Solutions to HW 22
Q1
The following standard redox couples are connected to make a galvanic cell.
Sn/Sn2+
E1o = -0.14
Hg/Hg2+
E2o = 0.85
Determine the following for the standard galvanic cell:
Eo =
Go =
log K =
0.99
V
-191
kJ
33.46
SOLN:
𝑬𝑬𝒐𝒐𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 = 𝑬𝑬𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 − 𝑬𝑬𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 = 𝟎𝟎. 𝟖𝟖𝟖𝟖 − (−𝟎𝟎. 𝟏𝟏𝟏𝟏) = 𝟎𝟎. 𝟗𝟗𝟗𝟗
∆𝑮𝑮 = 𝒏𝒏𝒏𝒏𝑬𝑬𝒐𝒐 = (𝟐𝟐)(𝟗𝟗𝟗𝟗. 𝟒𝟒𝟒𝟒𝟒𝟒)(𝟎𝟎. 𝟗𝟗𝟗𝟗) = 𝟏𝟏𝟏𝟏𝟏𝟏 𝒌𝒌𝒌𝒌
𝑬𝑬𝒐𝒐𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 =
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝑬𝑬𝒐𝒐𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 ∗ 𝒏𝒏
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 →
= 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 →
𝒏𝒏
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 =
𝟎𝟎. 𝟗𝟗𝟗𝟗 ∗ 𝟐𝟐
= 𝟑𝟑𝟑𝟑. 𝟓𝟓
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
Q2
1+
3+
The standard reduction potentials of the X /X and and Y /Y couples are 0.16 V and -0.14 V, respectively. What is
1+
3+
the cell potential to 0.01 V of the galvanic cell constructed from the two couples if [X ] = 0.179 M and [Y ] = 0.0270
M? Hint: you may wish to write a balanced equation for the reaction.
E=
0.29
V
The reaction allows us to determine n and Q:
𝟑𝟑𝑿𝑿+ + 𝟑𝟑𝒆𝒆− → 𝟑𝟑𝟑𝟑
𝒀𝒀 → 𝒀𝒀𝟑𝟑+ + 𝟑𝟑𝒆𝒆−
𝒏𝒏 = 𝟑𝟑
𝑸𝑸 =
[𝒀𝒀𝟑𝟑+]
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎
=
= 𝟒𝟒. 𝟕𝟕𝟕𝟕
[𝑿𝑿+]𝟑𝟑 (𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏)𝟑𝟑
𝑬𝑬 = 𝑬𝑬𝒐𝒐 −
Q3
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 = �𝟎𝟎. 𝟏𝟏𝟏𝟏 − (−𝟎𝟎. 𝟏𝟏𝟏𝟏)� −
𝒍𝒍𝒍𝒍𝒍𝒍(𝟒𝟒. 𝟕𝟕𝟕𝟕) = 𝟎𝟎. 𝟐𝟐𝟐𝟐
𝒏𝒏
𝟑𝟑
The standard reduction potentials of the X3+/X and and Y2+/Y couples are 0.10 V and -0.17
V, respectively. What is the cell potential to 0.01 V of the galvanic cell constructed from the
two couples if [X3+] = 0.240 M and [Y2+] = 0.0230 M?
0.31
E=
V
What would the cell potential be after the concentration of X3+ concentration had dropped to
0.010 M? Hint: If the concentration of X3+ drops, then the concentration of Y2+ must
increase! (Think reaction table.)
0.24
E=
V
2𝑋𝑋 3+ + 6𝑒𝑒 − → 2𝑋𝑋
3𝑌𝑌 → 3𝑌𝑌 2+ + 6𝑒𝑒 −
𝑛𝑛 = 6
[𝑌𝑌 2+ ]3 (0.023)3
𝑄𝑄 = 3+ 2 =
= 2.11𝑥𝑥10−4
[𝑋𝑋 ]
(0.24)2
𝑬𝑬 = 𝑬𝑬𝒐𝒐 −
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
−4
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 = �𝟎𝟎. 𝟏𝟏𝟏𝟏 − (−𝟎𝟎. 𝟏𝟏𝟏𝟏)� −
𝒍𝒍𝒍𝒍𝒍𝒍 �2.11𝑥𝑥10 � = 𝟎𝟎. 𝟑𝟑𝟑𝟑
𝒏𝒏
𝟔𝟔
2X3+
3Y

2X
3Y2+
0.24
0.23
-0.23
+(3/2)(0.23)
0.01
0.368
𝑄𝑄 =
[𝑌𝑌 2+ ]3 (0.368)3
=
= 498.36
[𝑋𝑋 3+]2
(0.01)2
𝑬𝑬 = 𝑬𝑬𝒐𝒐 −
Q4
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 = �𝟎𝟎. 𝟏𝟏𝟏𝟏 − (−𝟎𝟎. 𝟏𝟏𝟏𝟏)� −
𝒍𝒍𝒍𝒍𝒍𝒍(498.36) = 𝟎𝟎. 𝟐𝟐𝟐𝟐
𝒏𝒏
𝟔𝟔
What is the cathode pH in Co/Co2+(1.00 M)//H1+(x M), H2(1.0 atm)/Pt if E = 0.07 V? Give
the pH to 0.01.
pH =
+
Co + 2H -> Co
𝑛𝑛 = 2
𝑬𝑬 = 𝑬𝑬𝒐𝒐 −
2+
3.55
+ H2
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 → 𝟎𝟎. 𝟎𝟎𝟎𝟎 = 𝟎𝟎. 𝟐𝟐𝟐𝟐 −
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 → 𝑸𝑸 = 𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟎𝟎−𝟒𝟒
𝒏𝒏
𝟐𝟐
𝑄𝑄 =
[𝐶𝐶𝑜𝑜 2+]
[𝐻𝐻 +]2
→
1.24𝑥𝑥10−4 =
1
→
[𝐻𝐻 +]2
[𝐻𝐻+ ] = 2.83𝑥𝑥10−4
→
𝑝𝑝𝑝𝑝 = − log(2.83𝑥𝑥10−4 ) = 3.55
Q5
What is the standard reduction potential of the Y3+/Y redox couple if the cell potential of the
following cell at 25 oC is 0.80 V?
Y / Y3+ (1.2 M) // Br2(l), Br1-(0.20 M) / Pt
Eo =
0.330
V
One randomization shows this couple as Y3+, but the other shows it with Y+.
6Br- + 2Y3+ -> 3Br2 + 2Y
𝑸𝑸 =
[𝑩𝑩𝒓𝒓− ]𝟔𝟔
[𝒀𝒀𝟑𝟑+ ]𝟐𝟐
𝑬𝑬 = 𝑬𝑬𝒐𝒐 −
𝐸𝐸 𝑜𝑜 = 𝐸𝐸
=
(0.2)6
(1.2)2
= 4.44𝑥𝑥10−5
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍
𝒏𝒏
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
+
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 = 𝟎𝟎. 𝟖𝟖 +
𝒍𝒍𝒍𝒍𝒍𝒍(4.44𝑥𝑥10−5 )
𝒏𝒏
𝟔𝟔
= 0.76
𝐸𝐸𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐸𝐸𝑐𝑐𝑐𝑐𝑐𝑐 ℎ𝑜𝑜𝑜𝑜𝑜𝑜 − 𝐸𝐸 𝑜𝑜 = 1.09 − 0.76 = 0.33
What is the standard reduction potential of the Y1+/Y redox couple if the cell potential of the
following cell at 25 oC is 0.80 V?
Y / Y1+ (1.2 M) // Br2(l), Br1-(0.20 M) / Pt
Eo =
0.40
V
For example, if we have the same concentrations with the other cell couple then we have
2Br- + 2Y+ -> Br2 + 2Y
𝑸𝑸 =
[𝑩𝑩𝒓𝒓− ]𝟐𝟐
[𝒀𝒀𝟑𝟑+ ]𝟐𝟐
𝑬𝑬 = 𝑬𝑬𝒐𝒐 −
𝐸𝐸 𝑜𝑜 = 𝐸𝐸
=
(0.2)2
(1.2)2
= 0.0277
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍
𝒏𝒏
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
+
𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 = 𝟎𝟎. 𝟖𝟖 +
𝒍𝒍𝒍𝒍𝒍𝒍(0.0277)
𝒏𝒏
𝟐𝟐
= 0.69
𝐸𝐸𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐸𝐸𝑐𝑐𝑐𝑐𝑐𝑐 ℎ𝑜𝑜𝑜𝑜𝑜𝑜 − 𝐸𝐸 𝑜𝑜 = 1.09 − 0.69 = 0.40
Explanation on the setup of this problem. The question arises: how do you know which way to write the
redox couple? The cell is written, by convention with the anode reaction on the left and the cathode on
the right. The cathode is the electrode that attracts cations (hence the name). So metals plate out on
the cathode in electroplating (electrolytic cell). However, regardless of cell type the cathodic reaction
involves reduction and the anodic electrode involves oxidation. Hence, in the above cell the anodic
reaction is
Y  Y3+ + 3 eOr
Y  Y+ + eDepending on the randomization.