364 ❯ STEP 5. Build Your Test-Taking Confidence❯ Answers and Explanations for Exam 2, Section II
(Free Response)Question 1
(a) (i) PbO 2 (s) + 2 HSO 4 - (aq) + 2 H+(aq) + Pb(s) → 2 PbSO 4 (aq) + 2 H 2 O(l)
You get 1 point for this answer.
(ii) PbO 2 (s) + HSO 4 - (aq) + 3 H+(aq) + 2 e- → PbSO 4 (s) + 2 H 2 O(l) 1.68 V
Pb(s) + HSO 4 - (aq) → PbSO 4 (aq) + H+(aq) + 2 e-? V
PbO 2 (s) + 2 HSO 4 - (aq) + 2 H+(aq) + Pb(s) → 2 PbSO 4 (aq) + 2 H 2 O(l) 2.04 VIf the cell voltage is 2.04 V then the two half-reaction voltages must add to this. Therefore, 2.04 V
= (1.68 + ?) V making the voltage for the second half-reaction equal to 0.36 V. This is the voltage
for the oxidation half-reaction; to change this to a standard reduction potential it is necessary to
reverse the half-reaction, which means it is necessary to reverse the sign on the voltage. Thus, the
standard reduction potential for the second half-reaction is -0.36 V.
You get 1 point for calculating 0.36 V and 1 point for changing the sign to a negative value.
(b) It is necessary to use two of the equations provided on the exam to answer this problem. (If you already
know the equations, it will save you the time necessary to look them up.)
(i) The exam (and the appendixes at the back of this book) provides the equation ΔG° = - nFE°
plus the values of the Faraday constant, F = 96,485 coulombs per mole of electrons, and 1 volt =
1 joule
1 coulomb. For this problem, n = 2 moles of electrons and E° = + 2.04 V. Entering the values
into the given equation: ΔG° = - nFE° = −
(2 mol electrons)
96.485 coul
mol electrons(2.04 V)
J
(V)(coul)
= 393659 = 3.94 × 105 J or 394 kJ.
Give yourself 1 point for the correct setup of the equation including the conversion and 1 more
point for the correct answer.
(ii) It is possible to get the answer by two different means by using the equations given on the test (in
the back of this book). The values of the constants are also here. The important equations are ΔG° =- nFE° and ΔG° = - RT ln K. It is possible to combine these equations to ln K = nFE°
RT
. The first
equation was used in part b(ii), and it is acceptable to use the answer from there and go directly tothe second equation without recalculating the result. Using the third equation gives: ln K =
nFE°
RT=
(2 mol electrons)
96.485 coul
mol electrons(2.04 V)
J
(V)(coul)
8.314 J
mol K(298 K)
= 159 (which is the same answer youget from using the answer to (b)(ii) and the second equation. Finally, if ln K = 159, then K = e^159.
Give yourself 1 point for the correct setup of the equation, including the conversion, and 1 more
point for the correct answer.
(iii) At equilibrium ΔG° is always 0.
Give yourself 1 point for this answer.21-Moore_PE02_p341-370.indd 364 31/05/18 1:54 pm