Concept explainers
(a)
Interpretation:
The potential of the electrode with respect to a Ag/AgCl(sat’d) reference electrode after addition of different volumes of Cerium(IV) should be calculated.
Concept introduction:
Nernst equation gives the cell potential under non-standard conditions.
E − cell potential
E0 − standard cell potential
R − universal gas constant
T − temperature in Kelvin
n − number of electrons transferred
F − Faraday constant
Q − Reaction quotient

Answer to Problem 23.24QAP
Volume of Ce4+, mL | Ecell, V |
5.00 | 0.712754 |
10.00 | 0.723179 |
15.00 | 0.730108 |
20.00 | 0.735787 |
24.00 | 0.739971 |
24.90 | 0.740897 |
25.00 | 0.741 |
40.00 | 0.75882 |
45.00 | 0.769244 |
49.00 | 0.791022 |
49.50 | 0.800055 |
49.60 | 0.802945 |
49.70 | 0.806663 |
49.80 | 0.811887 |
49.90 | 0.820784 |
49.95 | 0.829627 |
49.99 | 0.849708 |
50.00 | 0.907667 |
50.01 | 1.020361 |
50.05 | 1.063076 |
50.10 | 1.081059 |
50.20 | 1.098961 |
50.30 | 1.109413 |
50.40 | 1.116822 |
50.50 | 1.122568 |
51.00 | 1.140405 |
55.00 | 1.181796 |
60.00 | 1.199619 |
75.00 | 1.223178 |
90.00 | 1.235262 |
Explanation of Solution
The electrode potentials for half cell reactions are as follows:
The overall reaction will be:
Overall reaction:
The system is at equilibrium all time. So, the electrode potential for the two half-cell reactions are always equal
(1) + (2) x 2
Since, the hydrogen ion concentration is at 1.00 M throughout the titration,one can simplify the above equation.
At equivalence point,
So,
Potential before the equivalence point can be determined by applying the Nernst equation for HNO2/NO3- half-cell reaction. And potential after the equivalence point can be determined by applying Nernst equation for the Ce4+/Ce3+ half-cell reaction.
Initial concentration of HNO2 =
Volume of Ce4+ spent at the equivalence point =
When 5.00 mL of Ce4+ solution is added,
Concentration of NO3- =
=
Concentration of HNO2 left =
=
When 50.01 mL of Ce4+ added
Concentration of Ce4+ =
=
Concentration of Ce3+ =
=
likewise, one can calculate the concentrations of the species when other volumes of Ce4+ added. And then the electrode potential can be calculated. The potential of the indicator electrode with respect to a Ag/AgCl reference electrode can be calculated by subtracting the Ag/AgCl standard electrode potential from the electrode potential for the redox reaction calculated.
Volume of Ce4+, mL | [NO3-] | [HNO2] | [Ce3+] | [Ce4+] | Ecell, V |
5.00 | 0.0025 | 0.0225 | 0.712754 | ||
10.00 | 0.004705882 | 0.018824 | 0.723179 | ||
15.00 | 0.006666667 | 0.015556 | 0.730108 | ||
20.00 | 0.008421053 | 0.012632 | 0.735787 | ||
24.00 | 0.00969697 | 0.010505 | 0.739971 | ||
24.90 | 0.00996997 | 0.01005 | 0.740897 | ||
25.00 | 0.01 | 0.01 | 0.741 | ||
40.00 | 0.013913043 | 0.003478 | 0.75882 | ||
45.00 | 0.015 | 0.001667 | 0.769244 | ||
49.00 | 0.015806452 | 0.000323 | 0.791022 | ||
49.50 | 0.015903614 | 0.000161 | 0.800055 | ||
49.60 | 0.015922953 | 0.000129 | 0.802945 | ||
49.70 | 0.015942261 | 9.64E-05 | 0.806663 | ||
49.80 | 0.015961538 | 6.43E-05 | 0.811887 | ||
49.90 | 0.015980785 | 3.22E-05 | 0.820784 | ||
49.95 | 0.015990396 | 1.62E-05 | 0.829627 | ||
49.99 | 0.01599808 | 3.4E-06 | 0.849708 | ||
50.00 | 0.907667 | ||||
50.01 | 0.031998 | 5.99952E-06 | 1.020361 | ||
50.05 | 0.031988 | 3.15874E-05 | 1.063076 | ||
50.10 | 0.031975 | 6.35492E-05 | 1.081059 | ||
50.20 | 0.031949 | 0.000127396 | 1.098961 | ||
50.30 | 0.031924 | 0.000191141 | 1.109413 | ||
50.40 | 0.031898 | 0.000254785 | 1.116822 | ||
50.50 | 0.031873 | 0.000318327 | 1.122568 | ||
51.00 | 0.031746 | 0.000634524 | 1.140405 | ||
55.00 | 0.03077 | 0.003076538 | 1.181796 | ||
60.00 | 0.02963 | 0.005925556 | 1.199619 | ||
75.00 | 0.026667 | 0.013333 | 1.223178 | ||
90.00 | 0.024243 | 0.019393636 | 1.235262 |
(b)
Interpretation:
A titration curve should be constructed
Concept introduction:
Potentiometric titration is similar to direct redox titrations but in potentiometric titrations no indicator is used. Instead, the potential is measured during the titration to obtain the equivalence point.

Answer to Problem 23.24QAP
Explanation of Solution
The dependent variable is the indicator electrode potential and the independent variable is volume of Ce4+ solution added.
(c)
Interpretation:
A first and second derivative curve for the data should be generated.
Concept introduction:
The first derivative curve of the potentiometric titration is plotted between

Answer to Problem 23.24QAP
First derivative curve:
Second derivative curve:
Explanation of Solution
ΔE | ΔV | ΔE/ΔV | Vave | Δ(ΔE/ΔV) | Δ(ΔV) | Δ(ΔE/ΔV)/Δ(ΔV) | Vave |
0.010425 | 5 | 0.002085 | 7.5 | ||||
0.006929 | 5 | 0.001386 | 12.5 | -0.0006992 | 5 | -0.00013984 | 10 |
0.005679 | 5 | 0.001136 | 17.5 | -0.00025 | 5 | -5E-05 | 15 |
0.004184 | 4 | 0.001046 | 22 | -8.98E-05 | 4.5 | -1.99556E-05 | 19.75 |
0.000926 | 0.9 | 0.001029 | 24.45 | -1.71111E-05 | 2.45 | -6.98413E-06 | 23.225 |
0.000103 | 0.1 | 0.00103 | 24.95 | 1.11111E-06 | 0.5 | 2.22222E-06 | 24.7 |
0.01782 | 15 | 0.001188 | 32.5 | 0.000158 | 7.55 | 2.09272E-05 | 28.725 |
0.010424 | 5 | 0.002085 | 42.5 | 0.0008968 | 10 | 8.968E-05 | 37.5 |
0.021778 | 4 | 0.005444 | 47 | 0.0033597 | 4.5 | 0.0007466 | 44.75 |
0.009033 | 0.5 | 0.018066 | 49.25 | 0.0126215 | 2.25 | 0.005609556 | 48.125 |
0.00289 | 0.1 | 0.0289 | 49.55 | 0.010834 | 0.3 | 0.036113333 | 49.4 |
0.003718 | 0.1 | 0.03718 | 49.65 | 0.00828 | 0.1 | 0.0828 | 49.6 |
0.005224 | 0.1 | 0.05224 | 49.75 | 0.01506 | 0.1 | 0.1506 | 49.7 |
0.008897 | 0.1 | 0.08897 | 49.85 | 0.03673 | 0.1 | 0.3673 | 49.8 |
0.008843 | 0.05 | 0.17686 | 49.925 | 0.08789 | 0.075 | 1.171866667 | 49.8875 |
0.020081 | 0.04 | 0.502025 | 49.97 | 0.325165 | 0.045 | 7.225888889 | 49.9475 |
0.057959 | 0.01 | 5.7959 | 49.995 | 5.293875 | 0.025 | 211.755 | 49.9825 |
0.112694 | 0.01 | 11.2694 | 50.005 | 5.4735 | 0.01 | 547.35 | 50 |
0.042715 | 0.04 | 1.067875 | 50.03 | -10.201525 | 0.025 | -408.061 | 50.0175 |
0.017983 | 0.05 | 0.35966 | 50.075 | -0.708215 | 0.045 | -15.73811111 | 50.0525 |
0.017902 | 0.1 | 0.17902 | 50.15 | -0.18064 | 0.075 | -2.408533333 | 50.1125 |
0.010452 | 0.1 | 0.10452 | 50.25 | -0.0745 | 0.1 | -0.745 | 50.2 |
0.007409 | 0.1 | 0.07409 | 50.35 | -0.03043 | 0.1 | -0.3043 | 50.3 |
0.005746 | 0.1 | 0.05746 | 50.45 | -0.01663 | 0.1 | -0.1663 | 50.4 |
0.017837 | 0.5 | 0.035674 | 50.75 | -0.021786 | 0.3 | -0.07262 | 50.6 |
0.041391 | 4 | 0.010348 | 53 | -0.02532625 | 2.25 | -0.011256111 | 51.875 |
0.017823 | 5 | 0.003565 | 57.5 | -0.00678315 | 4.5 | -0.001507367 | 55.25 |
0.023559 | 15 | 0.001571 | 67.5 | -0.001994 | 10 | -0.0001994 | 62.5 |
0.012084 | 15 | 0.000806 | 82.5 | -0.000765 | 15 | -5.1E-05 | 75 |
First derivative curve:
Second derivative curve:
The volume at which the second-derivative curve cross zero correspond to the theoretical equivalence point. Theoretical equivalence point is 50.00 mL.At the maximum of the peak in the first derivative curve has a slope of zero. So, in the second derivative curve crosses the x axis at the point, corresponding to the equivalence point.
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Chapter 23 Solutions
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