Transcript Catalyst - MAEDA AP Chemistry

Catalyst
 You create a Zn/H cell and the potential is 0.45 V at 25 oC when
[Zn2+] = 1.0 M and PH2 = 1.0 atm, what is the [H+]
 What is the pH of the solution in the cathode half – cell for a
Zn/H cell where PH2 = 1.0 atm, [Zn2+] = 0.10 M, and the cell
emf is 0.542 V.
End
Heartbeat and Electrochemistry
Lecture 8.6 – Concentration Cells, Free
Energy and Electrochemistry
Today’s Learning Targets
 LT 8.15 – I can calculate the non-standard electrode potentials
and/or concentrations for a concentration cell utilizing the
Nernst equation
 LT 8.16 – I can calculate the free energy of a reaction utilizing
the standard reduction potential for the reaction.
 LT 8.17 – I can calculate the equilibrium constant for a chemical
reaction utilizing the standard reduction potential for the
reaction.
Concentration Cells
 Cell emf depends on the concentration in each cell also
 We can construct a cell where both the anode and cathode have
the same ion, but there are different concentrations
 The more dilute solution goes in the anode
 The more concentrated solution goes in the cathode
 The driving force is the entropy of mixing
Electrochemistry Lab
 In the electrochemistry lab we studied the following
concentration cell:
Anode:
Ag (s) 
 Ag+ (aq,dilute) + e-
Cathode: Ag+ (aq,concentrated)+ e- 
 Ag (s)
o
E
= 0.80 V
Eo = 0.80 V
 Therefore, the Eo for this solution would be:
o
o
o
E cell
 E cathode
 E anode
 0.80  0.80  0 V
 If we add the reactions together we discover that the reaction is
the concentrated going to the dilute Ag+:
Ag+ (aq, concentrated)  Ag+ (aq, dilute)
 We can solve for E using the Nernst equation and Q is simply:

dilute
[Ag ]
Q

[Agconcentrated ]
Class Example
 A voltaic cell is constructed with two hydrogen electrodes.
Electrode 1 has a PH2 1.00 atm and an unknown concentration of
H+. Electrode 2 is a standard hydrogen electrode (PH2 =1 atm,
[H+] = 1 M). At 298 K, the measured cell potential is 0.211 V
and the electrical current is observed to flow from electrode 1
through the external circuit to electrode 2. Calculate [H+] for the
solution at electrode 1. What is the pH of the solution?
Table Talk
 A concentration cell is constructed with two Zn (s)-Zn2+ (aq)
half-cells. In one half cell [Zn2+] = 1.35 M, and the other [Zn2+]
= 3.75 x 10-4 M.
(a) Which half cell is the anode?
(b) What the emf of this cell?
Lab Calculations
 Complete the lab calculations for questions 1 – 4 on the handout
for Part II of the lab last week.
Spontaneity and Eo
 Voltaic cells use a positive Eo in order to produce cell potential
 Therefore, a Eo must be positive in order for a reaction to be
spontaneous
 This is why we can use standard
reduction potentials to predict
whether a metal will run in the
forward or reverse reaction.
 Needs to generate a +Eo when
reaction is ran.
ΔG and Eo
 Since the emf indicates whether a reaction is spontaneous we can
use it to calculate ΔG:
G  nFE
 Since n and F are always positive, the sign of E determines the
spontaneity of a reaction!
 Finally we can relate this to equilibrium constants

G
o
 n FE
G
o
 RT ln K
o
n FE  RT ln K
RT
o
E 
ln K
nF
o
Class Example
 Using the standard reduction potentials, calculate the values of
ΔG, Eo, and K for the reaction at 298 K:
4 Ag (s) + O2 (g) + 4 H+  4 Ag+ (aq) + 2 H2O
Table Talk
 Using the standard reduction potentials, calculate the values of
ΔG, Eo, and K for the reaction at 298 K:
3 Ni2+ (aq) + 2 Cr(OH)3 (s) + 10 OH- (aq)  3 Ni (s) + 2 CrO42+ + H2O (l)
Connecting Concepts!
 An electrochemical cell is constructed in which one ½ cell
contains a buffer of acetic acid (0.100 M) and sodium acetate
(0.050 M), a platinum electrode, and H2 bubbled at 1.0 atm. The
other half – cell contains a 0.050 M solution of AgNO3 and a
silver electrode. Calculate the cell potential for this galvonic cell.
Recall that Ka = 1.8 x 10-5 for acetic acid. NOTE – Eo for H+/H2
cell is 0.00 V and Eo for Ag+/Ag (s) cell is 0.960 V.
Connecting Concepts!
 You have the following voltaic cell:
Pb (s) | Pb2+ (sat. PbI2) || Pb2+ (aq, 0.150 M) |Pb (s)
 The Pb2+ cell that contains saturated PbI2 has lead ions in the
following equilibrium:
PbI2  Pb2+ (aq) + 2 I – (aq)
 Calculate the Ksp for PbI2
Practice Practice Practice!
Closing Time
 Read 20.5 and 20.6
 Homework: Lab Worksheet and calculations due
Thursday/Friday
 Unit 7 Exam Thursday/Friday
 Stations Review up on the Website!