Gen Chem II Exam 3 Study Guide
Gen Chem II Exam 3 Study Guide 603613
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This 5 page Study Guide was uploaded by Emmanuel Notetaker on Wednesday April 13, 2016. The Study Guide belongs to 603613 at University of Cincinnati taught by waddell, D in Winter 2016. Since its upload, it has received 125 views. For similar materials see Gen Chem II in Chemistry at University of Cincinnati.
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Date Created: 04/13/16
GEN CHEM II EXAM 3 STUDY GUIDE CHAP 19,20 and 21. Neutralization is the chemical process in which ____________. a. sodium ions react with chloride ions to form sodium chloride b. hydrogen ions react with chloride ions to form hydrogen chloride c. sodium ions react with hydroxide ions to form sodium hydroxide d. hydrogen ions react with hydroxide ions to form water Neutralization is the process in which hydrogen ions combine with hydroxide ions to form water. Which of the following cannot be a buffer? a.a mixture of a weak acid and a weak base b.a mixture of a weak acid and a strong base c.a mixture of a strong acid and a weak base d.a mixture of a strong acid and a strong base Because strong acids and strong bases dissociate completely, no equilibrium is attained. A mixture of a strong base and a strong acid can never be a buffer. . Which of the following salts will produce a basic solution? a. NaCl b. KBr c. KCN d. Ca(NO )3 2 + Find the parent acid by adding H to the salt anion. Find the parent base by adding OH to the salt cation. The combination of a strong base and a weak acid will produce a basic solution. 2. What is the net ionic equation for the neutralization reaction between HF and KOH? a.H + OH → H O 2 + - b.K + F → KF c. H + KOH → H O + K + 2 d.HF + OH → H O +2F - HF is a weak acid and does not dissociate completely. KOH is a strong base and does dissociate completely. K+ is the only spectator ion. The net ionic equation is HF + OH → H O + F . - 2 - Calculate the H ion and OH ion - concentrations in a 0.50M solution of HBr. a. [H ] = 0.50M and [OH ] = 0.50M - + - b. [H ] = 0.50M and [OH ] = 2.0M c. [H ] = 0.50M and [OH ] = 2.0 ´ 10 M -14 d. [H ] = 1.0 ´ 10 M and [OH ] = 1.0 ´ 10 M -7 + - Use the dissociation constant of water to find the [H ] and [OH ]. When is the oxidation number of oxygen not -2? a. in water b. in oxyacids c. in peroxide d. in sulfuric acid The oxidation of oxygen is always -2, except in peroxides, such as hydrogen peroxide (H O ), where 2 2 it is -1. In the following reaction, which chemical species is the oxidizing agent? - + 2+ 5H O2+ 2MnO + 6H → 4Mn + 8H O 2 5O 2 a. H O 2 2 b. MnO 4- + c. H d. Mn 2+ - - The species that is reduced is MnO ; therefore4 MnO is the oxidizing 4gent. In a redox reaction, the number of electrons lost by the reducing agent ____________. a. equals the number of protons in the reducing agent b. equals the number of protons in the oxidizing agent c. equals the number of electrons lost by the oxidizing agent d. equals the number of electrons gained by the oxidizing agent In a redox reaction, the number of electrons gained by the reduced species equals the number of electrons lost by the oxidized species Which of the following types of reactions do not usually involve redox? a. synthesis b. decomposition c. single-replacement d. double-replacement Double—replacement reactions involve the transfer of cations and anions but do not involve the transfer of electrons. Predict the products of the following single—replacement reaction. Fe(s) + CuSO (aq) 4 ? a. CuS(s) + Fe SO 2aq) 4 b. Fe(s) + Cu(s) + SO (aq) 4 c. Cu(s) + FeSO (aq)4 d. FeCuSO 4 Determine the net change in oxidation number of chlorine in the following reaction: - - - ClO (4q) + Br (aq) → Cl (aq) + Br (g) (in aci2 solution) a. -8 b. -7 c. 7 d. 8 Chlorine went from an oxidation number of +7 to -1, a change of -8. Complete and balance the following redox equation. When properly balanced with whole—number coefficients, the coefficient of S is _________. H 2 + HNO → S 3 NO + H O 2 a. 1 b. 2 c. 3 d. 4 3H 2 + 2HNO → 3S 3 2NO + 4H O ; 3 2 Balance the following ionic equation. What is the coefficient of the reducing agent in this reaction? Fe (aq) + MnO (aq) +4H (aq) → Fe (aq) + Mn (aq) + H O(l) 2+ 2 a. 5 b. 1 c. 8 d. 4 2+ - + 3+ 2+ Fe (aq) + MnO (aq) +48H (aq) → Fe (aq) + Mn (aq) + 4H O(l) The reducing agent2will be oxidized. The species that is oxidized is Fe ; 5 2+ In any electrochemical cell, the cathode is the ________. a. positive electrode b. negative electrode c. electrode at which some species gains electrons d. electrode at which some species loses electrons The cathode is the electrode that acts as the reducing site during a redox reaction. At this electrode, some species will always gain electrons. In a voltaic cell, the salt bridge __________. a. is not necessary for the cell to work b. acts as a mechanism to allow mechanical mixing of the solutions c. allows charge balance to be maintained in the cell Under standard conditions, what is the standard cell potential for the cell? Cd|Cu 2+ || Cu |Cu? a. +0.74 V b. -0.74 V c. 0.06 V d. -0.06 V The cathode is copper ion reducing to copper metal. The anion is cadmium metal oxidizing to cadmium ion. The cell potential is 0.34 V - (-0.40 V) = +0.74 V A voltaic cell in which the reaction involves the combustion of one reactant with oxygen to produce electric energy is a(n) _____________. a. primary battery b. secondary battery c. fuel cell d. electrolytic cell Voltais cells that rely on combustion to produce electric energy are known as fuel cells The pieces of metal that are placed on the outside of ships to help prevent corrosion are _________. a. galvanized metals b. sacrificial cathodes c. sacrificial anodes d. electrolyzed metals The pieces of metal are sacrificed to protect the hull. As oxidation takes place, the sacrificial metal corrodes. Oxidation occurs at the anode. Hence, these are sacrificial anodes. 4. The Hall-Héroult process, which is used to make aluminum metal, electrolyzes a molten mixture of aluminum oxide and cryolite (Na AlF ) to red3ce a6uminum cations to aluminum metal. Why is the cryolite used in this mixture? a. The cryolite lowers the cell potential for the reduction of aluminum cations. b. The cryolite raises the cell potential for the reduction of aluminum cations. c. The cryolite increases the conductivity of the mixture. d. The cryolite lowers the melting point of the mixture and thereby lowers the cost of manufacture by lowering the amount of heat energy needed to melt the mixture. The cryolite will lower the melting point of the mixture as a colligative property. The lower melting point will allow the aluminum manufacture to be more cost effective.
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