Electrochemistry ❮ 255Use the information on standard reduction
potentials in the following table to answer
questions 16–18.
E ç (V)
Cl 2 (g) + 2 e- → 2 Cl-(aq) +1.36
S 2 O 62 - (aq) + 4 H+(aq) + 2 e- →
2 SO 2 (g) + 2 H 2 O(l) +0.56
Cu^2 +(aq) + 2 e- → Cu(s) +0.34
Sn^2 +(aq) + 2 e- → Sn(s) -0.14
2 H 2 O(l) + 2 e- → H 2 (g) + 2 OH-(aq) -0.83
Ca^2 +(aq) + 2 e- → Ca(s) -2.87
K+(aq) + e- → K(s) -2.92
- A chemist constructs an electrolysis cell with two
platinum electrodes in a 1.00 M aqueous solution
of sodium chloride, NaCl. When she connects
the cell to a power source, an odorless gas forms
at one electrode, and a gas with a distinctive odor
forms at the other electrode. Choose the correct
statement from the following list.
(A) The odorless gas was oxygen.
(B) The odorless gas was the result of oxidation.
(C) The gas with the distinctive odor was the
result of oxidation.
(D) The odorless gas was evolved at the positive
electrode.
17. A student mixed an acidic 1.0 M sodium dithi-
onate, Na 2 S 2 O 6 , solution with a 1.0 M tin(II)
bromide, SnBr 2 , solution containing some tin
metal, Sn, and observed a gas forming. Which of
the following is the gas?
(A) H 2 (g)
(B) SO 2 (g)
(C) Br 2 (g)
(D) Sn(g)
18. A chemist constructs a galvanic cell with a tin,
Sn, electrode in a compartment containing a
1.0 M tin(II) perchlorate, Sn(ClO 4 ) 2 , solution
and a platinum, Pt, electrode in a compartment
containing a 1.0 M copper(II) chloride, CuCl 2 ,
solution. The salt bridge connecting the two
compartments contains a 1.0 M potassium sul-
fate, K 2 SO 4. Which of the following is the net
ionic equation for the cell?
(A) CuCl 2 (aq) + Sn(s) → SnSO 4 (aq) + Cu(s)
(B) Sn^2 +(aq) + Cu(s) → Cu^2 +(aq) + Sn(s)
(C) Cu^2 +(aq) + Sn(s) → Sn^2 +(aq) + Cu(s)
(D) 2 Cl-(aq) + 2 H 2 O(l) → H 2 (g) + 2 OH-(aq)- Cl 2 (g)
❯ Answers and Explanations
- A—The size of the electrode is not important
when determining the cell voltage. - B—The salt bridge serves as an ion source to
maintain charge neutrality. Deionized water
would not be an ion source, so the cell could not
operate. - A—The balanced equation is
2 MnO 4 - (aq) + 16 H+(aq) + C 2 O 42 - (aq) →
2 Mn^2 +(aq) + 8 H 2 O(l) + 10 CO 2 (g)
- C—The balanced equation is
S 2 O 32 - (aq) + 10 OH-(aq) →
2 SO 42 - (aq) + 5 H 2 O(l) + 8 e-
- C—It takes 4 moles of electrons (4 F ) to change
the platinum ions to platinum metal. The calcula-
tion would be
(0.80 F)
1 mol Pt
4 F
= 0.20 mole Pt.- D—The dichromate ion oxidizes the sulfide ion
to elemental sulfur, as the sulfide ion reduces
the dichromate ion to the chromium(III) ion.
Chromium goes from +6 to +3, while sulfur
goes from –2 to 0. The hydrogen remains at + 1
and the oxygen remains –2, so neither hydrogen
nor oxygen is oxidized nor reduced. - C—The balanced chemical equation is:
4 H+(aq) + NO 3 - (aq) + 3 e- → NO(g) + 2 H 2 O(l) - C—Using the equation:
log K =
nE °
0.0592 =^2 (0.12)
0.0592 = 4.05
You should realize that the log K = 4 gives
K = 104. The actual value is K = 1.1 × 104.