EC 319 GC Chemistry Worksheet

Experiment EC319 Electrochemical Cells: Post Lab QuestionsYou have experimental measurements of the voltage produced by three “known” electrochemical cells
created by the combination of your assigned half-cell with three other known half-cells.
In the questions below you will need to use standard reduction potentials. A table of these values for
the half-cells you used is shown below. An example of the calculations you need to perform for Q1 and
Q2 is shown on the next page.
1. For each of the three known electrochemical cells, calculate the theoretical voltage (cell potential)
under standard conditions (E0cell) as follows:
• show each of the balanced half-cell equations and the overall balanced redox reaction. Indicate
how many electrons are transferred in the overall reaction.
• indicate which half-cell is the cathode and which half-cell is the anode
• calculate the cell potential under standard conditions, E0cell = E0cathode – E0anode
2. For each of the three known electrochemical cells, calculate the theoretical voltage (cell potential)
under experimental conditions (Ecell). For each cell, use the standard cell potential E0cell (calculated in
Q1), the concentrations of solutions used to create the known half-cells, and a temperature of 25 °C.
3. Calculate the % error between your measured voltage and the theoretical voltage (from Q2) for
each of the three known electrochemical cell as follows:
% 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 =
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 − 𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
× 100
𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
Standard Reduction Potentials for selected half-reactions at 25 °C:
Half-reaction
E0 (V)
Cu2+ (aq) + 2 e- → Cu (s)
0.337
Fe3+ (aq) + 3 e- → Fe (s)
-0.036
Ag+ (aq) + e- → Ag (s)
0.799
Sn2+ (aq) + 2 e- → Sn (s)
-0.140
Example Calculations: An electrochemical cell is created by combining the following half-cells, using a
salt-bridge consisting of 0.1 M KNO3:
1. Gold metal in a solution of 0.100 M gold(III) nitrate
2. Zinc metal in a solution of 0.100 M zinc nitrate
The two half-cell reactions involved are as follows:
Half-reaction
E0 (V)
Au3+ (aq) + 3 e- → Au (s)
1.500
Reductive half-cell (cathode)
Zn2+ (aq) + 2 e- → Zn (s)
-0.760
Oxidative half-cell (anode)
Based on the standard reduction potentials of these two half-cells, the cell with the greater (more
positive) reduction potential (gold) will undergo reduction: Au3+ (aq) + 3 e- → Au (s). The cell with the
lesser (more negative) reduction potential (zinc) will undergo oxidation: Zn (s) → Zn2+ (aq) + 2 eIf these half-cells are connected together in an electrochemical cell, electrons will flow from the zinc half
cell (anode) to the gold half-cell (cathode). The balanced overall reaction for this transfer will be:
Reductive half-cell
2 [ Au3+ (aq) + 3 e- → Au (s) ]
cathode
Oxidative half-cell
3 [Zn (s) → Zn2+ (aq) + 2 e- ]
anode
Overall reaction
3 Zn (s) + 2 Au3+ (aq) → 3 Zn2+ (aq) + 2 Au (s)
Note that this is a 6 electron transfer
The standard cell potential for this electrochemical cell (E0cell ) can be calculated from:
E0cell = E0cathode – E0anode = 1.500 V – ( -0.760 V) = 2.260 V
Under the experimental conditions of this cell (0.100 M solutions and a temperature of 25 °C) the cell
potential (Ecell) can be calculated from:
0
𝐸𝐸𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝐸𝐸𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
−
0.0257
ln 𝑄𝑄
𝑛𝑛
where n = number of electrons transferred, and Q = reaction quotient
For the balanced overall reaction for this cell: 𝑄𝑄 =
Therefore
𝐸𝐸𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 2.260 𝑉𝑉 −
= 2.260 −
[𝑍𝑍𝑍𝑍2+ ]3
[𝐴𝐴𝐴𝐴3+ ]2
(0.100)3
0.0257
0.0257
ln (0.100)2 = 2.260 𝑉𝑉 −
ln(0.100)
6
6
0.0257
(−2.303) = 2.260 + 0.00986 = 𝟐𝟐. 𝟐𝟐𝟐𝟐𝟐𝟐 𝑽𝑽
6
So if the gold half-cell is connected to the positive (+ve) terminal (cathode) and the zinc half-cell is
connected to the negative terminal (anode) of a voltmeter, and the circuit completed by a KNO3 saltbridge connecting the two half-cells, the voltmeter will be expected to show that the electrochemical
cell produces a voltage of 2.270 V (2270 mV). Note that if the cells were connected the other way
around (gold connected to the –ve terminal while zinc connected to the +ve terminal), electrons would
flow in the opposite direction through the voltmeter and it would register a voltage of -2.270 V.
FC 319
Fe (Iron)
sn (Tin) 0.079 v 0.100 m Sncl concentration
Anode
Ag(s) Siver 0.922 V
cathocle
(U (s)
Data and observation
copper
Cathocle
Electrical cells
0.512
0.100 M AgNO3 concentration
0.100 (u(NO3)2 concentration
5/26/2
chem l