Nelson Chemistry Chapter 12 Quiz

Click on the correct answer



1 Reduction is indicated by
a loss of electrons
a decrease in oxidation number
a positive ion getting a higher positive charge
2OH-(aq) + SO2(g) ------> SO32-(aq) + H2O(l)

2 A solution of acidified potassium permanganate is stored in an iron container. The net ionic equation for a reaction that occurs is
MnO4-(aq) + 8H+(aq) + 5K(s) -----> Mn2+(aq) + 4H2O(l) + 5K+(aq)
2MnO4-(aq) + 16H+(aq) + 5Fe(s) -----> 2Mn2+(aq) + 5Fe2+(aq) + 8H2O(l)
MnO4-(aq) + 8H+(aq) + Fe2+(aq) ------> Mn2+(aq) + Fe3+(aq) + 4H2O(l)
MnO4-(aq) + 8H+(aq) + Fe(s) -----> Mn2+(aq) + 4H2O(l) + Fe2+(aq)

3 In the half-reaction Al3+(aq) + 3e- ----> Al(s), the aluminum ions are undergoing
oxidation
neutralization
electroysis
reduction

4 In the reaction 2Ag+(aq) + Cd(s) ------> Cd2+(aq) + 2Ag(s), the strongest oxidizing agent is
Ag+(aq)
Cd(s)
Cd2+(aq)
Ag(s)

5


Ir(s) + Tl+(aq) ------> no reaction

Ir3+(aq) + Re(s) ------> no reaction

3Tl+(aq) + Y(s) ------> 2Tl(s) + Y3+(aq)

In these reactions, the strongest oxidizing agent is
Tl+(aq)
Y3+(aq)
Ir3+(aq)
Re3+(aq)

6 Which reactants will result in a spontaneous reaction?
Fe2+(aq) + Pb2+(aq)
Cr2+(aq) + Zn2+(aq)
Sn2+(aq) + I2(s)
Na+(aq) + Pb(s)

7
Given the following spontaneous reactions,


X2+(aq) + Y(s) -----> X(s) + Y2+(aq)

X2+(aq) + Q(s) -----> X(s) + Q2+(aq)

Y2+(aq) + Z(s) -----> Y(s) + Z2+(aq)

Z2+(aq) + Q(s) ------> Z(s) + Q2+(aq)

the strongest oxidizing agent is
Q2+(aq)
Y2+(aq)
Z2+(aq)
X2+(aq)

8 The reducing agent capable of converting 1.0 mol/L Sn4+(aq) ions to Sn2+(aq) but not capable of converting 1.0 mol/L Sn2+(aq) to Sn(s) is
Pb(s)
Ni(s)
Cu(s)
Cr(s)

9 Using acidified 0.20 mol/L Cr2O72-(aq), a student titrated a solution of Sn(NO3)2(aq) of unknown concentration. In the titration, 34.0 mL of oxidizing agent were required to react with 43.0 mL of the reducing agent. What was the concentration of Sn(NO3)2(aq)?
0.17 mol/L
0.47 mol/L
0.53 mol/L
0.76 mol/L

10 In the methane-oxygen fuel cell reaction

CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g) Eo = 1.05 V

oxidation numbers show that
oxygen atoms lose electrons
hydrogen atoms gain electrons
O2(g) is the reducing agent
carbon atoms lose electrons