3.3E: Exercises

Practice Makes Perfect

Solve Coin Word Problems

In the following exercises, solve each coin word problem.

Exercise (PageIndex{1})

Jaime has $2.60 in dimes and nickels. The number of dimes is 14 more than the number of nickels. How many of each coin does he have? Answer 8 nickels, 22 dimes Exercise (PageIndex{2}) Lee has$1.75 in dimes and nickels. The number of nickels is 11 more than the number of dimes. How many of each coin does he have?

Exercise (PageIndex{3})

Ngo has a collection of dimes and quarters with a total value of $3.50. The number of dimes is seven more than the number of quarters. How many of each coin does he have? Answer 15 dimes, 8 quarters Exercise (PageIndex{4}) Connor has a collection of dimes and quarters with a total value of$6.30. The number of dimes is 14 more than the number of quarters. How many of each coin does he have?

Exercise (PageIndex{5})

A cash box of $1 and$5 bills is worth $45. The number of$1 bills is three more than the number of $5 bills. How many of each bill does it contain? Answer 10 at$1, 7 at $5 Exercise (PageIndex{6}) Joe’s wallet contains$1 and $5 bills worth$47. The number of $1 bills is five more than the number of$5 bills. How many of each bill does he have?

Exercise (PageIndex{7})

Rachelle has $6.30 in nickels and quarters in her coin purse. The number of nickels is twice the number of quarters. How many coins of each type does she have? Answer 18 quarters, 36 nickels Exercise (PageIndex{8}) Deloise has$1.20 in pennies and nickels in a jar on her desk. The number of pennies is three times the number of nickels. How many coins of each type does she have?

Exercise (PageIndex{9})

Harrison has $9.30 in his coin collection, all in pennies and dimes. The number of dimes is three times the number of pennies. How many coins of each type does he have? Answer 30 pennies, 90 dimes Exercise (PageIndex{10}) Ivan has$8.75 in nickels and quarters in his desk drawer. How many coins of each type does he have?

Exercise (PageIndex{11})

In a cash drawer there is $125 in$5 and $10 bills. The number of$10 bills is twice the number of $5 bills. How many of each are in the drawer? Answer 10 at$10, 5 at $5 Exercise (PageIndex{12}) John has$175 in $5 and$10 bills in his drawer. The number of $5 bills is three times the number of$10 bills. How many of each are in the drawer?

Exercise (PageIndex{13})

Carolyn has $2.55 in her purse in nickels and dimes. The number of nickels is nine less than three times the number of dimes. Find the number of each type of coin. Answer 12 dimes and 27 nickels Exercise (PageIndex{14}) Julio has$2.75 in his pocket in nickels and dimes. The number of dimes is 10 less than twice the number of nickels. Find the number of each type of coin.

Exercise (PageIndex{15})

Chi has $11.30 in dimes and quarters. The number of dimes is three more than three times the number of quarters. How many of each are there? Answer 63 dimes, 20 quarters Exercise (PageIndex{16}) Tyler has$9.70 in dimes and quarters. The number of quarters is eight more than four times the number of dimes. How many of each coin does he have?

Exercise (PageIndex{17})

Mukul has $3.75 in quarters, dimes and nickels in his pocket. He has five more dimes than quarters and nine more nickels than quarters. How many of each coin are in his pocket? Answer 16 nickels, 12 dimes, 7 quarters Exercise (PageIndex{18}) Vina has$4.70 in quarters, dimes and nickels in her purse. She has eight more dimes than quarters and six more nickels than quarters. How many of each coin are in her purse?

Solve Ticket and Stamp Word Problems

In the following exercises, solve each ticket or stamp word problem.

Exercise (PageIndex{19})

The school play sold $550 in tickets one night. The number of$8 adult tickets was 10 less than twice the number of $5 child tickets. How many of each ticket were sold? Answer 30 child tickets, 50 adult tickets Exercise (PageIndex{20}) If the number of$8 child tickets is seventeen less than three times the number of $12 adult tickets and the theater took in$584, how many of each ticket were sold?

Exercise (PageIndex{21})

The movie theater took in $1,220 one Monday night. The number of$7 child tickets was ten more than twice the number of $9 adult tickets. How many of each were sold? Answer 110 child tickets, 50 adult tickets Exercise (PageIndex{22}) The ball game sold$1,340 in tickets one Saturday. The number of $12 adult tickets was 15 more than twice the number of$5 child tickets. How many of each were sold?

Exercise (PageIndex{23})

The ice rink sold 95 tickets for the afternoon skating session, for a total of $828. General admission tickets cost$10 each and youth tickets cost $8 each. How many general admission tickets and how many youth tickets were sold? Answer 34 general, 61 youth Exercise (PageIndex{24}) For the 7:30 show time, 140 movie tickets were sold. Receipts from the$13 adult tickets and the $10 senior tickets totaled$1,664. How many adult tickets and how many senior tickets were sold?

Exercise (PageIndex{25})

The box office sold 360 tickets to a concert at the college. The total receipts were $4170. General admission tickets cost$15 and student tickets cost $10. How many of each kind of ticket was sold? Answer 114 general, 246 student Exercise (PageIndex{26}) Last Saturday, the museum box office sold 281 tickets for a total of$3954. Adult tickets cost $15 and student tickets cost$12. How many of each kind of ticket was sold?

Exercise (PageIndex{27})

Julie went to the post office and bought both $0.41 stamps and$0.26 postcards. She spent $51.40. The number of stamps was 20 more than twice the number of postcards. How many of each did she buy? Answer 40 postcards, 100 stamps Exercise (PageIndex{28}) Jason went to the post office and bought both$0.41 stamps and $0.26 postcards and spent$10.28. The number of stamps was four more than twice the number of postcards. How many of each did he buy?

Exercise (PageIndex{29})

Maria spent $12.50 at the post office. She bought three times as many$0.41 stamps as $0.02 stamps. How many of each did she buy? Answer 30 at$0.41, 10 at $0.02 Exercise (PageIndex{30}) Hector spent$33.20 at the post office. He bought four times as many $0.41 stamps as$0.02 stamps. How many of each did he buy?

Exercise (PageIndex{31})

Hilda has $210 worth of$10 and $12 stock shares. The numbers of$10 shares is five more than twice the number of $12 shares. How many of each does she have? Answer 15$10 shares, 5 $12 shares Exercise (PageIndex{32}) Mario invested$475 in $45 and$25 stock shares. The number of $25 shares was five less than three times the number of$45 shares. How many of each type of share did he buy?

Solve Mixture Word Problems

In the following exercises, solve each mixture word problem.

Exercise (PageIndex{33})

Lauren in making 15 liters of mimosas for a brunch banquet. Orange juice costs her $1.50 per liter and champagne costs her$12 per liter. How many liters of orange juice and how many liters of champagne should she use for the mimosas to cost Lauren $5 per liter? Answer 5 liters champagne, 10 liters orange juice Exercise (PageIndex{34}) Macario is making 12 pounds of nut mixture with macadamia nuts and almonds. Macadamia nuts cost$9 per pound and almonds cost $5.25 per pound. How many pounds of macadamia nuts and how many pounds of almonds should Macario use for the mixture to cost$6.50 per pound to make?

Exercise (PageIndex{35})

Kaapo is mixing Kona beans and Maui beans to make 25 pounds of coffee blend. Kona beans cost Kaapo $15 per pound and Maui beans cost$24 per pound. How many pounds of each coffee bean should Kaapo use for his blend to cost him $17.70 per pound? Answer 7.5 lbs Maui beans, 17.5 Kona beans Exercise (PageIndex{36}) Estelle is making 30 pounds of fruit salad from strawberries and blueberries. Strawberries cost$1.80 per pound and blueberries cost $4.50 per pound. If Estelle wants the fruit salad to cost her$2.52 per pound, how many pounds of each berry should she use?

Exercise (PageIndex{37})

Carmen wants to tile the floor of his house. He will need 1000 square feet of tile. He will do most of the floor with a tile that costs $1.50 per square foot, but also wants to use an accent tile that costs$9.00 per square foot. How many square feet of each tile should he plan to use if he wants the overall cost to be $3 per square foot? Answer 800 at$1.50, 200 at $9.00 Exercise (PageIndex{38}) Riley is planning to plant a lawn in his yard. He will need nine pounds of grass seed. He wants to mix Bermuda seed that costs$4.80 per pound with Fescue seed that costs $3.50 per pound. How much of each seed should he buy so that the overall cost will be$4.02 per pound?

Exercise (PageIndex{39})

Vartan was paid $25,000 for a cell phone app that he wrote and wants to invest it to save for his son’s education. He wants to put some of the money into a bond that pays 4% annual interest and the rest into stocks that pay 9% annual interest. If he wants to earn 7.4% annual interest on the total amount, how much money should he invest in each account? Answer$8000 at 4%, $17,000 at 9% Exercise (PageIndex{40}) Vern sold his 1964 Ford Mustang for$55,000 and wants to invest the money to earn him 5.8% interest per year. He will put some of the money into Fund A that earns 3% per year and the rest in Fund B that earns 10% per year. How much should he invest into each fund if he wants to earn 5.8% interest per year on the total amount?

Exercise (PageIndex{41})

Stephanie inherited $40,000. She wants to put some of the money in a certificate of deposit that pays 2.1% interest per year and the rest in a mutual fund account that pays 6.5% per year. How much should she invest in each account if she wants to earn 5.4% interest per year on the total amount? Answer$10,000 in CD, $30,000 in mutual fund Exercise (PageIndex{42}) Avery and Caden have saved$27,000 towards a down payment on a house. They want to keep some of the money in a bank account that pays 2.4% annual interest and the rest in a stock fund that pays 7.2% annual interest. How much should they put into each account so that they earn 6% interest per year?

Exercise (PageIndex{43})

Dominic pays 7% interest on his $15,000 college loan and 12% interest on his$11,000 car loan. What average interest rate does he pay on the total $26,000 he owes? (Round your answer to the nearest tenth of a percent.) Answer 9.1% Exercise (PageIndex{44}) Liam borrowed a total of$35,000 to pay for college. He pays his parents 3% interest on the $8,000 he borrowed from them and pays the bank 6.8% on the rest. What average interest rate does he pay on the total$35,000? (Round your answer to the nearest tenth of a percent.)

Everyday Math

Exercise (PageIndex{45})

As the treasurer of her daughter’s Girl Scout troop, Laney collected money for some girls and adults to go to a 3-day camp. Each girl paid $75 and each adult paid$30. The total amount of money collected for camp was $765. If the number of girls is three times the number of adults, how many girls and how many adults paid for camp? Answer 9 girls, 3 adults Exercise (PageIndex{46}) Laurie was completing the treasurer’s report for her son’s Boy Scout troop at the end of the school year. She didn’t remember how many boys had paid the$15 full-year registration fee and how many had paid the $10 partial-year fee. She knew that the number of boys who paid for a full-year was ten more than the number who paid for a partial-year. If$250 was collected for all the registrations, how many boys had paid the full-year fee and how many had paid the partial-year fee?

Writing Exercises

Exercise (PageIndex{47})

Suppose you have six quarters, nine dimes, and four pennies. Explain how you find the total value of all the coins.

Exercise (PageIndex{48})

Do you find it helpful to use a table when solving coin problems? Why or why not?

Exercise (PageIndex{49})

In the table used to solve coin problems, one column is labeled “number” and another column is labeled “value.” What is the difference between the “number” and the “value?”

Exercise (PageIndex{50})

What similarities and differences did you see between solving the coin problems and the ticket and stamp problems?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

How to Tighten Stomach Skin

This article was co-authored by Danny Gordon. Danny Gordon is an American College of Sports Medicine (ACSM) Certified Personal Trainer and Owner of The Body Studio for Fitness, a fitness studio based in the San Francisco Bay Area. With over 20 years of physical training and teaching experience, he has focused his studio on semi-private personal training. Danny received his Personal Trainer Certification from the California State University, East Bay and the American College of Sports Medicine (ACSM).

There are 9 references cited in this article, which can be found at the bottom of the page.

If you’ve recently lost a lot of weight – either from a weight loss program or after giving birth – you might have some loose skin around your belly. To tighten that skin, focus on exercises that will tighten your stomach. You can also do things like drink more water, eat protein-rich foods, and protect your skin to help tighten your skin. Be patient because it will take time to see results. However, keep in mind that you can only tighten the skin so much once it has stretched. [1] X Research source Along with offering some skin tightening benefits, building your stomach muscles will help to improve the appearance of your stomach by filling out the area, holding in your organs better, and preventing your back from arching out too much.

How to Know if You're Exercising Too Much

This article was co-authored by Joshua Grahlman, PT, DPT, FAFS. Dr. Joshua Grahlman, PT, DPT, FAFS, is the Founder and Chief Athlete Mechanic of Clutch PT + Performance, a private physical therapy clinic specializing in sports and orthopedics in New York City. With more than a decade of experience, Dr. Grahlman specializes in treating acute and chronic pain and injuries, sports performance optimization and post-operative rehabilitation. Dr. Grahlman earned his Doctorate of Physical Therapy (DPT) from Columbia University College of Physicians and Surgeons. He is one of just a few DPTs in New York City recognized as a Fellow in Applied Functional Science through the Gray Institute for Functional Transformation (GIFT). He is certified in Active Release Technique and Spinal Manipulation and is a TRX Suspension Training Specialist. Dr. Grahlman has spent his career treating athletes of all levels, from Ironman Champions and Olympians to marathoner moms. He consults for Triathlete, Men’s Health, My Fitness Pal and CBS News.

There are 19 references cited in this article, which can be found at the bottom of the page.

wikiHow marks an article as reader-approved once it receives enough positive feedback. In this case, 100% of readers who voted found the article helpful, earning it our reader-approved status.

There are many benefits to regular physical activity. Both cardio and strength training exercises can help improve your overall health. You can maintain your weight or even lose weight, improve your blood pressure, decrease your risk of diabetes, heart disease or stroke and even improve your mood. [1] X Trustworthy Source Centers for Disease Control and Prevention Main public health institute for the US, run by the Dept. of Health and Human Services Go to source However, if you're exercising too much or too hard, there can be negative side effects to your exercise routine. You may experience insomnia, increased heart rate and may even damage your heart in the long run. [2] X Research source Over exercising can also lead to a breakdown of muscles. This is in part due to the result of the release of cortisol, a stress hormone. The immune system can also become compromised, increasing your chances of getting sick. Make sure you pay attention to your body so you can take a rest if you're exercising too much.

Given the half-reduction reaction:

[FeCl_3 + e^- ightarrow Fe + 3Cl^-]

Calculate the Ksp of FeCl3 at 298 K

Given the following reaction at pH 7.0:

[CH_3COOH + CO_ <2(g)>+ 2H^+ + 2e^- ightarrow CH_3COCOOH + H_2O ]

1. Write down the reaction that produce s (H^+) i on in term of E o
2. Calculate the concentration o f (CH_3COCOOH) at 25 o C. Given the emf of the cell is -0.70 V.

A certain reaction in solution has provides H + ions and has an E º of .48V. If a calomel electrode is placed in solution (as part of a pH meter) and has an Eref of .11 V. Find the pH of the solution.

3.3E: Exercises

Answer the following to the best of your ability. Questions left blank are not counted against you. When you have completed every question that you desire, click the " MARK TEST " button after the last exercise. A new page will appear showing your correct and incorrect responses. If you wish, you may return to the test and attempt to improve your score. If you are stumped, answers to numeric problems can be found by clicking on "Show Solution" to the right of the question.

Do NOT type units into the answer boxes, type only the numeric values. Do NOT use commas or scientific notation when entering large numbers. Answer all non-integer questions to at least 3 significant figures. Correct answers MUST be within ± 1 unit of the third significant figure or they are scored as wrong.

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3.3E: Exercises

Example #1: FeS + NO3¯ ---> NO + SO4 2 ¯ + Fe 3+ [acidic sol.]

1) Separate out the half-reactions. The only issue is that there are three of them.

How did I recognize there there were three equations? The basic answer is "by experience." The detailed answer is that I know the oxidation states of all the elements on EACH side of the original equation. By knowing this, I am able to determine that there were two oxidations (the Fe going +2 to +3 and the S going 𕒶 to +6) with one reduction (the N going +5 to +2).

Notice that I also split the FeS apart rather than write one equation (with FeS on the left side). I did this for simplicity showing the three equations. I know to split the FeS apart because it has two "things" happening to it, in this case it is two oxidations.

Normally, FeS does not ionize, but I can get away with it here because I will recombine the Fe 2+ with the S 2 ¯ in the final answer. If I do everything right, I'll get a one-to-one ratio of Fe 2+ to S 2 ¯ in the final answer.

2) Balancing all half-reactions in the normal manner.

3) Equalize the electrons on each side of the half-reactions. Please note that the first two half-reactions (both oxidations) total up to nine electrons. Consequently, a factor of three is needed for the third equation, the only one shown below:

Adding up the three equations will be left as an exercise for the reader. With the FeS put back together, the sum of all the coefficients (including any that are one) in the correct answer is 15.

1) Remove all the spectator ions:

Notice that I did not write Sb 3+ . I did this to keep the correct ratio of Sb as reactant and product. It also turns out that it will have a benefit when I select factors to multiply through some of the half-reactions. I didn't realize that until after the solution was done.

2) Separate into half-reactions:

3) Balance as if in acidic solution:

6e¯ + Sb2 6+ ---> 2Sb
2e¯ + 4H + + CO3 2 ¯ ---> CO + 2H2O
H2O + C ---> CO + 2H + + 2e¯

Could you balance in basic? I suppose, but why?

4) Use a factor of three on the second half-reaction and a factor of six on the third.

6e¯ + Sb2 6+ ---> 2Sb
3 [2e¯ + 4H + + CO3 2 ¯ ---> CO + 2H2O]
6 [H2O + C ---> CO + 2H + + 2e¯]

The key is to think of 12 and its factors (1, 2, 3, 4, 6). You need to make the electrons equal on both sides (and there are 12 on each side when the half-reactions are added together). You get 12 H + on each side (3 x 4 in the second and 6 x 2 in the third). You get six waters with 3 x 2 in the second and 6 x 1 in the third.

Everything that needs to cancel gets canceled!

6) Here's a slightly different take on the solution just presented. (a) Write the net ionic equation:

(b) Notice that charges must be balanced and that we have zero charge on the right. So, do this:

(d) Add back the sodium ions and sulfide ions to recover the molecular equation.

7) Here's a discussion of a wrong answer to the above problem.

1) write the half-reactions:

I wrote the iodide as I3 3 ¯ to make it easier to recombine it with the chromium ion at the end of the problem.

2) Balance as if in acidic solution:

I used water as the product for the hydrogen peroxide half-reaction because that gave me a half-reaction in acid solution. It will all go back to basic at the end of the problem.

3) Recover CrI3 by combining the first two half-reactions from just above:

5) Add the half-reactions together. Strike out (1) electrons, (2) hydrogen ion and (3) water. The result:

6) Add 10 hydroxides to each side. This makes 10 more waters on the right, so combine with the water alreadyon the right-hand side to make 32:

1) Write the three half-reactions:

2) Balance each half-reaction as if in acidic solution:

3) Multiply the third half-reaction by three:

12H2O + 9YO + ---> 3Y3O7 2 ¯ + 24H + + 9e¯

This gives nine electrons on each side when the three half-reactions are added together.

4) Add the half-reactions together and eliminate like items (two H + and one H2O):

5) Add 22 hydroxides to each side and eliminate like items:

22OH¯ + 2XO2 + + 10YO + ---> X2O4 3 ¯ + Y¯ + 3Y3O7 2 ¯ + 11H2O

Eleven waters were eliminated after the addition of the 22 hydroxides.

1) Write the three half-reactions while also stripping out spectator ions (only the sodium ion!):

2) Balance the half-reations as if in acidic solution (we'll change to basic in a moment):

3) The first step in equalizing the electrons is to see that the Bi and the nitrate MUST be in a 1:3 ratio:

4) Now we balance the 27 electrons on the left with 27 on the right:

5) Multiply through and add the three half-reactions:

6) Eliminate duplicate water and hydrogen ion:

7) Change to basic solution and elminate the nine waters that result:

8) Add the sodium ion back in:

Comment: I could have written the nitrate in the same fashion as I wrote the iodide in a problem above, where I wrote I3 3 ¯, but I did not. If I had, then there would have been 24 electrons on the left in the final half-reaction.

The above problem was formatted the morning of February 20, 2011 while aboard the Queen Mary, in a stateroom on the "B" level.

1) Split into half-reactions:

2) Balance in acidic solution:

3) Combine first two half-reactions to recover Cu3P:

5) Add, then eliminate water and hydrogen ion to get:

6) Recover phosphoric acid:

7) Although not needed, here's a full molecular equation:

1) Write the half-reactions:

Note how I kept the As and the S together in the first two half-reactions. This is because I know I will recombine them in the final answer, so I wanted to easily preserve the 2:3 ratio of the As and the S.

If I had not done this, then I would have had to make sure the As half-rection was multiplied through by 2 and the S half-reaction multiplied through by three. As it is, all I need to do is make sure the two half-reactions are multiplied through by the same factor, if a factor is needed.

Also, I could have eliminated the hydrogen from H3AsO4. There's no need to do so, so I didn't.

2) Balance in acidic solution (because of the H3AsO4):

3) Multiply through by factors selected to balance the electrons:

This gives 30e¯ on each side, when the half-reactions are combined.

4) Show the three half-reactions with the factors applied:

5) Add everything, eliminating only electrons:

6) Eliminate hydrogen ion and water:

7) Recombine (and rearrange):

I put the water at the end to make the final answer correspond a bit closer to the order the substances were in the original problem statement.

The 10 hydrogens on the left-hand side did not magically appear out of nowhere. Keep in mind that this reaction is occuring in acidic solution. If I had removed the hydrogen from the arsenic acid at the beginning, I would have had 8 hydrogen ions on the right-hand side when all was said and done. I would have simply added 10 more H + on each side, recovering the HNO3 and the H3AsO4.

1) Write the half-reactions:

2) Balance in acidic solution (because of the H3AsO4):

3) Combine the first two half-reactions:

4) Multiply through by factors selected to balance the electrons:

4) Show the two half-reactions with the factors applied:

5) Add everything, eliminating only electrons:

6) Eliminate hydrogen ion and water:

7) Recombine (and rearrange):

I'm going to approach the solution in the normal way and let the difficulties with this equation appear during the solution.

1) Separate into half-reactions:

Here's the first problem: the S is oxidized and the O is reduced, but they are both in the sulfate. This difficulty is solved by splitting the sulfate between the two half-reactions.

S8 ---> S 6+

O2 ---> O4 8 ¯

Notice how the four oxides each have a 𕒶 charge, giving the 𕒼 total.

2) Balance the half-reactions:

Here arises the second problem and it lies in the number of electrons. If you equalize the number of electrons (multiply first half-reaction by eight, multiply second half-reaction by six), you create problems with the S 6+ and the O4 8 ¯.

What do I mean by this? Keep in mind that splitting the sulfate was done to separate the reduction from the oxidation, this type of split does not occur in nature. Therefore, we will be forced to re-unite the S 6+ and the O4 8 ¯ at the end of the balancing process.

That means the the S 6+ and the O4 8 ¯ MUST remain equal. We cannot do that using the above two half-reactions. This is because we are working under two constraints (the second one being unique to this problem):

3) The answer is to introduce a third equation, one that does not appear in the original problem. The third equation is this:

H2 ---> 2H + + 2e¯

The reason for this lies in our need to equalize the electrons. This half-reaction gives us what we need.

Also, go back to the original equation. Notice that there is zero charge on the left-hand side and 𕒶 on the right-hand side. Where did the two extra electrons come from? The answer: from the H2 that was not written in the problem. You may suspect that the original statement of the problem was deliberately incomplete. I think that would be a correct feeling. By the way, I did not write this problem.

4) All three half-reactions:

5) Since there are now eight electrons on each side we can add the three half-reactions together and recombine the S 6+ and the O4 8 ¯:

Notice how I combined the hydrogen ion and the sulfate. Do you recognize the equation? It is the formation equation for sulfuric acid, you can probably look up its enthalpy in an appendix in the back of your book. The value given here is � kJ/mol.

Bonus Example: Cr 3+ + I¯ + Cl2 ---> CrO4 2 ¯ + IO4¯ + Cl¯ (two answers given, both correct)

L | Standard Electrode (Half-Cell) Potentials

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Summary

We now know that a demand curve is constructed by considering a consumers willingness to pay and different quantities, and the aggregate demand curve is the sum of all consumers demand curves. While price changes influence our quantity demanded, shocks such as changes in income, price changes of related goods, changes in tastes, and expectations can shift our demand, resulting in a different willingness to pay at every level. Now that we understand demand, we can turn to supply and its determinants.

Key Concepts and Summary

Assigning the potential of the standard hydrogen electrode (SHE) as zero volts allows the determination of standard reduction potentials, , for half-reactions in electrochemical cells. As the name implies, standard reduction potentials use standard states (1 bar or 1 atm for gases 1 M for solutes, often at 298.15 K) and are written as reductions (where electrons appear on the left side of the equation). The reduction reactions are reversible, so standard cell potentials can be calculated by subtracting the standard reduction potential for the reaction at the anode from the standard reduction for the reaction at the cathode. When calculating the standard cell potential, the standard reduction potentials are not scaled by the stoichiometric coefficients in the balanced overall equation.