REVIEW QUESTIONS Chapter 19

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1 Chemistry 10 ANSWER KEY REVIEW QUESTIONS Chapter For each of the following unbalanced equations, (i) write the half-reactions for oxidation and reduction, and (ii) balance the overall equation in acidic solution using the half-reaction method a) MnO + Cl Mn + + Cl Cl Cl + e (oxid) MnO + e Mn + (red) No action necessary (with H O) Cl Cl + e MnO + e Mn + + H O (with H + ) Cl Cl + e H + + MnO + e Mn + + H O (multiply by factor) No action necessary H + + MnO + Cl Cl + Mn + + H O b) FeS + NO NO + SO + Fe + FeS SO + Fe e (oxid) + e NO (red) No action necessary (with H O) H O + FeS SO + Fe e + e NO + H O (with H + ) H O + FeS SO + Fe e + 8 H + H e NO + H O (multiply by factor) 1 H O + FeS SO + Fe + + e + H + H e 8 NO + 1 H O FeS H + SO + Fe NO + H O 1

2 . Balance the following redox reaction in acidic solution, and determine the oxidizing and reducing agents Mg (s) + (aq) Mg + (aq) + NH + (aq) reducing oxidizing agent agent Mg Mg + + e (oxid) + 8 e NH + (red) Mg Mg + + e + 8 e + NH Mg Mg + + e (with H O) + 8 e NH + + H O Mg Mg + + e (with H + ) +10 H e NH + + H O Mg Mg e (multiply by factor) +10 H e NH + + H O Mg H + Mg + + NH + + H O. Balance the following redox reaction in basic solution, and determine the oxidizing and reducing agents Al (s) + (aq) Al(OH) + NH (aq) reducing oxidizing agent agent Al Al(OH) + e (oxid) + 8 e NH (red) Al Al(OH) + e + 8 e NH (with OH and H O) Al + OH Al(OH) + e + H O 8 e NH + OH (with H O and OH ) Al + OH Al(OH) + e + H O + 8 e NH + 9 OH (multiply by factor) 8 Al + OH 8 Al(OH) + e +18 H O + e NH +7 OH 8 Al + OH H O 8 Al(OH) + NH

3 . Balance the following redox reaction in basic solution, and determine the oxidizing and reducing agents Fe(OH) (aq) + OCl (aq) FeO (aq) + Cl (aq) reducing oxidizing agent agent Fe(OH) FeO + e (oxid) OCl + e Cl (red) No action necessary (with OH and H O) OH + Fe(OH) FeO + e + H O H O + OCl + e Cl + OH (with H O and OH ) OH + Fe(OH) FeO + e + H O H O + OCl + e Cl + OH (multiply by factor) 10 OH + Fe(OH) FeO + e + 8 H O H O + OCl + e Cl + OH Fe(OH) + OCl + OH FeO + Cl + H O

4 . The diagram below shows a voltaic cell with the anode on the left side and the cathode on the right side. Given that this is a magnesium and aluminum cell, identify metals A and B, identify solutions A and B, write half-reactions for each electrode, direction of electron flow, the polarities of the anode and the cathode, calculate the cell potential, and write a shorthand cell notation. electrons Anode ( ) Mg Al Cathode (+) + Al + Mg Mg Mg + + e Al + + e Al Anode: Mg(s) Mg + (aq) + e E = +.7 V Cathode: Al + (aq) + e Al(s) E = 1. V Overall: Mg(s) + Al + (aq) Mg + (aq) + Al(s) E cell = 0.71 V Mg(s) Mg + (aq) Al + (aq) Al(s)

5 . A voltaic cell employs the reaction: Sn (s) + Ag + (aq) Sn + (aq) + Ag (s) Calculate the voltage produced by this reaction under standard conditions at C. (Use Table 18.1 in your textbook for standard reduction potentials). Anode: Sn Sn + + e E = +0.1 volts Cathode: Ag + + e Ag E = volts Overall: Sn + Ag + Sn + + Ag E cell = = +0.9 volts 7. The standard voltage ( ) for the voltaic cell shown below is +0.8 volts. In In + Cu + Cu Determine the standard reduction potential for: In + + e In Anode: In + + e In E =??? Cathode: Cu + + e Cu E = +0. volts E cell = E cat E an = +0.8 volts E an = E cat E cell = = 0. volts 8. A voltaic cell used the reaction shown below: Sn + (aq) + Hg + (aq) Sn + (aq) + Hg + (aq) a) Calculate the voltage for this reaction under standard conditions. (Use Table 18.1 in your textbook for standard reduction potentials) Anode: Sn + Sn + + e E = 0.1 volts Cathode: Hg + + e + Hg E = volts Overall: Sn + + Hg + Sn Hg E = = +0.7 volts b) Calculate Gº for this reaction. G = w max = nfe = mol e (900 C/mole e )(0.7 volts)= 1.x10 J

6 9. A voltaic cell uses the reaction shown below, with a measured standard cell potential of 1.19 V: Tl + (aq) + Cr + (aq) Tl + (aq) + Cr + (aq) a) Write the two half-cell reactions. Oxidation Cr + (aq) Cr + (aq) + e E = +0.0 V Reduction Tl + (aq) + e Tl + (aq) E =??? b) What is the E for the reduction of Tl +? E cell = E (Tl + ) = 1.19 V E (Tl + ) = = 0.9 V c) Sketch the voltaic cell, label the anode and the cathode, and indicate the direction of the electron flow. 10. For each pair of substances below, use Reduction Potential in your textbook to determine the one that is the stronger oxidizing agent: The more positive the reduction potential, the easier the substance is reduced and the stronger the oxidizing agent it will be. a) Br (l) or I (s) (1.09 vs. 0.) b) Ag + (aq) or Cu + (aq) (0.80 vs. 0.) c) Cl (g) or Au + (aq) (1. vs. 1.) d) Mg(s) or K(s) (.7 vs..9) e) H O (aq) or Cr O 7 (aq) (1.78 vs. 1.)

7 11. The standard cell potential for the reaction shown below is 0. V: Eu + (aq) + e Eu + (aq) Use table of reduction potentials in your textbook to suggest two substances capable of reducing Eu + to Eu +. Any species with a standard reduction potential of less than 0. can reduce Eu + to Eu +, since it would be a stronger reducing agent that Eu +. Two examples are Zn and H (g). 1. The standard cell potential for the reaction shown below at 98 K is.0 V: Al (s) + I (s) Al + (aq) + I (aq) Calculate the emf generated by this cell when [Al + ] =.0 x 10 M [I ] = M Writing the two half-reactions: Using the Nernst equation: Al Al + + e I + e I 0.09 E = E log Q n = n + - Q = [Al ] [I ] 0.09 E =.0 log (.0x10 E =.0 ( 0.1) =.7 ) (0.010) 7

8 1. Use the standard reduction potentials listed in your textbook to determine the equilibrium constant for each of the following reactions: a) Zn(s) + Sn + (aq) Zn + (aq) + Sn(s) Anode: Zn(s) Zn + (aq) + e E = +0.7 V Cathode: Sn + (aq) + e Sn(s) E = 0.1 V Overall: Zn(s) + Sn + (aq) Zn + (aq) + Sn(s) E = 0. V At equilibrium, Nernst equation can be simplified to: = E log K n ne (0.) log K = = = K = 10 = 8.9 x b) Cd(s) + H + (aq) Cd + (aq) + H (g) Anode: Cd(s) Cd + (aq) + e E = +0.0 V Cathode: H + (aq) + e H (g) E = 0.00 V Overall: Cd(s) + H + (aq) Cd + (aq) + H (g) ne (0.0) log K = = = E = 0.0 V K = 10 =. x

9 1. A voltaic cell utilizes the reaction shown below at 98 K: Al(s) + Mn + (aq) Al + (aq) + Mn(s) a) Calculate the emf for this cell under standard conditions. Anode: Al(s) Al + (aq) + e E = +1. V Cathode: Mn + (aq) + e Mn(s) E = 1.18 V Overall: Al(s) + Mn + (aq) Al + (aq) + Mn(s) E = 0.8 V b) Calculate the emf for this cell when [Al + ] = 1. M and [Mn + ] = 0.10 M. E = E E = 0.8 E = log Q n = n 0.09 (1.) log (0.10) (0.0) = 0. V 1. Use the standard reduction potentials listed in your textbook to determine the equilibrium constant the following reactions: O (g) + H + (aq) + Fe + (aq) Fe + (aq) + H O(l) Anode: Fe + (aq) Fe + (aq) + e E = 0.77 V Cathode: Overall: H + (aq) + O (g) + e H O(l) E = +1. V ne (0.) log K = = = E = 0. V K = 10 = 1. x

(aq) 5VO2 + (aq) + Mn 2+ (aq) + 10H + + 4H 2 O. (aq) 5VO2 + (aq) + Mn 2+ (aq) + 2H + (aq) basic solution. MnO2 + 2H 2 O) 3H 2 O + I IO 3

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