Redox Reactions in Acidic Basic Solutions Chemistry Lab Report

Balancing Redox Reactions Practice (– remove spectator ions)1. Balance each of the following acidic solution:
a. NaClO3(aq) + NO(g) à Cl2(aq) + NaNO3(aq)
b. CrBr3(aq) + LiMnO4(aq) à MnBr2(aq) + Li2Cr2O7(aq)
2. Balance in basic solution:
a. N2O4(g) + NaBrO à NaNO3(aq) + NaBr(g)
b. PbO2(s) + NaNO2(aq) à Pb(OH)2(s) + NaNO3(aq)
7. balance in acidic solutions:
a. Na2S2O3 + I2 à NaI + Na2S4O6
b. MnBr2(aq) + NaBiO3(s) → BiBr3(aq) +NaMnO4 (aq)
Concentration for Cu2+/Cu(s) is 0.10 M
Literature Values
Experiment 12: Voltaic Cells
Reduction-oxidation (redox) reactions are chemical reactions in which electrons are transferred.
An element in one species is oxidized (loses electrons) and an element in another species is
reduced (gains electrons). This is easiest to visualize in metal/metal ion redox reaction.
Oxidation reaction:
Reduction reaction:
Zn(s) → Zn 2+(aq) + 2 eCu 2+(aq) + 2 e- → Cu(s)
Reduction and oxidation always occur in tandem; you cannot have one without the other. The
sum of these two half-reactions give us the net reaction: Zn(s) + Cu 2+(aq) → Zn 2+(aq) + Cu(s). No
electrons are included in this overall reaction, but it is understood that they have been
transferred. If the number of electrons in the oxidation reaction does not match the number in the
reduction reaction, each equation is multiplied by some number so that all electrons cancel out in
the overall reaction.
Oxidation reaction:
Reduction reaction:
Overall reaction:
2 (Al(s) → Al 3+(aq) + 3 e-)
3 (Cu 2+(aq) + 2 e- → Cu(s))
2 Al(s) → 2 Al 3+(aq) + 6 e3 Cu 2+(aq) + 6 e- → 3 Cu(s)
2 Al(s) + 3 Cu 2+(aq) → 2 Al 3+(aq) + 3 Cu(s)
Sometimes, the oxidized and reduced elements are contained within compounds or polyatomic
ions. In this next example, nitrogen in nitrogen monoxide has an oxidation number of +2; this
changes to +5 in the nitrate ion. Manganese in permanganate has an oxidation number of +7
which is reduced to +2 in the manganese(II) ion.
Oxidation reaction:
Reduction reaction:
Overall reaction:
2 H2O (l) + NO(g) → NO3–(aq) + 4 H+ (aq) + 3 eMnO4 -(aq) + 8 H-(aq) + 5 e- → Mn 2+(aq) + 4 H2O (l)
5 (2 H2O (l) + NO(g) → NO3– (aq) + 4 H+ (aq) + 3 e- )
3 (MnO4 -(aq) + 8 H-(aq) + 5 e- → Mn 2+(aq) + 4 H2O (l))
3 MnO4 -(aq) + 5 NO(g) + 4 H-(aq) → 5 NO3–(aq) + 3 Mn 2+(aq) + 2 H2O (l)
An electrochemical cell is a device used to separate a chemical reaction into two component
half- reactions so that the electrons are transferred through an external circuit, rather than
transferred directly in solution. This allows us to harness the energy of the moving electrons into
useful work. When the cell is constructed so that electrons flow spontaneously, this is called a
galvanic cell, which is also known as a voltaic cell.
The anode is the solid in the anodic half-cell where oxidation occurs. In the example above,
metallic zinc is the anode. The cathode is the solid in the cathodic cell where reduction occurs;
copper metal in the diagram. As electrons flow through the wire, there would be a build of
negative charge in the cathodic half-cell leading to impedance, the pushback of electron flow due
to excess negative charge. To have a complete circuit, electrons must be able to flow through the
wire and anions through a salt bridge or porous barrier. This allows the system to maintain
charge neutrality.
The diagram above can be summarized using the following electrochemical cell notation where
the single vertical line denotes a phase boundary and double vertical lines denote a salt bridge.
anode | anodic solution || cathodic solution | cathode
Zn(s) | Zn(NO3)2(aq) || Cu(NO3)2 (aq) | Cu(s)
Because the tendency to gain electrons, or reduction potential, can only be determined by being
compared to another substance, chemists have agreed to assign hydrogen ion a reduction
potential of zero and compare every other reduction to it. Species with a great tendency to gain
electrons have a positive number compared to hydrogen; species with a lesser tendency, a
negative number. Several standard reduction potentials are given in a table at the end of this
theory section.
To determine the overall difference in potentials in a cell, we usually find the difference in
potentials by subtracting values that we find on a table.
Ecell = Ecathode – Eanode
(Eq 1)
The table values reflective potentials for reductions under standard conditions: 1 atm, 1M, 298K.
To adjust for variations in these we use the Nernst equation:
= table value (Eq 1)
R = 8.3144 J/molK
T = temperature in Kelvin
N = number of mols of electrons transferred
F = Faraday constant, 96485 C/mol eQ = equilibrium quotient
𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑒𝑙𝑙
− 𝑛𝐹 ln 𝑄
In our lab, we will construct half-cells and with them, build voltaic cells and measure voltages.
We will reuse half-cells whenever possible to cut down on waste. Be sure to rinse the outside of
the porous cup well when you reuse it, to minimize contamination. Each time you measure a
voltage, you should record a positive number; if your voltage is negative, you have the leads
backwards. Pay attention to concentrations. If the concentration is not 1.0 M, you will need to
use the Nernst equation. (But do use the concentrations that you are instructed to use. Not every
cell should be 1.0 M.)
To determine the KSP of an insoluble salt, write the solubility equation and plug in 0.10 M for the
concentration of the anion. Use the Nernst equation to determine the concentration of the metal
ion; plug that value in.
For example, one cell containing 20.0 mL of a 0.20 M sodium carbonate solution and a few
drops of 0.10 M zinc nitrate is compared with another half-cell, a reference, containing 0.10 M
copper(II) nitrate. The resulting potential is 1.143 V when zinc is the anode. The small amount of
zinc carbonate formed is an insoluble precipitate. We can assume that all the zinc ion has been
Zn2+ + CO3 2- → ZnCO3(s)
Our goal is to use our data to determine the KSP for zinc carbonate.
Zn2+ + CO3 2-
KSP = [Zn2+][CO32-]
Because there were only a few drops of zinc ion in our half-cell, the concentration of carbonate is
still effectively 0.20 M. This can be plugged into the KSP expression. The amount is zinc ion that
is not tied up in the precipitate will be receiving electrons from the copper in the other half-cell.
We use that voltage and the Nernst equation to determine the zinc ion concentration.
𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑒𝑙𝑙

ln 𝑄
Q comes from the redox reaction that generates the current:
Zn(s) + Cu 2+(aq) → Zn 2+(aq) + Cu(s)
[𝑍𝑛2+ ]
𝑄 = [𝐶𝑢2+]
= 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒
− 𝐸𝑎𝑛𝑜𝑑𝑒
= 0.337𝑣 − (−.0763𝑉) = 1.100𝑉
1.143V = 1.100V −
[Zn2+] = 0.00352 M
KSP = [Zn2+][CO32-]= (0.00352)(0.20)= 7.0 x 10-4
0.0257𝑉 [𝑍𝑛2+ ]
Table of Reduction Potentials
Table of Reduction Potentials
F2(aq) + 2 e→ 2 F-(aq)
Br2(i) + 2 e→ 2 Br-(aq)
Ag (aq) + 1 e
→ Ag(s)
Fe (aq) + 1 e
→ Fe 2+(aq)
I2(aq) +
→ 2 I-(aq)
O2(g) + 2 H2O(l) + 4 e- → 4 OHCu 2+(aq) + 2 e→ Cu(s)
2 H +(aq) + 2 e→ H2(g)
Pb2+(aq) + 2 e→ Pb(s)
Sn2+(aq) + 2 e→ Sn(s)
Fe (aq) + 2 e
→ Fe(s)
Zn (aq) + 2 e
→ Zn(s)
Li +(aq) + 1 e→ Li(s)
E (V)
Part 1 – Construction of Voltaic Cells.
The class will work as a large group as your instructor directs.
1. Each pair will combine two half-cells to make a voltaic cell. Use two 10-mL beakers
inside a 250-mL beaker with a salt bridge as your cell. One cell will contain the reference
cell: Copper electrode and 0.1 M copper(II) solution. The other 10-mL beaker will
contain the assigned half-cell. The other half cell will be one of the following:
a. Zn2+/Zn
b. Pb2+/Pb
c. Ag1+/Ag
d. Fe2+/Fe
e. Fe+3/ Fe+2 (graphite)
f. I2/ I- (graphite)
g. Br2/ Br- (graphite)
2. Connect the Pasco voltmeter to the cell. The voltage reading for each cell needs to be a
positive value. (Switch the alligator clips if you read a negative voltage.)
3. Measure and record the voltage of each cell. Be sure to record which is the anode and
which is the cathode. (When the voltage reading is positive, the anode is attached to the
black lead; the cathode, the red lead.)
4. Be sure to note the concentrations for each of the solutions in your data.
Part 2 – Effect of Concentration of Cells on Cell Potential
1. Set up a voltaic cell using Cu+2/Cu in each half-cell. One half cell will contain 0.050 M
Cu+2 and the other half-cell will contain 2.75 M Cu+2. The voltage reading for each cell
needs to be a positive value. (Switch the alligator clips if you read a negative voltage).
Record the voltage.
2. Set up a voltaic cell using Fe+3/Fe in each half-cell. One half cell will contain 0.050 M
Fe+3 and the other half-cell will contain 2.75 M Fe+3. The voltage reading for each cell
needs to be a positive value. (Switch the alligator clips if you read a negative voltage.).
Record the voltage.
Part 3 – Corrosion of nails
1. Note observations of prepared nails in petri dishes; identify the anode region of each nail
and the cathode region.
a. Note: the pink coloring is due the reduction of oxygen, forming hydroxide.
There is phenolphthalein in the agar solution, which turns pink in basic
b. Note: the dark blue is the formation of a metal complex called prussion blue,
indicating the presence of iron (III) ion due to the oxidation of the iron nail.
Part 4 – Cell current (instructor demo)
1. Your instructor will prepare two zinc-copper voltaic cells; one cell will connect two halfcells with a salt bridge, while the other will contain one half-cell in a porous cup that is
placed inside the second half cell. Record the cell potential and the cell current of each
voltaic cell.
Part 5 – Determining KSP using a Voltaic Cell
1. Each pair of students will be assigned one of the following insoluble salts of Pb2+ or Ag+
from the list below.
8.0 mL of 0.10 M
One drop of 0.10 M
2. Use Cu2+/Cu as the reference half-cell. Create this half-cell in the 10-mL beaker (as in
Part 1) using about 8 mL of 0.10 M solution.
3. In the beaker half-cell, add the listed sodium solution and nitrate solution for your
assigned salt from the table above, using the listed electrode.
4. Record the potential and note which is the anode and which is the cathode.
Part 1 – Voltaic Cells
1. Draw a complete diagram of the zinc/copper cell. Be sure to include anode/cathode,
anodic/cathodic solutions, direction of electron flow, salt bridge, and the direction of
cation/anion flow.
2. Write the electrochemical cell notation for each cell constructed
3. Write the balanced NIE for each cell constructed
4. Calculate the cell potential for each cell using the Nernst Eqn
5. Determine the standard half-cell reduction potential for each half-call tested
6. Determine the % error for each standard half-cell reduction potential.
Part 2 – Concentration on Cell Potential
1. Calculate the cell potential for each concentration cell.
Part 3 – Corrosion of nails
1. Explain your observations of the nails….what is being oxidized in each nail, and why is
the oxidation occurring where you observed?
Part 5 – Experimental Determination of KSP
➢ Write the redox equation for the reaction between the metals and metal ions in the
standard state; calculate Eo for this reaction.
➢ Using the Eo and the measured E, calculate the metal ion concentration using the Nernst
➢ Calculate the KSP for your insoluble salt using the metal ion concentration and the correct
KSP expression.
Be sure to include the following results:
➢ From part 1, create a summary table for all 7 reduction potential including your
experimental reduction potential, the theoretical reduction potential, and the % error
(with sign).
➢ From part 2, qualitatively summarize and explain your results.
➢ From part 4, Compare the cell potential of the two cells and explain your data. Also,
compare the cell current of the two cell and explain your data.
➢ From part 5, compare your determined KSP with a known value (cite reference) and report
your % error. Discuss possible sources of experimental error.
(als) 7(a laght be
121 Agt land the -> 2Agls
Zagina) + Cult – Cat Cayl + 2Ag (s)
Q=[ Cuzt [

2 2
0.0257in (10)
0.95- E cel
E cena
FO -0.31 de enero = 0.791
= E cathode
a. Brz/ Br” (graphite)
Cu(s) →
→ Cuaq) + 2e-
Br26 +2e → 2Br-(aq)
+ Cu(s)
– Eºcell
→ 2Br-
[Br]^2 [0.10]
= 0.02
[Br2] [0.050]
Ecell = E
0.75 V
E° = Eºcathode – E°anode
0.75 = Eºcathode – 0.31
0.80 = Eºcell

= 1.06 V
Percent Error
Vlit – Vobs
x 100 = Percent error
11.07 – 1.06
x 100 = 0.935%

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