Study Guide Electrochemistry
Name______
Study Guide Electrochemistry
Terms to review:
Reduction
Oxidation
Reducing agent
Oxidizing agent
Half- reaction
Galvanic cell
Cathode
Anode
Salt bridge
Electromotive force (emf)
Volt = 1 joule/coulomb
Review:
Mg(s) + 2H+(aq) → Mg+2(aq) + H2(g)
- Write half reactions for the following including electron transfer
- Identify the oxidation and the reduction
- Where does the oxidation occur? Where does the reduction occur?
Standard Reduction Potentials
The emf of a galvanic cell is a combination of the potentials of two half-reactions. The potentials are described relative to one absolute standard, the standard hydrogen electrode.
2H+ + 2e- → H2 Eo = 0.00 V
- Galvanic cells require Eo> 0 V ( sum of oxidation and reduction potential)
- One of the reduction potentials will have to be reversed (to form an oxidation half-reaction) in every Eo calculation. It is the one with a lower reduction potential
- When a reaction is reversed, Eo gets the opposite sign
- When you multiply a reaction by a coefficient, the Eo is not changed
Cell Emf
Calculate the emf values. State which are galvanic.
a) Mg(s) + 2H+(aq) → Mg+2(aq) + H2(g)
b) Cu+2(aq) + 2Ag(s) → Cu (s) + 2Ag+(aq)
Reduction Strength
Place in order of increasing strength as oxidizing agents
Fe+2, ClO2 , F2 , AgCl
Composing Galvanic Cells
Given the following, decide which is the anode and cathode, balance, write the overall cell reaction and calculate Eo cell if the cells are galvanic
Zn+2 and Sn+2
Line Notation Is a method of representing electrochemical cells . Each side will have a “phase boundary” separating cells. A single vertical line ( ׀ ) denotes a boundary with the solid phase on the outside. (׀׀ ) indicated a porous disk or salt bridge
Zn (s) Zn+2(aq) Cu + 2(aq) Cu (s)
Try writing half reactions, then line notations
a) Hg+2 + Cd → Hg + Cd2+
b) Pb + 2Cr+3 → Pb2+ + 2Cr2+ ( a platinum electrode is used at the cathode)
Describing Galvanic Cells
Describe a galvanic cell based on the half reactions below. Draw it
Cu2+ + 2e= → Cu
Cr2O72- + 14H+ + 6e - → 2Cr3+ + 7H2O
Which metal will be on the anode side?
Will the electron flow be from copper to platinum or from platinum to copper?
Cell Potential, Electrical Work and Free Energy
Actual work achieved is always less than the theoretical work available.
At standard conditions ∆Go = -nFEo
∆Go = free energy (in joules)
n = moles of electrons exchanged in the redox reaction
F = the Faraday, a constant ( 96486 Coulombs per mole of electrons)
E = cell voltage ( Joules/Coulomb)
∆Go and E have opposite signs. For a spontaneous process, ∆Go is “-“and E is “+”
Example:
Calculate the ∆Go for the reaction: Zn2+ (aq) + Cu(s) → Zn(s) + Cu2+ (aq)
Will the zinc ions plate out on a copper strip?
Dependence of Cell Potential on Concentration
In a concentration cell, current flows due only to a difference in concentration of an ion in tow different compartments of a cell. LeChatelier’s principle is applicable. In a cell where there is an equal concentration of metal ion on both sides, Eocell = 0.
Example
A cell has on its left side a 0.20 M Cu+2 solution. The right side has a 0.050 M Cu+2 solution. The compartments are connected by Cu electrodes and a salt bridge. Designate the cathode, anode and direction of current.
Solution
The current will flow in this cell until the concentration of Cu+2 is equal in both compartments. This means that the concentration of Cu+2 in the left-hand side (0.20 M ) must be reduced by the reaction:
Cu+2+ 2e- → Cu
The left hand-side will be the ______. The right-hand side ( 0.050M Cu2+ ) will be the ______. Current will flow from right to left ( ______to ______)
Concentration cells produce a very small voltage. Nonstandard concentrations produce a cell voltage that is different from that at standard concentrations.
Ecell (nonstandard) does not equal Eocell
Use the Nernst equation to calculate cell voltage at nonstandard concentrations
E cell = Eocell - 0.0592 log Q at 25 oC
n
The thought is that if reactants concentrations are greater than product concentration, log Q will be < 0
And Ecell will be ≥Eocell
The Nernst Equation
Example
Calculate Ecell for a galvanic cell based on the following half-reactions at 25oC
Cd2+ + 2e- → Cd Eo = -0.40V
Pb2+ + 2e- → Pb Eo = -0.13V
Where [Cd2+] = 0.010 M and [Pb2+] = 0.100M
To be galvanic, Eo must be greater than 0V. Cadmium must undergo the oxidation. Write the net cell reaction and Eo cell
Use the Nernst equation and solve for Ecell
Example 2
Calculate the Ecell for a galvanic cell based on the following half reaction at 25o C
(Equation 1) FeO4-2 + 8H+ + 3e- → Fe+3 + 4H2O Eo = +2.20V
(Equation 2) O2 + 4H+ + 4e- → 2H2OEo = +1.23V
Where [FeO4-2 ] = 2.0 x 10-3 M[O2] = 1.0 x 10-5 atm
[Fe+3] = 1.0 x 10-3 MpH = 5.2
Which equation has a higher reduction potential?
It will be reduced
Which equation is oxidized?
Reverse it and change the sign
Balance the equations electrically
Write the net equation and find the Eocell
Use the pH to find the [H+]
Use the Nernst equation to find the Ecell
*since your calculator may flip out at trying to take 6.31 x 10-6 to the 20th power, you may want to split it up
(6.31)20 x (10-6) 20
Equilibrium Constants and Cell Potential
Remember that both E and ∆ G = 0 at equilibrium. Therefore a new equation can be derived
To relate E to K
log (K) = __ nEo__ at 25oC at equilibrium conditions
0.0592
Example:
Calculate the equilibrium constant for the reaction
Cu2+ + 2e= → CuEo = 0.34 V
Cr2O72- + 14H+ + 6e - → 2Cr3+ + 7H2OEo = 1.33 V
Find which equation is reduced
Find which equation is oxidized and reverse it
Balance the equations electrically
Find the net and the Eocell
How many moles of electrons are transferred?
Use log (K) = __ nEo__ at 25oC at equilibrium conditions
0.0592
Example 2:
Consider the reaction
Ni+2(aq) + Sn(s) → Ni(s) + Sn2+(aq)
Calculate the following: Eocell, ∆G, and K at 25oC.
a) Find the half reactions:
b) Is the reaction spontaneous under standard conditions?
c) Find ∆G
d) Calculate K using either log K = nEo/0.059 or ln K = - ∆G/ RT use R = 8.3128 J/K mol
Electrolytic Cells
Electrolysis is the process of forcing a current through a cell to produce a chemical change for which the cell potential is negative. In order for electrolysis to occur, you must apply an external voltage that is greater than the potential of the galvanic cell if you want to force the reaction in the opposite (electrolytic) direction.
Example
What voltage is necessary to force the following electrolysis reaction to occur?
2I-(aq) + Cu+2(aq) → I2(s) + Cu (s)
Write half reactions with Eo Find the Eocell
Which process would occur at the anode? Cathode? Assuming the iodine oxidation takes place at a platinum electrode, what is the direction of electron flow in this cell?
More than 0.20 V must be externally applied to make this reaction proceed. The direction of electron Flow is always from anode to cathode (“fat cat”),so it is from the platinum electrode to the copper electrode.
If you put a potential on a system which contains one metal ion, and the potential is above that at which the metal ion will reduce, you will plate out the metal. The amount of metal reduced is directly related to the current, in amps (Coulombs/sec), that flows in the system. In practice, this is not the case because electrochemistry is not a perfect process. (But we will pretend that it is) Use dimensional analysis:
Moles of e- = 1 mole e- x C x s
96,486 C s
↑ ↑ ↑
Faraday amps time
Electrolysis Example:
How many grams of copper can be reduced by applying a 3.00 A current for 16.2 min to a solution containing Cu+2 ions?
Pattern
Grams of Cu → moles of Cu → moles of e- → Coulombs → current → time
Or
Time – current-coulombs- moles of e- moles of Cu – grams
If reduction potentials are far apart, electrolysis can be used to separate a mixture of ions.
Remember the metal ion with the highest reduction potential is the easiest to reduce
Example: Electrolysis of Water
What volume of H2 (g) and O2(g) is produced by electrolyzing water at a current of 4.00 A for 12.0 minutes.
2H2O(l) → 2H2 (g) + O2(g)
In practice, the actual ratio is not exactly 2:1 ( oxygen solubility)
Find the moles of each gas
mol H2 x 1 mole e- x C x s = mol H2
2 mol e- 96,486 C s
↑ ↑ ↑
Faraday amps time
Then use proportion ( mol ratio) to determine O2 moles
At standard conditions, 1 mol of gas = 22.4 L, use this to convert moles to liters.
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