sodium atoms have a single valence electron.
- B Fe2+ is oxidized in the redox reaction. To get the standard oxidation
potential for Fe2+ → Fe3+ + e−, just take the opposite of the standard
reduction potential. So E^0 ox for Fe2+ → Fe3+ + e− is −0.77 volts. Cl 2 is
reduced for the reduction half-reaction, and E^0 red is 1.36 volts. The
potential difference for the overall reaction (E^0 cell) is −0.77 volts + 1.36
volts = 0.59 volts.
- D Remember that many metals react with acids to produce hydrogen gas.
Your first step should be to write out the reaction. Here it is: Zn(s) +
2HCl(aq) → ZnCl 2 (aq) + H 2 (g). Notice that the products of this reaction
are Zn2+(aq), Cl−(aq) (from ZnCl 2 ), and H 2 (g).
- E This is a Le Chatelier’s principle question. How can we produce
crowding on the left side of the equation and drive equilibrium to the
right (increasing the SO 2 concentration)? Don’t be fooled by (C):
Catalysts do not affect equilibrium, but changing concentrations does
influence equilibrium. Increasing the concentration of O 2 will produce
crowding on the left side and lead to an increase in the concentration of
SO 2.
- A Decreasing the volume of the system will increase the concentration of
reactants. Why? Because the same number of molecules now exists in a
smaller space, and this increases the ratio of molecules per volume. How
will this affect the reaction rate? It will increase reaction rate because
decreasing the volume makes it more likely that molecules will collide.
What about equilibrium? Reducing the volume will force the equilibrium
to shift in the direction that produces fewer moles of gas. This means the
equilibrium concentration of reactants will decrease because equilibrium
will shift to the right (4 moles of gas on the right versus 5 on the left).
Only a temperature change will affect the value of Keq, so this will stay
the same. Among the three, only item I will increase, so the answer is
(A).
- B When quantities are given for more than one reactant, you must see
which is limiting. Fifty-six grams of CO (molecular weight = 28 amu) are
