1. International University for Science & Technology College of Pharmacy
General Chemistry
(Students of Dentistry)
Prof. Dr. M. H. Al-Samman

2. CHEMISTRY
Chapter 4 2

3. Solutions and Solution Stoichiometry
Solution is a homogeneous distribution of the components of a substance in the components of another .
we call the substance with the higher amount : a solvent (the substance doing the dissolving) and, the substance with the lesser amount : a solute ( the substance being dissolved)
As an exception :We consider water a solvent in any system regardless of it’s amount.

4. Types of Electrolytes
We can divide the solutions according to the conductivity into two types:
-Electrolytes
-Nonelectrolytes
Electrolytes : which are formed by ionic solute in ionic or polar solvent such as the solution of NaCl in water.
Electrolytes are divided into two types:
A)-Strong Electrolytes :which disassociate completely in water to give ions (anions and cations) and they are good conductors.

5. Types of Electrolytes Strong electrolytes include:
-Strong acids (HCl, HBr, HI, HNO3, H2SO4,HClO3, HClO4)
-Strong bases (IA and IIA hydroxides)
-Most water-soluble ionic compounds
B)-A weak electrolytes :which dissociate partially.
-Weak electrolyte solutions are poor conductors.
-Different weak electrolytes dissociate to different extents.
A nonelectrolyte : which is formed by a nonionic substance in a nonionic solvent , such as fat in diethyl ether .

6. Types of Electrolytes Nonelectrolyte does not dissociate.
--A nonelectrolyte is present in solution almost exclusively as molecules. (example: fat in chloroform).
-Nonelectrolyte solutions do not conduct electricity.

7. Acids … -Taste sour, if diluted with enough water to be tasted safely.7
Acids …
-Taste sour, if diluted with enough water to be tasted safely.
-May produce a pricking or stinging sensation on the skin.
-Turn the color of litmus or indicator paper from blue to red.
-React with many metals to produce ionic compounds and hydrogen gas.
-Also react with bases, thus losing their acidic properties.

8. 8
Bases …
-Taste bitter, if diluted with enough water to be tasted safely.
- Feel slippery or soapy on the skin.
-Turn the color of litmus or indicator paper from red to blue.
- React with acids, thus losing their basic properties.

9. Acids and Bases: The Arrhenius Concept9
Acids and Bases: The Arrhenius Concept
There are several definitions which may be used to describe acids and bases.
An Arrhenius acid is a compound that ionizes in water to give hydrogen ions H+ .strong acids give high concentration of hydrogen ions.
An Arrhenius base is a compound that ionizes in water to give hydroxyl ions OH– . Strong bases give high concentration of hydroxyl.
Arrhenius’s theory in electrolytes:
Certain substances dissociate into cations and anions when dissolved in water.

10. Acids and Bases: The Arrhenius Concept10
Acids and Bases: The Arrhenius Concept
- HClO₄ ,HClO₃,HBr, HI ,HCl, H₂SO₄ , HNO₃ , are strong acids.(strong acids ionize completely in diluted solutions).
- All bases of group one and group two are strong bases.(strong bases ionize completely in diluted solutions).
Neutralization it is the reaction between acid and base to give salt and water.
A salt is the product of the neutralization reaction between acid and base.
HCl + NaOH → NaCl + H₂O
Acid Base Salt Water

11. Acids and Bases: The Arrhenius Concept11
Acids and Bases: The Arrhenius Concept
Strong acid: it is the acid that ionizes completely in diluted solution.
Weak acid: it is the acid that ionizes partially in a diluted solutions.
Strong base: it is the base that ionizes completely in diluted solutions.
Weak base : it is the base that ionizes completely in diluted solutions.

12. Solutions and Solution Stoichiometry
Concentration of a solution: the quantity of a solute in a given quantity of solution (or solvent).
A concentrated solution contains a relatively large amount of solute vs. the solvent (or solution).
A dilute solution contains a relatively small concentration of solute vs. the solvent (or solution).
“Concentrated” and “dilute” aren’t very quantitative …

13. 13
Molar Concentration
Molarity (M), or molar concentration: It is the number of moles of the solute in one liter of the Solution.
Example:
A solution that is 0.35 M sucrose contains 0.35 moles of sucrose in each liter of solution.
-Keep in mind that molarity signifies moles of solute per liter of solution, not liters of solvent.

14. 14
Molar Concentration
Molarity (M), or molar concentration: It is the number of moles of the solute in one liter of
the Solution.
Example:
What is the molarity of ammonium sulphate salt in a solution containing 26.4 g of the salt in 4 lit. of the solution?.

15. We can read formulas in terms of moles of atoms or ions.

16. 16
Molarity
Q- Find the number of ammonium ions NH₄⁺ in a solution of ammonium sulphate containing g. in 4 Lit.? atomic masses: H=1, N=14, O=16, S=32, thin calculate the molarity of the compound. A-
-The formula of ammonium sulphate= ( NH₄ )₂SO₄
-The molar mass = 132 g.
-The number of moles= 10.56/132 = 0.08 mol.
The number of moles of NH₄⁺ ions is twice the number of moles of ammonium sulphate

17. 17
Molarity
The number of moles of NH₄⁺ ions : 2x0.08=0.16 mol. - The number of ammonium ions NH₄⁺ = 0.16 x 6.023x 10²³ = x 10²² ion -The molarity of ammonium sulphate ( NH₄ )₂SO₄= M= 0.08/4 = 0.02 mol/Lit

18. Example:
What is the molarity of a solution in which 333 g potassium hydrogen carbonate ( KHCO₃ ) is dissolved in enough water to make 10.0 L of solution?
Answer:
Number of moles = 333/100 = 3.33 mol.
The molarity = 3.33/10 = mol/Lit
We want to prepare a 6.68 molar solution of NaOH (6.68 M NaOH).
How many moles of NaOH are required to prepare L of 6.68 M NaOH?
How many liters of 6.68 M NaOH can we prepare with 2.35 kg NaOH?

19. Example: Answer: (a) 1 lit of NaOH contains 6.68 mol
0.500 lit needs x mol
Number of NaOH moles required :
0.500x6.68/1 = 3.34 mol
(b) Number of moles in 2.35 kg ( 2350 g):
2350/40 = mol.
number of liters: 58.75/6.68 = lit
Example:
The label of a stock bottle of aqueous ammonia indicates that the solution is 28.0% NH3 by mass and has a density of g/mL. Calculate the molarity of the solution.

20. Example
A chemical reaction familiar to geologists is that used to identify limestone. The reaction of hydrochloric acid with limestone, which is largely calcium carbonate, is seen through an effervescence—a bubbling due to the liberation of gaseous carbon dioxide:
CaCO3(s) + 2 HCl(aq)  CaCl2 (aq) + H2O(l) + CO2(g)
How many grams of CaCO3(s) are consumed in a reaction with 225 mL of 3.25 M HCl?

21. Example cont. Answer: Number of moles of hydrochloric acid:
0.225 x 3.25 = mol.
Mass of hydrochloric acid:
0.731 x 36.5 = g
CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
1 mol mol
100 g x36.5 x
Mass of the consumed CaCO3(s) :
x = g

22. Dilution of Solutions
Dilution: is the process of adding more solvent to the solution to lower or decrease the concentration.
- Addition of solvent does not change the amount of solute in a solution but does change the solution concentration.
-It is very common to prepare a concentrated stock solution of a solute, then dilute it to other concentrations as needed.

23. Conductivity Example
Example: Which of the following is the best conducting solution: 0.1M HCl, 0.2M CH₃COOH, 4M CH₃CH₂OH, 0.6M NH₄OH, 0.09M NaOH Answer: 0.1 M HCl as strong acid with high molarity

24. Conductivity Example
Example: Which of the following is the best conducting solution: 0.1M HCl, 0.2M CH₃COOH, 4M CH₃CH₂OH, 0.6M NH₄OH, 0.09M NaOH, 0.1M H₂SO₄ Answer: 0.1M H₂SO₄ as strong acid with high molarity(the number of ions provided by H₂SO₄ molecule are 3 ions while the number of ions provided by HCl molecule are 2).

25. Dilution Calculations …
… couldn’t be easier.
Moles of solute does not change on dilution.
Moles of solute = M × V
Therefore …
Mconc × Vconc = Mdil × Vdil

26. Indicators
Indicators : are organic compounds commonly used in analytical analysis to tell:
that a neutralization is complete, or:
if a solution is acidic or basic. Phenol red is …
… yellow in acidic solution …
… orange in neutral solution …
… and red in basic solution.

27. Dilution Calculations …
Example: How many ml of a concentrated solution of sodium hydroxide 0.4 M , are needed to prepare 1000 ml of 0.1 diluted solution. Answer: Mconc × Vconc = Mdil × Vdil 0.4 x Vconc = 0.1 x1000 Vconc= (0.1x1000)/0.4 = 250 ml

28. Dilution Calculations …
Example: If we diluted 200 ml of 0.2M solution of hydrochloric acid up to 1000 ml , what is the molarity of the diluted solution. Answer: Mconc × Vconc = Mdil × Vdil 0.2 x 200 = Mdil x 1000 Mdil = 0.04 mol/lit

29. 29
PH of Solutions
The PH:
Is the minus logarithm of the hydrogen ions concentration(in moles).
PH= -log [ H⁺ ]
Example:
The hydrogen ions concentration of hydrochloric acid solution is mole , calculate the PH value of the solution.
Solution:
PH= - log [10⁻³]
PH = 3  

30. PH of Solutions Example:30
PH of Solutions
Example:
15- What the PH of M solution of hydrochloric acid?
Solution:
PH= - log [ H⁺ ]
PH= - log [ 10³⁻ ]
PH= 3

31. PH of Solutions Example:31
PH of Solutions
Example:
16- Calculate the PH value of a solution of sulphuric acid containing 4.9 g per liter of the acid?
solution
-the molar concentration of the acid = 4.9/98=0.05 mol/lit
-the molar concentration of hydrogen ions:
[ H⁺ ]=2 [ M ] = 0.05 x 2 = 0.1 mol/lit
PH = - log [ 10⁻¹]
PH= 1

32. PH of Solutions Example:32
PH of Solutions
Example:
17- Calculate the PH value of a solution of hydrochloric acid containing 0.02 M of the acid?
solution
-since the hydrochloric acid is a strong acid, so the molar concentration of acid is equal to the molar concentration of hydrogen ions:
[ H⁺ ]=0.02
PH = - log [ 2x10⁻²]
PH=(-) [log2 + log10⁻²]
PH=(-) [ 0.30 – 2 ]
PH=(-) [ -1.7]
PH = 1.7

33. PH of Solutions - Since the ionic constant of water at 25C⁰ is :33
PH of Solutions
- Since the ionic constant of water at 25C⁰ is :
[H⁺] . [OH⁻] = 10⁻¹⁴
- And by writing the expression of ( - log ) for the above equation, we get :
PH + POH = 14
Where:
PH = - log[H⁺]
POH = - log[OH⁻]
14 = - log10⁻¹⁴

34. PH of Solutions Example:34
PH of Solutions
Example:
18- Calculate the PH value of M solution of sodium hydroxide?
Solution:
[ OH⁻ ] = 10⁻³
POH = - log [ OH⁻ ]
POH = - log [ 10⁻³ ]
POH = 3
We know that : PH + POH = 14
So we write:
PH + 3 = 14
PH = 11

35. 35
Hydrates
A hydrate is an ionic compound in which the formula unit includes a fixed number of water molecules associated with cations and anions
EOS
Examples:
BaCl2 . 2 H2O
CuSO4 . 5 H2O

36. 36
Hydrates
Notice:
-The water molecules in the hydrate compounds are part of the formula, so when we calculate the molecular mass of the hydrate compound, we should add the mass of the water molecules to the molar mass of the compound.
We name here some of the hydrates:
BaCl₂. 2H₂O : barium chloride dihydrate.
CuSO₄. 5H₂O : copper sulphate penta hydrate.
CoCl₃. 6H₂O : cobalt (III)chloride hexa hydrate.

37. Acid–Base Reactions: Net Ionic Equations
HCl + NaOH  H2O + NaCl + Q
In the reaction above, the HCl, NaOH, and NaCl all are strong electrolytes and dissociate completely.
The actual reaction occurs between ions.
Na+ and Cl– are spectator ions.
H+ + Cl– + Na+ + OH–  H2O + Na+ + Cl–
Q+ H+ + OH–  H2O
Q= 57.3 J/lit , it is the same for all neutralization reactions.

38. Calculating Ion Concentrations in Solution
In M Na2SO4:
-two moles of Na+ ions are formed for each mole of Na2SO4 in solution, so [Na+] = M.
-one mole of SO42– ion is formed for each mole of Na2SO4 in solution, so [SO42–] = M.
An ion can have only one concentration in a solution, even if the ion has two or more sources.

39. Calculating Ion Concentrations in Solution
How many moles of Na+ ions are in a solution containing 0.24 mole of Na2SO4 and 0.46 mole of NaCl.
Na2SO4  2 Na+ + SO42–
1 mol mol
0.24 mol x mol
X= 0.48 mol
NaCl  Na+ + Cl –
1 mol mol
0.46 mol y mol
Y= 0.46 mol
The number of moles of Na+ : = 0.94 mol

40. Reactions that Form Precipitates
There are limits to the amount of a solute that will dissolve in a given amount of water.
-If the maximum concentration of solute is less than about 0.01 M, we refer to the solute as insoluble in water.
When a chemical reaction forms such a solute, the insoluble solute comes out of solution and is called a precipitate.

41. Guide Line in Precipitation
-all compounds of nitrate ion(NO₃⁻) , acetate (ch₃coo⁻ ) and perchlorate (ClO₄⁻) are soluble.
-all compounds of group (l) in periodic table and ammonium ion (NH₄⁺) are soluble.
- most compounds of chloride (Cl⁻), bromide (Br⁻) and iodide (I⁻) are soluble except of Ag⁺, Pb²⁺, and Hg₂²⁺
- most compounds of sulfate are soluble, except of group ll and Ag⁺, Pb²⁺, and Hg₂²⁺ are insoluble.
- most compounds of hydroxide ( OH⁻), sulfide (S²⁻) and phosphate ( PO₄³⁻ ) are insoluble.

42. Guide Line in Precipitation
-With these guidelines we can predict precipitation reactions: Example: -When solutions of sodium carbonate and iron(III) nitrate are mixed, a precipitate of Fe(CO₃)₃ will be formed .

43. Guide Line in Precipitation
-With these guidelines we can predict precipitation reactions: Example: -When solutions of lead acetate and calcium chloride are mixed, a precipitate of PbCl₂ will be formed.

45. Guide Line in Precipitation
Example: Assign the insoluble compounds out of the following compounds: (a)Ca(OH)₂ , (b)Na₂SO₄, (c)K₂S, (d)PbCl₂, (e)HgCl₂ Answer: (a) and (d) only

46. Guide Line in Precipitation
:Example
One cup (about 240 g) of a certain clear chicken broth yields g AgCl when excess AgNO3(aq) is added to it. Assuming that all the Cl– is derived from NaCl, what is the mass of NaCl in the sample of broth?
Answer:
AgNO₃ + NaCl → AgCl + NaNO₃
1 mol mol mol
X /143.32
Moles of NaCl : x= 0.03 mol
Mass of NaCl = 0.03 x 58.5 = g

47. Reactions Involving Oxidation and Reduction
Oxidation: it is loosing one electron or more (Loss of electrons) , and the positive increase of the oxidation number.
Reduction: it is gaining one electron or more (Gain of electrons), and the decrease of the oxidation number.
Both oxidation and reduction must occur simultaneously. .
Historical: “oxidation” used to mean “combines with oxygen”; the modern definition is much more general.

48. Oxidation Numbers
An oxidation number is the charge on an ion, or a hypothetical charge assigned to an atom in a molecule or polyatomic ion.
Examples: in NaCl, the oxidation number of Na is +1, that of Cl is –1 (the actual charge).
In CO2 (a molecular compound, no ions) the oxidation number of oxygen is –2, because oxygen as an ion would be expected to have a 2– charge.
The carbon in CO2 has an oxidation number of +4 (Why?)

49. Rules for Assigning Oxidation Numbers
- The oxidation number of any atom in the neutral state is 0
- The oxidation number of any compound is 0 and it is equal to the sum of the oxidation numbers of the atoms in the formula.
- The sum of the oxidation numbers of the atoms in an ion is equal to the charge on the ion. Such as NH₄+
- In compounds, the group 1A metals all have an oxidation number of +1 and the group 2A metals all have an oxidation number of +2.
- In binary ionic compounds, the oxidation number of fluorine is –1.( F, Cl, Br, I )
- In compounds, hydrogen has an oxidation number of +1.(except in the Hydrides like CaH₂ , Hydrogen has an oxidation number -1 ).

50. Rules for Assigning Oxidation Numbers
-In most compounds, oxygen has an oxidation number of
–2. ( except in hydrogen peroxide and its compounds where oxygen has oxidation number -1, such as H₂O₂ , Na₂O₂ )
-In binary compounds with metals, group 7A elements have an oxidation number of –1, group 6A elements have an oxidation number of –2, and group 5A elements have an oxidation number of –3.
- The maximum oxidation number of a nonmetal is equal to the group number.
For nitrogen, +5.
For sulfur, +6.
For chlorine, +7.
The minimum oxidation number is equal to the (group number – 8).

51. Examples in assigning the oxidation number
What are the oxidation numbers assigned to the under lined atoms of the following:
(a)KClO4 (b) Cr2O72– (c) CaH (d) Na2O2 (e) Fe3O4
Answer:
(a)K+1,O-2, so Cl = (d) Na+-O-O-Na+ just like H2O2
(b) 2Cr + ( 7x-2 ) = (e) Fe3O4 = FeO+ Fe2O3
2Cr-14 = Fe=2 ,Fe=3
(c) Ca = +2 , so H = Cr = +12
Cr = 6

52. Oxidizing and Reducing Agents
An oxidizing agent causes another substance to be oxidized.
The oxidizing agent is reduced.
A reducing agent causes another substance to be reduced.
The reducing agent is oxidized.
Mg + Cu2+  Mg2+ + Cu
What is the oxidizing agent? What is the reducing agent? Mg  Mg²⁺ e- oxidation
Cu²⁺ +2e = Cu reduction

53. Oxidation Numbers of Nonmetals
The maximum oxidation number of a nonmetal is equal to the group number.
For nitrogen, +5.
For sulfur, +6.
For chlorine, +7.
The minimum oxidation number is equal to the (group number – 8).
ClO4-: what is the ox number of chlorine? Is this an ox agent or a red agent?
Cl-? NO3-?

54. Activity Series of Some Metals
In the activity series, any metal above another can displace that other metal.
Mg metal can react with …
Will lead metal react with Fe3+ ions?
Will iron metal dissolve in an acid to produce H2 gas?
… Cu2+ ions to form Cu metal.
If a silver coin falls against a nail on a sunken ship, the coin will remain bright for a long time, while the nail will rust badly. Explain.

55. Oxidation–Reduction Oxidation- Reduction Example:
Balance the following equation using the redox method:
FeSO4 + KMnO4 + H2SO4 → K2SO4 + MnSO4 + H2O+Fe2 (SO4) 3
Answer :
To balance the equation , let’s display first the oxidation and the reduction in the equation :
- then we have to multiply each oxidized or reduced element by the number of oxidation change of the other, then we balance the equation:

56. Oxidation–Reduction Oxidation- Reduction
- then we have to multiply each oxidized or reduced element by the number of oxidation change of the other, then we balance the equation:
-Then :

57. Typical Example Example:
Balance the following equation using the redox method:
+ NO + H2O HNO Zn → Zn(NO3) 2
Answer:
1- We have to assign the elements where the oxidation and reduction are occurred, we notice that nitrogen element changed from oxidation number 5 to oxidation number 2 , and zinc changed from oxidation number 0 to oxidation number 2.
2- we have to multiply the number of change of the zinc by the nitrogen monoxide NO , and the number of change of the nitrogen by the zinc nitrate Zn(NO3) 2 :

58. Typical Example Oxidation–Reduction
3- we have to multiply the zinc in the left side of the equation by 3 to get it equal to the zinc in the right side of the equation, and since the total number of the nitrogen atoms in the right side of the equation is 8 we should multiply nitric acid by 8.
4- we have to multiply the water in the right side of the equation by four to get the number of hydrogen atoms equal to the number of
hydrogen atoms in the left side.

59. Class Practice Problem
Balance the following equations:
(a) Na(s) + H2O(l) NaOH(aq) + H2(g)
2Na(s) + 2H2O(l) NaOH(aq) + H2(g)
(b) Al(s) + HCl(aq) AlCl3(aq) + H2(g)
2Al(s) + 6HCl(aq) AlCl3(aq) + 3H2(g)
(c) C2H4(g) + O2(g) CO2(g) + H2O(l)
C2H4(g) + 3O2(g) CO2(g) + 2H2O(l)

60. Half-Reactions
In any oxidation–reduction reaction, there are two half-reactions
Oxidation takes place when a species loses electrons to another species
Cu(s)  Cu2+(aq) + e–
EOS
Reduction takes place when a species gains electrons from another species
Zn2+(aq) + e–  Zn(s)
Redox Rxn

61. Half-Reaction Method of Balancing Redox Equations
Separate an oxidation–reduction equation into two half-equations, one for oxidation and one for reduction
Cu(s)  Cu2+(aq) + e– oxidation
Ag+(aq) + e–  Ag(s) reduction
EOS
As with all chemical equations, one must have mass AND charge balance

62. Half-Reaction Method of Balancing Redox Equations
Balance the atoms and the electric charge in each half-reaction
Cu(s)  Cu2+(aq) e– oxidation
Ag+(aq) e–  Ag(s) reduction
EOS
Adjust the coefficients in the half-equations so that the same number of electrons appears in each half-equation 2

63. Half-Reaction Method of Balancing Redox Equations
Add together the two adjusted half-equations to obtain an overall oxidation–reduction equation
Cu(s)  Cu2+(aq) e– oxidation
2 Ag+(aq) e–  2 Ag(s) reduction
Cu(s) + 2 Ag+(aq) e–  Cu2+(aq) e– + 2 Ag(s)
EOS
Cu(s) Ag+(aq)  Cu2+(aq) Ag(s)
Write the net balanced chemical equation …

64. A Qualitative Description of Voltaic Cells
- A voltaic cell uses a spontaneous redox reaction to produce electricity.
- A half-cell consists of an electrode (strip of metal or other conductor) immersed in a solution of ions.
This Zn2+ becomes a Zn atom.
Both oxidation and reduction occur at the electrode surface, and equilibrium is reached.
This Zn atom leaves the surface to become a Zn2+ ion.

65. Important Electrochemical Terms
- An electrochemical cell consists of two half-cells with the appropriate connections between electrodes and solutions.
-Two half-cells may be joined by a salt bridge that permits migration of ions, without completely mixing the solutions.
-The anode is the electrode at which oxidation occurs.
- The cathode is the electrode at which reduction occurs.
- In a voltaic cell, a spontaneous redox reaction occurs and current is generated.
- Cell potential (Ecell) is the potential difference in volts between anode and cathode.
Ecell is the driving force that moves electrons and ions.

66. A Zinc–Copper Voltaic Cell
Positive and negative ions move through the salt bridge to equalize the charge.
… the electrons produced move through the wire …
… to the Cu(s) electrode, where they are accepted by Cu2+ ions to form more Cu(s).
Zn(s) is oxidized to Zn2+ ions, and …
Reaction: Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)

67. Activity Series of Some Metals
In the activity series, any metal above another can displace that other metal.
Mg metal can react with …
Will lead metal react with Fe3+ ions?
Will iron metal dissolve in an acid to produce H2 gas?
… Cu2+ ions to form Cu metal.
If a silver coin falls against a nail on a sunken ship, the coin will remain bright for a long time, while the nail will rust badly. Explain. 67

68. Cell Diagrams
- A cell diagram is “shorthand” for an electrochemical cell.
- The anode is placed on the left side of the diagram. (oxidized) element
- The cathode is placed on the right side (reduced one)
- A single vertical line ( | ) represents a boundary between phases, such as between an electrode and a solution.
- A double vertical line ( || ) represents a salt bridge or porous barrier separating two half-cells.

69. Standard Electrode Potentials
Since an electrode represents only a half-reaction, it is not possible to measure the absolute potential of an electrode.
The standard hydrogen electrode (SHE) provides a reference for measurement of other electrode potentials.
The SHE is arbitrarily assigned a potential of V.

70. Standard Electrode Potentials (cont’d)
-The standard electrode potential, E°, is based on the tendency for reduction to occur at an electrode.
- E° for the standard hydrogen electrode is arbitrarily assigned a value of V.
- All other values of E° are determined relative to the standard hydrogen electrode.
- The standard cell potential (E°cell) is the difference between E° of the cathode and E° of the anode.
E°cell = E°(cathode) – E°(anode)

71. Criteria for Spontaneous Change in Redox Reactions
- If Ecell is positive, the forward reaction is spontaneous.
- If Ecell is negative, the forward reaction is nonspontaneous (the reverse reaction is _____).
- If Ecell = 0, the system is at equilibrium.
- When a cell reaction is reversed, Ecell and DG change signs.

72. Measuring the Standard Potential of the Cu2+/Cu Electrode
The voltmeter reading and the direction of electron flow tell us that …
… Cu2+ is more easily reduced than H+, by volts.
Standard hydrogen electrode
Cu2+ + 2e  Cu E° = V

73. Measuring the Standard Potential of the Zn2+/Zn Electrode
The voltmeter reading and the direction of electron flow tell us that …
… Zn2+ is harder to reduce than H+, by volts.
Standard hydrogen electrode
Zn2+ + 2e  Zn E° = – V

74. Voltic Cells Examples Example:
Will copper metal displace silver ion from aqueous solution? That is, does the reaction
Cu(s) + 2 Ag+(1 M)  Cu2+(1 M) + 2 Ag(s)
occur spontaneously from left to right?
{Ag e Ag}……….reduction
{ Cu Cu e}……oxidation
E = Ecathod - E anode
= E Ag+/Ag - E Cu+2 /Cu
= 0.800v – 0.340v = v because the E positive
So displacement is spontaneous

75. Typical Voltic Cell Example
Draw the short hand diagram of the cell which does the reaction:
Cu(s) + 2 Ag+(1 M)  Cu2+(1 M) + 2 Ag(s)
If the electrode potential of Ag⁺/Ag=0.800v and the electrode potential of Cu/Cu⁺⁺=0.340v :
1- Calculate the cell potential of the cell
2-Does the reaction occur spontaneously from left to right?
Answer: 1-
E = E cathode - E anode
E= E Ag+/Ag - E Cu+2 /Cu
E= 0.800v – 0.340v = v
3- because the E positive value, So displacement is spontaneous

76. Other Secondary Cells
The nickel–cadmium (NiCd) cell uses a cadmium anode and a cathode containing Ni(OH)2.
- A NiCd cell can be recharged hundreds of times. It produces 1.2 V (a Leclanché cell produces 1.5 V).
Nickel–metal hydride cells (NiMH) use a metal alloy anode that contains hydrogen.
- In use, the anode releases the hydrogen, forming water. Like the NiCd cell, a NiMH cell produces 1.2 V.
Lithium-ion cells use a lithium–cobalt oxide or lithium–manganese oxide material as the anode. The electrolyte is an organic solvent containing a dissolved lithium salt.
- Many modern laptop computers and cellular phones use lithium-ion cells.