

pH = 14 - 6.9667 = 7.0333

That's not exactly 7.0000000000 neutral, but it's pretty darn close, especially if we're thinking about this in terms of selecting a visual acid-base indicator for a titration. 

Quiz 2 Hint

Quiz 2 has a heat capacity problem, and the heat capacity value given in the problem is in units of "joules per (mole Kelvin)". Remember, for heat capacity problems, we're looking at changes in temperature. Whether it's Celsius or Kelvin, the ΔT is the same because 1 degree C and 1 K are exactly the same size. You can use either, but make sure you're doing everything in the right order. A temperature that changes from 3.45°C to 12.71°C is changing by:

ΔT = 12.71°C - 3.45°C = 9.26°C

If you prefer Kelvin, convert BOTH individual temperatures to K, THEN subtract:

ΔT = 285.86K - 276.60K = 9.26K

Other questions, let me know...

Quiz 1 questions...

I've gotten a few questions about Quiz 1. Here are a few hints...

Question 1: How many moles of nitrogen atoms are in a given mass of iron(III) nitrate?
Start with a balanced chemical formula. How many moles of N are in each mole of iron(III) nitrate? Use the mole ratio to get moles of N from the moles of iron(III) nitrate.

Question 2: How many moles of chlorine atoms are in a given volume of a given concentration vanadium(IV) chlorate solution?
This is almost the same as question 1, but you need to use concentration and volume to determine the moles of vanadium(IV) chlorate instead of mass and molar mass.

Two other helpful hints:
1. Don't wait until the day before a quiz is due to work on it. This is especially true because...
2. When you have a question, show me your work. If you have the problem all set up, it's much easier for me to look at what you've done and help you work through the problem correctly. It's always easier to help you when you can give me an idea of where you're getting confused.

Spring Semester 2014 is here!

We're 18 hours away from the first Gen Chem II class of 2014! Watch this blog for general info about the topics we'll be exploring and answers to specific questions I get by email. I'll also be tweeting links and daily summaries using #GenChem2014. And I might try a few other things this semester…

A couple tips that are always useful:
1. If there's a way to calculate moles of a substance, that might be a good start.
2. Balance your chemical formulas. Then balance your chemical reactions. Balance is the key to chemistry. And skiing. If you don't balance your formulas and reactions, you're gonna have a bad time.
3. Be curious. That what science is all about.
4. Ask questions. Unless you're the only person in the room, it's pretty likely that someone else in the room has the same question you do.
5. Be bold. It's OK to answer a question incorrectly in class, we're learning chemistry here not juggling chainsaws. The other side of this is to be supportive of your classmates, we're all here to learn (including me).

Enjoy the beautiful weather today and I'll see you tomorrow.

Email question 2013-12-12

------------------------------
Hello Dr. Bodwin. I've been working on some old exams for studying and on the exam 1a from fall 2011, the last question is asking about the empirical formula, and I was just wondering when solving for each part, where do the last two numbers come from?
For example,
C -> (71.98 g)/(12.011 g/mol) = 5.993 mols -> 4.5 -> 9
------------------------------

Let me pull up the whole problem from the exam key:

{from:  http://www.drbodwin.com/teaching/exams/c150fe1ak.pdf}

When determining empirical formulas from percent composition data, the first step is to assume that you have 100g of sample. You can assume any amount of sample you like, but 100g simplifies things a little because if I have 100g of a sample and I know that 71.98% of that sample is carbon, then there must be 71.98g of carbon in the sample. Using the percentages given in the problem, we know that the "100g" sample contains 71.98g of carbon, 6.711g of hydrogen, and 21.31g of oxygen. Grams are great, but we want to count the number of different atoms, so we need to convert grams to moles... That's the first step in the calculation that is shown.
Once we have moles, we know the relative amounts of each element present in the sample, and we can write a balanced chemical formula:

C_5.993H_6.658O_1.332

Hmm, that doesn't look quite right... But at this point, we have the correct relationship between the moles of each element, so we can force those relationships to be whole numbers by dividing all of them by the smallest one. Essentially, we're saying "what if 1.332 actually represents 1 oxygen atom?" Dividing them all gives the forumla:

C_4.5H₅O₁

Still not perfect, but it's a lot more "normal" looking than the first formula. Again, since we know that the relationship between moles is correct here, we can multiple all the subscripts by something that gives us a nice, round, whole number ratio. If we double everything, we get:

C₉H₁₀O₂

And now we have a good empirical formula for this elemental analysis.

Other questions? Let me know.

Suggested problems posted

Chapter 5 (and 4, sort of) suggested problems are posted:
http://www.drbodwin.com/teaching/genchem.php

Enjoy. I also tweaked the calendar on that page to reflect the changes we've had.

Problem Set 2

There's a key posted for the problem set we did in class that was titled "Problem Set #2" on my Gen Chem webpage:
http://www.drbodwin.com/teaching/genchem.php
Direct link to the key:  http://www.drbodwin.com/teaching/problemsets/c150gps02k.pdf

Don't forget, there are also old exams posted at:
http://www.drbodwin.com/teaching/examarchive.php

Other questions, let me know…

In-Class Problem Set #01

If you want to check your answers to the first problem set we looked at in class, here's the link:

http://www.drbodwin.com/teaching/problemsets/c150gps01k.pdf

I hope studying is going well, let me know if you have any additional questions. And don't forget, if you need a little study break tomorrow afternoon, the Dragon Soccer team has a home game at 3pm. After today's rain, tomorrow looks like it will be a lovely day to watch some soccer.

Carbons in a propane sample

Today in class we looked at a problem that some of you didn't quite get to the end of by the time class ended. Here it is. If you haven't already worked it through, give it a good try before you jump ahead to the answer...

How many carbon atoms are in a 37.43L sample of propane gas at 17.52°C and 1.472atm?

This starts out as an Ideal Gas Law problem with a single set of conditions.

PV = nRT

Plugging in the values from the problem:

(1.472atm)(37.43L) = n(0.08206^L.atm/_mol.K)((17.52+273.15)K)

n = 2.3099mols of C₃H₈(g)

{NOTE: I'm in the middle of the problem, so I'm not rounding for significant figures yet, 

but it looks like 4 sig figs would be good at this point…}

Each propane molecule contains 3 carbon atoms, so each mole of propane molecules contains 3 moles of carbon atoms.

(2.3099mols C₃H₈) (3mols C/1mol C₃H₈) = 6.92974mols C

(6.92974mols C) (6.022x10^{23 C atoms}/_{mol C}) = 4.173x10²⁴ C atoms in the sample.

DISCLAIMER: This problem assumes that propane is behaving as an ideal gas under these conditions. There's a pretty good chance that it would not be ideal under these conditions, but for the purposes of this problem let's assume it is.

Welcome to Fall 2013!

Welcome to the first day of Fall 2013 at MSUM!

 If you're looking for the textbook for class, here's the link on Amazon. http://www.amazon.com/Chemistry-Science-Context-Third-Edition/dp/0393934314/ref=sr_1_1?ie=UTF8&qid=1377519676&sr=8-1&keywords=chemistry+3rd+edition+gilbert
This is not an endorsement or recommendation of amazon.com, it's just the quickest way I could find a fairly universal link to the correct textbook. The book is available at the MSUM Bookstore and many other places.

In-Class problems 2013-07-18

Today was buffers day in class so we looked at 294 different ways to use the Henderson-Hasselbalch equation. Remember, the Henderson-Hasselbalch equation is just a rearrangement of the K_a expression we use for understanding any acid... You can always plug directly into the K_a expression and get the same result you will get using Henderson-Hasselbalch. So the problems...

1. Combine 12.642g of chlorous acid with 15.372g of lithium chlorite, dilute to 500.0mL. What is the expected pH of this buffer? (Chlorous acid K_a = 1.12x10^-2)

Plug into the Henderson-Hasselbalch equation...

And solve...

2. Prepare 500.0mL of a 0.650M HN₃/N₃^-1 buffer at pH = 5.10 from HN₃(s) and NaN₃(s). (Hydrazoic acid K_a = 1.93x10^-5)

Start off by solving for the ratio of conjugate acid to conjugate base using either the Henderson-Hasselbalch equation or just an unmodified K_a expression. I'll use H-H...

[N₃^-1] / [HN₃] = 2.4297

[N₃^-1] = 2.4297[HN₃]

Now that we know the ratio of these concentrations, we can solve for the actual concentrations by using the relationship...

0.650M = [HN₃] + [N₃^-1]

0.650M = [HN₃] + 2.4297[HN₃] = 3.4297[HN₃]

[HN₃] = 0.1895M

[N₃^-1] = 0.650 – [HN₃] = 0.650 – 0.1895 = 0.4605M

To make a solution that's 0.1895M HN₃ at 500.0mL, we need...

(0.5000L)(0.1895M) = 0.09475mols HN₃

(0.09475mols HN₃)(43.029g/mol) = 4.077g HN₃(s)

To make a solution that's 0.4605M N₃^-1 at 500.0mL, we need...

(0.5000L)(0.4605M) = 0.23025mols N₃^-1

(0.23025mols N₃^-1)(65.011g/mol) = 14.969g NaN₃(s)

So we should be able to make the target buffer by combining 4.077grams of HN₃ and 14.969g of NaN₃ in enough water to make 500.0mL of solution.

NOTE: If you're actually making a buffer, be very careful about the order of addition of the components. Whenever you're combining a solid or concentrated solution with a solvent, it's usually a good practice to add the solid or concentrated stock to the larger volume of solvent slowly with very good mixing. Dissolving and/or mixing can liberate a LOT of heat in some cases that could be dangerous if the order of addition is reversed.

NOTE2: Hydrazoic acid is not a solid at room temperature, and the pure liquid is a non-trivial safety risk... We can talk about this type of a buffer on paper, but there's very little chance you (or I) will ever prepare or use a hydrazoic acid-based buffer.

3. What is K_a of a weak acid, “HA”, if a solution made by dissolving 0.316mol HA and 0.327mol A^-1 in water and diluting to 750.0mL has a pH of 9.374?

Plug in to Henderson-Hasselbalch or the generic Ka expression...

9.374 = pK_a + log (0.327 / 0.316)

pK_a = 9.359

K_a = 4.374x10^-10

4. What is K_b of a weak base, “B”, if a solution made by dissolving 0.143mol B and 0.158mol HB⁺¹ in water and diluting to 400.0mL has a pH of 5.975?

Similar to the previous problem, plug in to Henderson-Hasselbalch or the generic Ka expression...

5.975 = pK_a + log (0.143 / 0.158)

pK_a = 6.018

pK_b = 14 – 6.018 = 7.892

K_a = 1.043x10^-8

Good luck.

Rate Law and Integrated Rate Law Problem {2013-07-03}

Under some set of conditions, ammonia gas and fluorine gas react to form nitrogen trifluoride gas and hydrogen gas at 8.73ºC. What is the correct rate law expression (including the rate law constant) given the following data:

Rxn #	[NH3]0	[F2]0	Rate (M/min)
1	0.274	0.218	1.841x10-2
2	0.274	0.436	7.362x10-2
3	0.822	0.436	2.209x10-1

The rate law expression for this reaction is:

Rate₀ = k[NH₃]₀^x [F₂]₀^y

Comparing Rxn #1 and #2:

y = 2, the reaction is 2^nd order with respect to fluorine concentration

Comparing Rxn #3 and #2:

x = 1, the reaction is 1^st order w.r.t. ammonia concentration

Plugging in values from Rxn #1 and solving for k...

1.841x10^-2 M/min = k (0.274M)¹ (0.218M)²

k = 1.42 M^-2min^-1

In Rxn #2, how much time must pass before [F₂] = 0.324M?

Plugging in to the 2^nd order integrated rate law expression...

t = 0.558 minutes

In Rxn #1, how much time must pass before [NH₃] = 0.217M?

Plugging in to the 1^st order integrated rate law expression...

t = 0.164 minutes

Know Your Polyatomic Ions

I expect you to know polyatomic ions. They're part of the vocabulary of chemistry and if you have to look them up every time one pops up it will slow everything down. I was about to type up a post when I thought "Hmm, this seems like something I would have typed up before..." Sure enough just about a year ago...
http://chemistryingeneral.blogspot.com/2012/06/polyatomic-ions.html

Enjoy!

Colligative properties and gas law problems 2013-07-02

1. 12.64g of sodium sulfate is dissolved in 400.0mL of water. What are the boiling point and freezing point of the solution?

Na₂SO₄ = 142.041g/mol

12.64g / 142.041g/mol = 0.0889884mols Na₂SO₄

{NOTE: don't round sig figs in the middle of a problem...}

(0.0889884mols Na2SO4) / 0.4000kg water = 0.22247m Na₂SO₄

{NOTE: density of water is 1.0000g/mL...}

ΔT_bp = (0.512 ºC/m)(0.22247m)(3mol particles / 1mol Na₂SO₄) = 0.342ºC

{NOTE: I think I used a different value for the boiling point elevation constant in class. This one is correct.}

{NOTE: When Na₂SO₄ dissolves in water, the result is 2 Na⁺(aq) ions and 1 SO₄^-2(aq) ion. 3 particles.}

T_bp = 100.000ºC + 0.342ºC = 100.342ºC

ΔT_fp = (1.858 ºC/m)(0.22247m)(3mol particles / 1mol Na₂SO₄) = 1.240ºC

T_fp = 0.000ºC – 1.240ºC = -1.240ºC

2. A weather balloon is filled with 23.65L of an ideal gas at 28.73ºC and 1.042atm pressure. It is released and rises to an altitude where the temperature is -6.35ºC the pressure is 0.842atm. How many moles of gas are in the balloon and what is its volume when it reaches the described altitude?

Plugging in to the Ideal Gas Law:

PV = nRT

(1.042atm)(23.65L) = n(0.08206 L.atm/mol.K)(301.88K)

n = 0.9948 moles

{NOTE: Convert to kelvins! If you pay attention to the units on R you'll be less likely to forget...}

Now that we know how many moles of gas are present, the next part could be solved by plugging in to the regular Ideal Gas Law again:

PV = nRT

(0.847atm)V = (0.9948mols)(0.08206 L.atm/mol.K)(266.80K)

V = 25.7L

Or by using the comparative form of the Ideal Gas Law:

P₁V₁/n₁T₁ = P₂V₂/n₂T₂

{NOTE: Since “n” is not changing, we can drop it from the equation...

(1.042atm)(23.65L) / 301.88K = (0.847atm)V₂ / 266.80K

V₂ = 25.7L

Practice, practice, practice...

Begin again...

Summer 2013 Gen Chem II starts bright and early tomorrow morning! Kinetics, equilibrium, acids & bases, thermodynamics, redox, and nuclear chemistry all wrapped up in a neat little 5-week package. Let the games begin!
http://www.drbodwin.com/teaching/genchem.php

Moderation...

For the past couple weeks, I've been getting hammered by spam in the comments so I have turned on comment moderation (for now). If you have any comments that are on-topic, please be sure that they will be allowed, I'm just trying to avoid a flood of spam-bot garbage in the comments section of my posts. Hopefully this will be a short-term problem...

Have a great day.

Email question - van't Hoff factor

----------
Can you please explain how to determine the i factor.  Now matter how much I read the book and look at examples I can't figure it out.  Thanks,
----------
There are a couple things that could be "the i factor", but I'll assume this question is about the van't Hoff factor that's used when determining changes due to colligative properties. Remember, colligative properties are properties of a solution that depend only upon the number of solute particle, not the identity of those solute particles. It doesn't matter if those solute particles are molecules, or cations, or anions, or a big random mixture of them all. When an ionic compound dissolves in water, it dissociates (at least somewhat) into its component ions. The number of ions (particles) that each formula unit breaks into when it dissolves is the van't Hoff factor. For example, if NaCl(s) is dissolved in water, it forms 1 Na⁺¹(aq) ion and 1 Cl^-1(aq) ion, so each formula unit (NaCl) forms 2 particles (ions) in solution; the van't Hoff factor is "2". For polyatomic ions, the ions don't break down into their individual atom, they stay polyatomic ions, so if K₂SO₄(s) is dissolved in water, it forms 2 K⁺¹(aq) ions and 1 SO₄^-2(aq) ion in solution; the van't Hoff factor is "3".

Exam strategies...

How do you take an exam? The most important thing is, of course, knowing the material, but there are also some "tricks" that can work in your favor, especially on a standardized multiple-choice exam. Some of this comes down to game theory, so it can be useful in settings outside of exams.

1. Know the rules - In a standardized exam, there can be different rules and formulas used for calculating your score. Some of the formulas have the effect of penalizing you for guessing if you truly don't know the answer. Always check to determine whether or not there is a penalty for guessing. Personally, I prefer to just take the total number of correct answers and not mess around with an elaborate formula that penalizes you for guessing.

2. Improve your odds - When you look at a series of multiple choice answers, there are often answers included that have to be obviously wrong if you think about them a little bit. If you're looking at a 4-option multiple choice question, your odds of randomly guessing the correct answer without even reading the answers is 25%. Eliminate one wrong answer, and your odds increase to 33%. Eliminate two, you're at 50%. Keep this in mind when you get to a question that has you stumped; rather than hunt for the right answer, try to eliminate wrong answers. Be like Sherlock Holmes; when you have eliminated all other possibilities, the remaining possibility, no matter how unlikely, must be correct.

3. Build momentum - Don't get hung up on a question you don't know when there are others you know the answer to. Answering a few questions correctly builds confidence and helps you tackle the more challenging questions. This is a challenge for "linear" test-takers who naturally want to start with #1 and work through in order to the end of the exam. {I was a very linear test-taker, and to some extent I still am...} It's OK to skip a question and come back to it later. This also helps with time management on the exam.

4. Answer all the questions - This is related to #1 and #2 above. If there is no penalty for guessing, then make sure you answer every question, even if you have no idea what the correct answer is. For a 5-option multiple choice question, there's a 1 in 5 (20%) chance that you'll random guess it correctly. If you don't answer the question at all, there's a 100% chance that you'll get it wrong.

Good luck.

Redox Lab question...

A few people have asked about the chemical reactions for the Redox lab hand-in {Redox Hand-in}, and this is often a source of confusion for students, so let me answer it here. First of all, I think a number of people get confused because you're trying to over-think the question and make it more complex than it really is.

So you need to draw a voltaic cell... you can draw this by hand, or you can draw it electronically. DO NOT just find one online and copy-paste it into your hand-in, if that's all I wanted I would have pasted this {voltaic cell} into the hand-in myself before I posted it.

The part that causes some confusion is the "...write a correctly balanced net ionic equation for the spontaneous process..." Don't over-read that! The metal cation solutions you were using in lab were probably nitrate salts, but nitrate (or whatever anion might have been present) was a spectator in all of your reactions. Net ionic equations are actually easier than full-formula equations because they're not cluttered up with with a bunch of extra stuff, and net ionic equations actually just describe the CHEMISTRY that's happening rather than distracting you with a bunch of spectator ions and species.

Now, I'm not going to write out a net ionic equation that's the exact answer to one that you have to write, but here's an example. Let's say I made up the voltaic cell Fe|Fe⁺³||Cd⁺²|Cd with the black/negative lead of my meter hooked up to the Fe(s) electrode and the red/positive lead connected to the Cd(s). The potential I measure is -0.32V. Because the measured potential is negative, the cell is running backwards, so the spontaneous cell reaction is Cd|Cd⁺²||Fe⁺³|Fe. Translating that into a reaction, we can write the two half-reactions as:

Cd(s) <=> Cd⁺²(aq) + 2 e^- 

3e^- + Fe⁺³(aq) <=> Fe(s)

Adding those up gives the overall (or "net") reaction:

3Cd(s) + 2 Fe⁺³(aq) <=> 2 Fe(s) + 3 Cd⁺²(aq)

{Remember to multiply each half-reaction by an appropriate integer to make all the electrons cancel...}

That's a "correctly balanced net ionic equation for the spontaneous process" in this case. Now go do that for all the cells you measured in the experiment.