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Determining rate order from concentration vs. time data

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As a reaction proceeds, we can measure the concentration of a reactant at various times. Then we use this data to determine the reaction order. For example, here’s the data for some arbitrary reaction during its first 30 seconds:

Time (s) [A] (M)
0 0.100
5 0.0632
10 0.0400
15 0.0253
20 0.0160
25 0.0101
30 0.0064

 

To find the reaction order, we create three graphs. One is [A] vs. time, a second is ln [A] vs. time and the third is 1/[A] vs. time. If the first graph is linear, the reaction is zeroth order with respect to [A]. If the ln of concentration graph is linear, the reaction is first order, and if the reciprocal concentration graph is linear, the reaction is second order. In the old days (before calculators), we would calculate all the numbers and graph the data by hand and hope that one was obviously linear. Here, I used my graphing calculator to quickly generate three scatter graphs, shown below:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

It should be pretty obvious that the red graph (ln [A] vs time) is the only linear graph, and that means the reaction is first order with respect to [A]. The rate equation is thus: rate = k[A] and the absolute value of the slope of the line (found using a linear regression) is the rate constant k.

What if the data has a bit of variability in it and the data doesn’t fit perfectly on a straight line on any of the three graphs? With your graphing calculator, create the three scatter graphs as I did above, then perform a linear regression on each graph. The graph with r2 closest to 1.0 will be the correct graph.

Determining rate order from initial reaction rate

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When you are studying chemical kinetics, a typical problem will look something like this:

Compounds A and B react together according to the following reaction:

A + 2 B → C + D

The following initial reaction rate data was collected. Determine the rate equation for this reaction.

Rxn [A] [B] Initial rxn rate
1 0.0500 M 0.0500 M 1.2 x 10-4 M/s
2 0.100 M 0.0500 M 2.4 x 10-4 M/s
3 0.100 M 0.100 M 9.6 x 10-4 M/s

 

These problems are pretty easy to solve if you have learned exponential equations in algebra 2. The trick is to compare two different reactions where only one compound changes its concentration. For example, if you compare reaction 1 with reaction 2, you will see that [A] changes and [B] is the same. If you compare reactions 2 and 3, you see that [B] changes and [A] is held constant. However, when you compare reactions 1 and 3, both [A] and [B] change.  So do not compare reactions 1 and 3.

Next, remember that your answer will be in the form

Rate = k [A]m [B]n

Therefore, your job is to determine the values of m, n, and k. You do this by taking the values from the table and plugging them into this equation.

Let’s find the value of m first. You need two equations where [A] changes and [B] does not—that is, equations 1 and 2.

\begin{matrix} (2) \quad & 2.4 \cdot 10^{-4} = k(0.100)^m(0.0500)^n \\ (1) \quad & 1.2 \cdot 10^{-4} = k(0.0500)^m(0.0500)^n \end{matrix}

Now divide the first equation by the second. The magic here is that k and the term with the n exponent both cancel out, leaving you with a simple expression:

 2.0 = 2^m

It should be pretty obvious that m = 1. What if the numbers don’t divide evenly and you end up with an equation such as

2.03 = 1.982^m?

Unless told otherwise, you can assume that the exponent is supposed to be an integer, and therefore m= 1 in this problem too.

Now, let’s solve for n. We need two equations where [B] changes and [A] does not. That would be equations 2 and 3. Set it up the same way as before:

\begin{matrix} (3) \quad & 9.6 \cdot 10^{-4} = k(0.100)^m(0.100)^n \\ (2) \quad & 2.4 \cdot 10^{-4} = k(0.100)^m(0.0500)^n \end{matrix}

And divide as before:

4.0 = 2^n

Thus n = 2. Now we can solve for k by taking any of the three original equations and plugging in all the given values. Let’s use equation 3.

9.6 \cdot 10^{-4} = k(0.100)^1(0.100)^2

Solving this equation gives k = 0.96. Therefore, our final answer is

Rate = 0.96 [A] [B]2

Now here’s an additional trick I give you at no extra charge: If you are pretty good with exponents and scientific notation, you should see that you can often solve for the values of m and n in your head. In our first step, the rate doubled when [A] doubled, so the rate with respect to A is first order (m = 1). In our second step, the rate quadrupled when [B] doubled, so the rate with respect to B is second order (n = 2). (That was certainly a lot easier than the work we did above!) Then plug the known values into the rate equation to solve for k.

Differential rate law vs. integrated rate law

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When we study rates of chemical reactions, we look for an equation that describes the reaction rate. The equation can be in one of two forms, known as the differential rate law and the integrated rate law.

If you have taken calculus, you’ll recognize these names and you even know how to derive the integrated law from the differential law. If you haven’t taken calculus, don’t worry! You probably won’t be asked to derive the equations. However, you should recognize each form of the law for zeroth, first and second order reactions.

The differential rate law gives the reaction rate as a function of the concentration of the reactants. The integrated rate law gives the concentration as a function of time. Both equations are useful, and you will probably use both in your chemistry class.

The table below shows the two rate laws for each reaction order in terms of a single reactant A.

Order  Differential rate law  Integrated rate law 
 Zeroth  Rate = k  [A] = -kt + [A]_0
 First  Rate = k[A]  \ln{[A]} = -kt + \ln{[A]_0}
 Second  Rate = k[A]2  \dfrac{1}{[A]} = kt + \dfrac{1}{[A]_0}

 

You might have noticed that each of the integrated law equations is in the form y = mx + b, the equation for a line. In the old days (before graphing calculators), this is the method that chemists used to determine the reaction order. If the graph of concentration vs. time is a straight line, the reaction is zeroth order. If the graph of ln(concentration) vs. time is a straight line, the reaction is first order. And if the graph of 1/concentration vs. time is a straight line, the reaction is second order.

 

Calorimetry calculations II—measuring heats of reactions

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In my previous post, I showed you how to work a calorimetry problem when you drop a hot block of metal into the calorimeter. In this post, we will look at the heat of reaction problem. The principles are the same: all of the heat released or absorbed by the reactants is absorbed or released by the water. In formula terms, this is written as ΔH = -q. From the M-CAT equation we examined in the previous post, we can write this as

\Delta H_{rxn} = -m_{solution} c_{solution} \Delta T_{solution}

We use this equation to find the molar heat of reaction as shown here:

Example: The temperature of a beaker containing 150 mL of water drops by 3.85°C when 8.00 g of ammonium nitrate are added. Determine the molar heat of solution for ammonium nitrate. Assume the specific heat capacity of the solution is 4.18 J/g·°C.

Answer: The heat of solution is defined as the ΔH for the following reaction:

NH_4NO_3 \; (s) \rightarrow NH_4^+ \; (aq) + NO_3^- \; (aq)

The mass of 150 mL of water is 150 g. Therefore the total mass of the solution is 158 g. So

\Delta H_{rxn} = -(158)(4.18)(-3.85) = 2540 \;  J

The molar mass of ammonium nitrate is 80.0 g, so the molar heat of solution is

\dfrac{2540 \; J}{8.00 \; g \; NH_4NO_3} \times \dfrac{80.0 \; g}{1 \; mol} = 25.4 \; kJ/mol

Notes: We must include the mass of the ammonium nitrate in the M-CAT equation because it cools down along with the water. We also assumed the specific heat capacity of the solution was the same as the specific heat capacity of pure water. For dilute solutions, this is a reasonable assumption.

Calorimetry calculations I—dropping hot blocks into water

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Calorimetry refers to the measure of how much heat is transferred in a reaction or phase change. We measure this heat with a device called a calorimeter. Although serious chemists will use serious calorimeters costing hundreds or thousands of dollars, you can duplicate much of their work by pouring some water into a Styrofoam cup. Your answers won’t be as precise as the serious chemists’ answers, but for two or three sig figs, this method is good enough.

There are two kinds of calorimetry problems; we’ll look at one here and the other in the next post. The first involves dropping a heated block of metal into the cup of water. The block cools down and the water warms up until the block and water are at the same temperature (thermal equilibrium). In the second type of problem, a chemical reaction takes place in the water. If the reaction is exothermic, the water absorbs the heat and warms up. If the reaction is endothermic, heat is absorbed from the water and the water cools down.

Let’s examine the first problem: There are two guiding principles we observe. The first is that heat will flow from the hot block into the water until the block and the water are at the same temperature. The second principle is that (we assume) there is no loss of heat to the air. In other words, the heat lost by the block exactly equals the heat absorbed by the water.

Both the heat lost by the block and the heat the water absorbs are governed by a simple equation:

q = mc \Delta T

Here q is the amount of heat gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature. Many teachers call this the M-CAT equation because the Δ looks like an A if you squint really hard. If this helps you remember it, you can call it that too. The value of c, the specific heat capacity, is a constant for a given substance. It measures how much heat must be absorbed by 1 g of a material to raise its temperature by 1°C. You might be asked to memorize the specific heat capacity of water. It is 1.00 cal/g·°C or 4.18 J/g·°C. You probably will not need to memorize specific heat capacity values for any other substance.

With the equation, calorimetry problems of this type are simple algebra problems. We have q_{water} = m_{water} c_{water} \Delta T_{water} and  q_{block} = m_{block} c_{block} \Delta T_{block} . Also, since the heat lost by the block equals the heat gained by the water, we can write q_{water} = -q_{block} . Therefore,

m_{water} c_{water} \Delta T_{water} = -m_{block} c_{block} \Delta T_{block} .

This is our magic equation. There are two kinds of questions you will answer with this equation. In one, we tell you the initial and equilibrium temperatures and you calculate the specific heat capacity of the block substance. In the other, you are given the initial temperatures and the specific heat of the block and you calculate the equilibrium temperature. Here are examples that show both types:

Example 1: A 25 g block of aluminum is heated to 85°C and placed into a calorimeter with 80 g of water at 25°C. The final temperature of the calorimeter and block is 28.8°C. Calculate the specific heat of aluminum.

Solution 1: We calculate the ΔT of the water (3.8°C) and the aluminum block (-56.2°C) and plug everything into the equation above.

(80)(4.18)(3.8) = -(25)(c)(-56.2). Solving gives c = 0.90 J/g·°C.

Example 2: A 25 g block of aluminum is heated to 85°C and placed into a calorimeter with 80 g of water at 25°C. The specific heat of aluminum is 0.90 J/g·°C. What is the equilibrium temperature of the calorimeter?

Solution 2: This problem (really just a rewording of example 1) is only a little more complicated to solve than example 1. First note that ΔT = Tf – Ti. We make this substitution and our magic equation becomes

(80)(4.18)(Tf – 25) = -(25)(0.90)(Tf – 85)

Distribute and simplify to get 334.4 Tf – 8360 = 1912.5 – 22.5 Tf. Solving gives Tf = 28.8°C.

Note: This is one of the rare times you do not need to convert the temperature from °C to kelvins. This is because the temperature term in the equation is a change in temperature and not an absolute temperature. Changing the temperature by 1°C is the same as changing it by 1 K.

SI unit prefixes

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Even with scientific notation, very large numbers and very small numbers can be awkward to work with. That is why scientists have developed a set of standard prefixes that are attached to a unit to indicate multiples of that unit. Each of the prefixes represents a particular power of ten. There are currently 20 prefixes in use:

 

Prefixes greater than 1

Factor

Name

symbol

1024

yotta

Y

1021

zetta

Z

1018

exa

E

1015

peta

P

1012

tera

T

109

giga

G

106

mega

M

103

kilo

k

102

hecto

h

101

deka

da

 

Prefixes less than 1

Factor

Name

symbol

10-1

deci

d

10-2

centi

c

10-3

milli

m

10-6

micro

μ

10-9

nano

n

10-12

pico

p

10-15

femto

f

10-18

atto

a

10-21

zepto

z

10-24

yocto

y

 

Do you have to memorize all of these? No. If you are a high school chemistry or physics student, you should know the prefixes in blue above. If you are an AP chemistry or AP physics student, you should also learn the prefixes in red. If you plan to be a scientist or an engineer, it wouldn’t hurt you to learn the rest, but the extreme prefixes are not so common.

Why do the prefixes skip three powers of 10? This allows us to use numbers that are always between 1 and 999. This is easiest to explain with an example. Let’s say we are measuring very small electrical currents. The following table shows how we would state the values without scientific notation, with scientific notation, and with prefixes. You can see how convenient this prefix method is:

Current Scientific notation Prefix notation
0.50 A 5.0 x 10-1 A 500 mA
0.050 A 5.0 x 10-2 A 50 mA
0.0050 A 5.0 x 10-3 A 5 mA
0.00050 A 5.0 x 10-4 A 500 μA
0.000050 A 5.0 x 10-5 A 50 μA
0.0000050 A 5.0 x 10-6 A 5 μA
0.00000050 A 5.0 x 10-7 A 500 nA
0.000000050 A 5.0 x 10-8 A 50 nA
0.0000000050 A 5.0 x 10-9 A 5 nA
0.00000000050 A 5.0 x 10-10 A 500 pA

How to use ICE charts

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When you are given the initial concentrations of the components in a reaction and you want to know the final concentrations at equilibrium, an ICE chart can be very useful. The ICE chart (some teachers call it an OCE chart or an IΔE chart) is a tool that helps you set up the proper algebraic expression in order to calculate the concentration change of each compound. It’s based on a very simple concept: the initial concentration of a compound plus the change in concentration of that compound must equal its final (equilibrium) concentration. To set up an ICE chart, follow these easy rules:

  1. Write out the balanced equilibrium equation.
  2. On the first line (I), put down the initial concentrations of each compound. If a compound is not in the initial mix, write 0. Be sure that you calculate the molarity, and not just the number of moles.
  3. Pick a direction for the reaction to proceed. It’s usually pretty obvious, but if you aren’t sure which way the reaction will go, just choose one direction. It doesn’t matter if you guess wrong.
  4. On the second line (C), use a variable to write the change in concentration for each compound using the balanced equation as a guide. For each compound that is consumed, write –Ax, where A is the coefficient of the compound in the balanced reaction. For each compound that is being formed, write +Ax. See the example below.
  5. On the third line (E), write the sum of the first two lines. This must be the final concentration of each compound.
  6. Use the equilibrium law and the value of Kc to write an expression with the equilibrium concentrations. Then solve this equation for x.
  7. Use this value to determine the equilibrium concentrations of each compound. [Note: if the value of x is negative, this means you guessed wrong in step 3, and the reaction actually proceeds in the opposite direction. You will still get the correct final values of each compound if you add or subtract this negative value in this step.]

Example 1: 1.0 mole of H2 and 1.0 mole of I2 are placed in a 2.0 L flask. The following reaction takes place at 400°C: H2 (g) + I2 (g) → 2 HI (g). What are the concentrations of the three compounds at equilibrium? The Kc for the formation of hydrogen iodide (HI) from hydrogen and iodine gases is 64.1 at 400°C.

Solution: Start by writing the equation and the initial concentrations. Don’t forget to divide by the volume to find the molarities of all the compounds.

 

 

 

Next, put in the changes in concentration in terms of x. This reaction must go to the right, and we get two moles of HI for each mole of the reactants:

 

 

 

 

Then add the two columns to calculate to equilibrium concentrations:

 

 

 

 

Finally, write the equilibrium expression and solve:

K_C=64.1= \dfrac{(2x)^2}{(0.5-x)^2}

\sqrt{64.1} = \sqrt{\dfrac{(2x)^2}{(0.5-x)^2}}

8.0= \dfrac{2x}{0.5-x}

4-8x=2x

4=10x

x=0.4

\therefore [ \text{H}_2]=[ \text{I}_2]=0.1 \, M; \, [ \text{HI}]=0.80 \, M

 Note that if you are given the change in concentration or the initial and final concentrations of a compound, you do not need to use an ICE chart. You can calculate the equilibrium values with a little bit of arithmetic. See the following example:

Example 2: The reaction in Example 1 is repeated at a different temperature with the same initial concentrations, and the final concentration of HI is determined to be 0.60 M. Determine the value of Kc at this new temperature.

Solution: To solve for Kc, we need to know the final concentrations of all three components. We are given the final value of HI, and knowing that we produced 0.60 M HI means that we must have lost 0.30 M of each of the reactants (use the balanced reaction to determine this). Since we started with 0.50 M of each, the final concentrations of each reactant must be 0.20 M (just subtract the numbers). So the equilibrium expression gives us:

K_C= \dfrac{(0.6)^2}{(0.2)(0.2)}=9.0

Balancing redox reactions in basic solutions

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When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how to balance a redox reaction in a basic solution.

Example : Iodide ions are oxidized in a basic solution by permanganate ions to produce iodine and Mn2+ ions. Show the balanced reaction.

Solution: I’ve made this example a little harder than in my other redox reaction posts because we first need to figure out from the example what the half reactions look like. This is not as complicated as it sounds. Just follow the English and turn it into chemical formulas. Our problem should therefore look like this:

I + MnO4 → I2 + Mn2+

Now comes the tricky part. It turns out it isn’t very easy to balance a redox reaction in basic solutions by adding OH ions, because when we balance the hydrogen atoms, we throw off the balance of the oxygen atoms. So we employ a clever technique. We first balance the reaction as though it were an acidic solution, then we change the solution from acidic to basic. This is a five-step process, and you’ll notice the first four steps are identical to the process for balancing a redox reaction in an acidic solution:

  1. Determine the number of electrons transferred in each half reaction, and balance the electrons.
  2. Balance the oxygen atoms by adding water molecules.
  3. Balance the hydrogen atoms by adding H+ ions.
  4. Check you answer by verifying that the charges are balanced.
  5. Cancel out the H+ ions by adding an equal number of OH ions to both sides, which changes the H+ ions into water.

Let’s see how this works on our example. The iodine atom goes from an oxidation number of 1- on the left to 0 on the right, so each iodine atom loses one electron. Since there are two iodine atoms in I2, we think of this half-reaction as losing two electrons. The manganese goes from an oxidation number of 7+ in the permanganate ion to an oxidation number of 2+. It has gained five electrons. To balance the electrons, we need a total transfer of ten electrons. Multiply the iodine compounds by five and the manganese compounds by two. (Note that the iodide ion was multiplied by 10 rather than 5 to balance the number of iodine atoms):

10 I + 2 MnO4 → 5 I2 + 2 Mn2+

Now we balance the oxygen atoms. There are 8 oxygen atoms on the left and none on the right, so we add 8 water molecules to the right side:

10 I + 2 MnO4 → 5 I2 + 2 Mn2+ + 8 H2O

Next we balance the hydrogen atoms by adding 16 H+ ions to the left (we pretend we have an acidic solution):

10 I + 2 MnO4 + 16 H+ → 5 I2 + 2 Mn2+ + 8 H2O

The equation should be fully balanced, and we check our work by verifying the charges balance. 10- (the iodide ions), 2- (the permanganate ions) and 16+ (the hydrogen ions) gives a net charge on the left of 4+. On the right side, we also have a total charge of 4+.

Finally, we are ready to change the reaction from acidic to basic. We cancel out the 16 H+ ions by adding 16 OH ions to both sides:

10 I + 2 MnO4 + 16 H+ + 16 OH → 5 I2 + 2 Mn2+ + 8 H2O + 16 OH

Because H+ and OH ions react to form water molecules, we simplify the left side:

10 I + 2 MnO4 + 16 H2→ 5 I2 + 2 Mn2+ + 8 H2O + 16 OH

Our last step is to simplify the equation by subtracting the 8 extra water molecules from the right side. We have to subtract the same number from the left side to keep the reaction balanced:

10 I + 2 MnO4 + 8 H2O → 5 I2 + 2 Mn2+ + 16 OH

(Look for other tips on this website if you need help with balancing simple redox reactions or redox reactions in acidic solutions.)

Balancing redox reactions in acidic solutions

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When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how to balance a redox reaction in an acidic solution.

Example 1: Balance the following reaction:

MnO4 + NO + H+ → Mn2+ + H2O + NO3

Solution: To balance reactions like this, use the following four-step procedure:

  1. Determine the number of electrons transferred in each half reaction, and balance the electrons.
  2. Balance the oxygen atoms by adding water molecules.
  3. Balance the hydrogen atoms by adding H+ ions.
  4. Check you answer by verifying that the charges are balanced.

In this reaction, the Mn atom goes from an oxidation number of 7+ in the permanganate ion to an oxidation number of 2+. It has gained five electrons. The nitrogen atom goes from an oxidation number of 2+ in the nitric oxide to 5+ in the nitrate ion. It has lost three electrons. To balance the electrons, we need to make the total number of electrons transferred equal 15 (because that’s the least common multiple of three and five!). We multiply the Mn ions by 3 and the N compounds by 5:

3 MnO4 + 5 NO + H+ → 3 Mn2+ + H2O + 5 NO3

The next step is to balance the oxygen atoms. There are 17 oxygen atoms on the left side of the reaction and 15 on the right side (before you include the water). Therefore we need two water molecules on the right:

3 MnO4 + 5 NO + H+ → 3 Mn2+ + 2 H2O + 5 NO3

Finally, we balance the hydrogen atoms (there are four of them in the water molecule on the right) by adding 4 hydrogen ions on the left:

3 MnO4 + 5 NO + 4 H+ → 3 Mn2+ + 2 H2O + 5 NO3

We check our work by verifying the charges balance. 3- (the permanganate ions) and 4+ (the hydrogen ions) gives a net charge on the left of 1+. 6+ (the manganese ions) and 5- (the nitrate ions) on the right also gives a net charge of 1+. We have succeeded!

Example 2: Complete and balance the following reaction. Assume it takes place in an acidic solution:

Cr2O72 + HSO3 → SO42 + Cr3+

Solution: In this reaction, the Cr atoms go from an oxidation number of 6+ in the dichromate ion to an oxidation number of 3+ in the chromate ion. Therefore, each chromium atom gains three electrons. But since there are two chromium atoms in the dichromate ion, we must think of this half-reaction as gaining six electrons overall. The sulfur atom goes from an oxidation number of 4+ to 6+, so it loses two electrons.

In order to balance the electrons, we need a net transfer of six electrons in each half-reaction. We multiply the ions with the sulfur atoms by three. Note that we also multiply the chromium ion on the right by two to balance the number of Cr atoms:

Cr2O72 + 3 HSO3 → 3 SO42 + 2 Cr3+

The next step is to balance the oxygen atoms.  There are 16 oxygen atoms on the left and 12 on the right, so we add 4 water molecules to the right side:

Cr2O72 + 3 HSO3 → 3 SO42 + 2 Cr3+ + 4 H2O

Finally, we balance the hydrogen atoms by adding H+ ions. There are three H atoms on the left and 8 H atoms on the right, so we need five H+ ions on the left:

5 H+ + Cr2O72 + 3 HSO3 → 3 SO42 + 2 Cr3+ + 4 H2O

We check our answer by checking the charge on each side. The left side has charges of 5+ (the H ions), 2- (the dichromate ion) and 3- (the hydrogen sulfite ions) for a net of zero. The right side has charges of 6- (the sulfate ions) and 6+ (the chromate ions) also for a net of zero.

(Look for other tips on this website if you need help with balancing simple redox reactions or redox reactions in basic solutions.)

Balancing redox reactions– the basics

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When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how the process works when both the oxidizing agent and the reducing agent are single elements.

Example (a very simple example): Balance the following reaction:

… Mg(s) + … Al3+(aq) → … Mg2+(aq) + … Al(s)

Solution: At first glance, it may appear that the reaction is already balanced because there is one magnesium atom and one aluminum atom on each side. But notice that the charges do not balance. Because aluminum gains three electrons in this reaction, but magnesium atoms lose only two electrons, you will need more magnesiums than aluminums. To find the right ratio, determine the least common multiple of 2 and 3, which is 6. Then multiply the magnesiums by 3 and the aluminums by 2 so that there is a total transfer of 6 electrons in this reaction:

3 Mg(s) + 2 Al3+(aq) → 3 Mg2+(aq) + 2 Al(s)

The charges balance, as required.

(Look for other tips on this website if you need help with balancing redox reactions in acidic or basic solutions.)

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