Chapter 23, Problem 23.24QAP

Interpretation Introduction

(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.

$\text{E}={\text{E}}^0-\frac{2.303RT}{nF}\log Q$

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

${\text{E}}_{cell}={\text{E}}_{cathode}^0-{\text{E}}_{anode}^0$

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:

${\text{Ce}}^{4+}+\text{ e }\to {\text{ Ce}}^{3+}{\text{ E}}^0=1.44\text{ V}$

${\text{NO}}_3^{-.}{\text{ + 3 H}}^{+}\text{ + 2e }\to {\text{ HNO}}_2{\text{ + H}}_2{\text{O E}}^0=0.94\text{ V}$

$\text{AgCl(s) + e }\to {\text{ Ag(s) + Cl}}^{-}{\text{ E}}^0=0.199$

The overall reaction will be:

Overall reaction: ${\text{2 Ce}}^{4+}{\text{ + HNO}}_2{\text{ + H}}_2\text{O }\to {\text{ 2 Ce}}^{3+}{\text{ + NO}}_3^{-}{\text{ + 3 H}}^{+}$

$\begin{array}{l}{\text{E}}_{eq}={\text{E}}_{{\text{Ce}}^{\text{4+}}{\text{/Ce}}^{3+}}^0-\frac{0.0592}{1}\log \frac{\left[{\text{Ce}}^{3+}\right]}{\left[{\text{Ce}}^{4+}\right]}\to \left(1\right)\\ {\text{E}}_{eq}={\text{E}}_{{\text{NO}}_{\text{3}}^{\text{-}}{\text{/HNO}}_{\text{2}}}^0-\frac{0.0592}{2}\log \frac{\left[{\text{HNO}}_2\right]}{\left[{\text{NO}}_3^{-}\right]{\left[{\text{H}}^{+}\right]}^3}\\ {\text{E}}_{eq}=2{\text{E}}_{{\text{NO}}_{\text{3}}^{\text{-}}{\text{/HNO}}_{\text{2}}}^0-0.0592\log \frac{\left[{\text{HNO}}_2\right]}{\left[{\text{NO}}_3^{-}\right]{\left[{\text{H}}^{+}\right]}^3}\to \left(2\right)\end{array}$

The system is at equilibrium all time. So, the electrode potential for the two half-cell reactions are always equal

(1) + (2) x 2

${\text{3E}}_{\text{eq}}={\text{E}}_{{\text{Ce}}^{\text{4+}}{\text{/Ce}}^{\text{3+}}}^0+2{\text{E}}_{{\text{NO}}_{\text{3}}^{\text{-}}{\text{/HNO}}_{\text{2}}}^{\text{0}}-0.0592\log \frac{\left[{\text{Ce}}^{3+}\right]\left[{\text{HNO}}_2\right]}{\left[{\text{Ce}}^{4+}\right]\left[{\text{NO}}_3^{-}\right]{\left[{\text{H}}^{+}\right]}^3}$

Since, the hydrogen ion concentration is at 1.00 M throughout the titration,one can simplify the above equation.

${\text{3E}}_{\text{eq}}={\text{E}}_{{\text{Ce}}^{\text{4+}}{\text{/Ce}}^{\text{3+}}}^{\text{0}}+2{\text{E}}_{{\text{NO}}_{\text{3}}^{\text{-}}{\text{/HNO}}_{\text{2}}}^{\text{0}}-0.0592\log \frac{\left[{\text{Ce}}^{3+}\right]\left[{\text{HNO}}_2\right]}{\left[{\text{Ce}}^{4+}\right]\left[{\text{NO}}_3^{-}\right]}$

At equivalence point, $\left[{\text{Ce}}^{4+}\right]=2\left[{\text{HNO}}_2\right]\text{ and }\left[{\text{Ce}}^{3+}\right]=2\left[{\text{NO}}_3^{-}\right]$

So,

$\begin{array}{l}{\text{3E}}_{\text{eq}}={\text{E}}_{{\text{Ce}}^{\text{4+}}{\text{/Ce}}^{\text{3+}}}^0+2{\text{E}}_{{\text{NO}}_{\text{3}}^{\text{-}}{\text{/HNO}}_{\text{2}}}^{\text{0}}\\ {\text{E}}_{\text{eq}}=\frac{{\text{E}}_{{\text{Ce}}^{\text{4+}}{\text{/Ce}}^{\text{3+}}}^{\text{0}}+2{\text{E}}_{{\text{NO}}_{\text{3}}^{\text{-}}{\text{/HNO}}_{\text{2}}}^{\text{0}}}{3}\end{array}$

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 = $\frac{0.05\times 40.00}{75.00}=0.026667\text{ M}$

Volume of Ce⁴+ spent at the equivalence point = $\frac{0.026667\text{ mol}}{\text{1000 mL}}\times 75.00\text{ mL}\times \frac{2\text{ mol}}{1\text{ mol}}\times \frac{1000\text{ mL}}{0.0800\text{ mol}}=50.00\text{ mL}$

When 5.00 mL of Ce⁴+ solution is added,

Concentration of NO₃- = $\frac{0.0800\text{ mol}}{1000\text{ mL}}\times 5.00\text{ mL}\times \frac{1\text{ mol}}{2\text{ mol}}\times \frac{1000\text{ mL}}{\left(75.00+5.00\right)\text{mL}}$

= $0.0025\text{ M}$

Concentration of HNO₂ left = $\frac{\frac{0.026667\text{ mol}}{1000\text{ mL}}\times 75.00\text{ mL - }\frac{0.0800\text{ mol}}{1000\text{ mL}}\times 5.00\text{ mL}\times \frac{1\text{ mol}}{2\text{ mol}}}{\left(75.00+5.00\right)\text{ mL}}\times 1000\text{ mL}$

= $0.0225\text{ M}$

When 50.01 mL of Ce⁴+ added

Concentration of Ce⁴+ = $\frac{\frac{0.0800\text{ mol}}{1000\text{ mL}}\times 50.01\text{ mL - }\frac{0.026667\text{ mol}}{1000\text{ mL}}\times 75.00\text{ mL}\times \frac{\text{2 mol}}{\text{1 mol}}}{\left(75.00+50.01\right)\text{ mL}}\times 1000\text{ mL}$

= $5.999\times {10}^{-6}\text{ M}$

Concentration of Ce³+ = $\frac{\frac{0.026667\text{ mol}}{1000\text{ mL}}\times 75.00\text{ mL}\times \frac{2\text{ mol}}{1\text{ mol}}}{\left(75.00+50.01\right)\text{ mL}}\times 1000\text{ mL}$

= $0.031997\text{ M}$

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

Interpretation Introduction

(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.

Interpretation Introduction

(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 $\frac{\Delta E}{\Delta V}$ and average volume of titrant added. The second derivative curve is plotted between $\frac{\left(\Delta E/\Delta V\right)}{\Delta V}$ and average volume of titrant added. By generating derivative curved one can obtain the equivalence point accurately.

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.