1. The document provides steps to convert between solubility in grams per liter, molar solubility in mol/L, molar concentrations of ions, and Ksp. 2. It gives the definitions of a reducing agent, oxidizing agent, and half-reaction. A reducing agent causes reduction, an oxidizing agent causes oxidation. Half-reactions involve electrons. 3. An anode is where oxidation occurs and a cathode is where reduction occurs in a voltaic cell. Electrons flow from the anode through the external circuit to the cathode.

1. KEY
GENERAL CHEMISTRY-II (1412)
S.I. # 26
1. The Solubility of an y compound in grams per liter can be converted to molar
solubility. The molar solubility can be used to determine the concentrations of ions
in a solution. The concentrations of ions can be used to calculate Ksp. The steps
can be reversed and solubility calculated from Ksp. Pg 741
2. Fill in the boxes below:
Compound Solubility (g/L)  Molar Solubility (mol/L) Molar Concentration of ions  Ksp
3. Ksp for AgI is 8.3x10-17 (appendix D) and Kf for Ag(CN)2- is 1x1021 (pg750 table
17.1) calculate the equilibrium constant for the reaction
AgI(s) + 2CN-(aq)  Ag(CN)2-(aq) + I-(aq)
AgI(s)  Ag+(aq) + I-(aq)
Ag (aq) + 2CN-(aq)  Ag(CN)2-(aq)
+
AgI(s) + 2 CN-(aq)  Ag(CN)2-(aq) + I-(aq)
K = Ksp x Kf = [Ag+][I-] x [Ag(CN)2-] = (8.3x10-17)(1x1021) = 8x104
[Ag+][CN-]2
4. Define reducing agent and oxidizing agent and give their functions.
A reducing agent is the substance that is oxidized and causes the reduction of some other
substance.
An oxidizing agent is reduced and causes the oxidation of some other substance.
5. What is a half-reaction?
A half reaction is a balanced chemical equation that includes electrons.
Oxidation half has the electrons on the product side
Reduction half has the electrons on the reactants side.
6. Define anode and cathode in terms of a voltaic cell.
A solid surface called an electrode where oxidation occurs is called the anode
The electrode where the reduction occurs is called the cathode.
The electrons released at the anode flow through an external circuit (where they
do work) to the cathode.
7. What is the function of a salt bridge?
Electrical neutrality in the solution is maintained by the migration of ions between the
two compartments through a device such as a salt bridge.

2. KEY
8. What is emf?
The emf is the electromotive force generated by a voltaic cell, which moves the
electrons from the anode to the cathode through the external circuit.
9. The emf of a cell is called the cell potential Ecell. And when it is under standard
conditions it is called the standard emf E° cell. For a half-reaction, E° red is a measure
of the tendency of the reduction to occur. The value of E° cell is positive for a voltaic
cell and can be found by the equation: E° cell = E° red(cathode) - E° red(anode). Fluorine
(F2) has the most positive value for E° red and is the strongest oxidizing agent.
10. Indicate whether the following balanced equations involve oxidation-reduction.
If they do, identify the elements that undergo changes in oxidation number.
A. PBr3(l) + 3 H2O (l)  H3PO3(aq) + 3 HBr (aq)
No redox
B.NaI (aq) + 3 HOCl(aq)  NaIO3 (aq) + 3 HCl (aq)
I is oxidized from -1 to +5; Cl is reduced from +1 to -1.
C. 3 SO2(g) + 2 HNO3 + H2O(l)  3 H2SO4(aq) + 2 NO(g)
S is oxidized from +4 to +6; N is reduced from +5 to +2.
D. 2 H2SO4(aq) + 2 NaBr(s)  Br2(l) + SO2(g) + Na2SO4(aq) 2 H2O (l)
S is reduced from +6 to +4; Br is oxidized from -1 to 0
11. complete and balance the half reactions and indicate whether the half is
reduction or oxidation.
A. Mo3+  Mo(s)
Mo3+ + 3e-  Mo(s) (reduction)
B. H2SO3(aq)  SO42-(aq)
H2SO3(aq) + H2O(l)  SO42-(aq) + 4H+(aq) + 2e- (oxidation)
C. NO3-(aq)  NO(g)
NO3-(aq) + 4H+(aq) + 3e-  NO(g) + 2 H2O(l) (reduction)
D. Mn2+(aq)  MnO2(s)
Mn2+(aq) + 4 OH- (aq)  MnO2(s) + 2 H2O(l) + 2e- (oxidation)
E. Cr(OH)3(s)  CrO42- (aq)
Cr(OH)3(s) + 5 OH- (aq)  CrO42- (aq) + 4 H2O(l) + 3e- (oxidation)