For the electrochemical cell Zn (s) | ZnCl 2 (aq, 0.350 M) ? HCl (aq, pH=3.20) |
ID: 1045984 • Letter: F
Question
For the electrochemical cell
Zn(s) | ZnCl2(aq, 0.350 M) ? HCl(aq, pH=3.20) | H2(g), 1.00 atm) | Pt(s)
a) Write the net ionic equation that describes this cell's reaction.
b) What is the equilibrium constant for the cell reaction?
c) What is the cell potential under the conditions described?
d) Describe the current that flows in the salt bridge while the cell reaction is proceeding.
Possibly Useful Information: 0°C = 273.15 K R = 8.314 J/(mol × K) = 0.08206 L × atm/(mol × K)
1 F = 96,500 C/mol e– 1 ampere = 1 Coulomb/second
Substance
?Hfº(kJ/mol)
?Gfº(kJ/mol)
Sº(J/(mol × K))
H2O(l)
–285.8
–237.2
69.91
H2O(g)
–241.82
–228.6
188.83
HCN(g)
+135.1
+124.7
201.78
CH4(g)
–74.81
–50.8
186.26
NH3(g)
–46.11
–16.5
192.45
CO2(g)
–393.5
–386.0
213.6
O2(g)
0
0
205.138
Standard Reduction Potentials at 25°C
Reduction Half-reaction
E°, Volts
Cl2(g) + 2 e– ® 2 Cl–(aq)
O2(g) + 4 H+(aq) + 4 e– ® 2 H2O(l)
Br2(g) + 2 e– ® 2 Br–(aq)
NO3–(aq) + 3 H+(aq) + 2 e– ® HNO2(aq) + H2O(l)
Hg2Cl2(s) + 2 e– ® 2 Hg(l) + 2 Cl–(aq)
AgCl(s) + e– ® Ag(s) + Cl(aq)
SO42–(aq) + H2O(l) + 2 e– ® SO32–(aq) + 2 OH–(aq)
2 H+(aq) + 2 e– ® H2(g)
Fe2+(aq) + 2 e– ® Fe(s)
Zn2+(aq) + 2 e– ® Zn(s)
+1.35827
+1.2291
+1.078
+0.940
+0.26808
+0.22233
+0.17
0
–0.44
–0.7618
Substance
?Hfº(kJ/mol)
?Gfº(kJ/mol)
Sº(J/(mol × K))
H2O(l)
–285.8
–237.2
69.91
H2O(g)
–241.82
–228.6
188.83
HCN(g)
+135.1
+124.7
201.78
CH4(g)
–74.81
–50.8
186.26
NH3(g)
–46.11
–16.5
192.45
CO2(g)
–393.5
–386.0
213.6
O2(g)
0
0
205.138
Explanation / Answer
Zn(s) | ZnCl2(aq, 0.350 M) ? HCl(aq, pH=3.20) | H2(g), 1.00 atm) | Pt(s)
Write the net ionic equation that describes this cell's reaction.
The net ionic equation will be
Zn(s) + 2H+ ---> H2(g) + Zn+2(aq)
What is the equilibrium constant for the cell reaction?
From Nernst equation
Ecell = E0cell - 0.0592 / n logQ
When Ecell = 0, Q = Keq
E0cell = E0cathode - E0anode
Here
E0cathode = E0H2 = 0 V
E0anode = E0Zn = -0.7618
E0cell = 0 - (-0.7618) = 0.7618 V
What is the cell potential under the conditions described?
Again in the given conditions
From Nernst equation
Ecell = E0cell - 0.0592 / n logQ
Q = [Zn+2] / [H+]2
pH = 3.20 = -log[H+]
[H+] = 0.00063
Q = 0.350 / (0.00063)2 = 8.81 X 105
Ecell = 0.7618 - 0.0592/2 log (8.81 X 105) = 0.586 V
Describe the current that flows in the salt bridge while the cell reaction is proceeding.
the electrons flow from cathode to anode in the salt bridge
The direction of current will be from anode to cathode