Electrochemical cell can be used to measure the concentrations of certain specie
ID: 564182 • Letter: E
Question
Electrochemical cell can be used to measure the concentrations of certain species in solution. For example, the voltage of an electrochemical cell based on the reaction
H2 (g) + CU2+(aq) ---- > 2 H+ + Cu(s) is sensitive to both the CU2+ concentration and the H+ concentration in solution. If the H+ concentration is held constant, then the voltage only depends on the CU2+ concentration, and we can use the cell to measure the CU2+ concentration in an unknown solution. The tabulated data shows the measured voltage the hydrogen/copper electrochemical cell just discussed for several CU2+ concentration. Examine the data and answer the question that follow.
[CU2+] Voltage (V)
0.100 0.310
0.200 0.319
0.300 0.325
0.400 0.328
0.500 0.331
0.700 0.335
1.000 0.340
a. Construct a graph of the measured voltage versus the copper concentration. Is the graph linear?
b. Determine how you might manipulate the data to produce the linear graph. (Hint: See the Nernst equation)
c. Reconstruct a graph of the data using the method to produce the linear graph from part(b). Determine the slope and y-intercept of the best-fitting line to the points in your graph. Could you have predicted the slope and intercept from the Nernst equation?
d. The voltage of two unknown solutions are measured and recorded. Use the slope and intercept from part(C) to determine the CU2+ concentration of the unknown solutions.
[CU2+] Voltage (V)
i 0.303
ii 0.338
No pressure is given in the question
Explanation / Answer
a) Plot the voltage vs Cu2+ as below.
Plot of voltage vs [Cu2+]
The graph is clearly not linear.
b) The Nernst equation is given as
E = E0 – 0.0592/n*log Q where Q = ratio of the concentrations of the products and the reactants.
For the given reaction, we have
Q = [H+]2/[H2][Cu2+] (the concentration of solid Cu is not included in the equilibrium constant expression).
For a fixed concentration of H+ and H2, we have
Q’ = Z/[Cu2+] where Z = [H+]2/[H2]
Plug in the Nernst equation and obtain
E = E0 – 0.0592/n*log Z/[Cu2+] = E0 – (0.0592/2)*log Z + (0.0592/2)*log [Cu2+]
= A + 0.0296*log [Cu2+] …….(1)
where A = E0 – (0.0296)*log Z is a constant.
c) Plot E (voltage) against log [Cu2+]. Find out log [Cu2+] first as before.
[Cu2+], M
log [Cu2+]
Voltage, V
0.100
-1.000
0.310
0.200
-0.699
0.319
0.300
-0.523
0.325
0.400
-0.398
0.328
0.500
-0.301
0.331
0.700
-0.155
0.335
1.000
0.000
0.340
Plot of voltage vs log [Cu2+]
The slope of the plot is 0.0298 while the intercept is 0.3399.
E0 can be easily determined from the standard reduction potential values. If we know the fixed concentrations of H+ and H2, we can easily determine the intercept.
d) Put (i) y = 0.303 in the equation for the line and obtain
0.303 = 0.0298x + 0.3399
====> x = -1.2382
Compare the equation of the line with (1) and get -1.2382 = log [Cu2+]
=====> [Cu2+] = antilog (-1.2382) = 0.05778 0.0578 M (ans).
(ii) Put y = 0.338 and get
0.338 = 0.0298x + 0.3399
====> x = -0.06376
Again, log [Cu2+] = -0.06376
=====> [Cu2+] = antilog(-0.06376) = 0.8634 M (ans).
[Cu2+], M
log [Cu2+]
Voltage, V
0.100
-1.000
0.310
0.200
-0.699
0.319
0.300
-0.523
0.325
0.400
-0.398
0.328
0.500
-0.301
0.331
0.700
-0.155
0.335
1.000
0.000
0.340
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