I. TITTLEElectrochemical Analysis

II. AIMS1. Determine chemical potential of Cu2+

2. Analysis Cu2+ with potentiometryIII. BASIC THEORY

The principle potentiometry analysis is interaction between analyte with electrode. Electric potential that produced will measured by potentiometer. The important part in potentiometer is electrode. The function of electrode is catch signal of electric that caused by analyte in the solution.electrode that used in measurement of chemical electro consist of work electrode and comparing electrode. The electric signal that produced proportional with analyte activity. The equation to calculate chemical potential from substance that experience oxidation-reduction reaction is Nert equation.

aA+bB+ ... ⟷ cC+dD+ ...G = GΔ Δ o+RT ln KG = GΔ Δ o+RT ln aCc aDd/aBb aAa

Because a ≅ c for diluted solution, so G = GΔ Δ o+RT ln [C]c [D]d/[A]a [B]b

According of equation G = -nFE then substituted into equation above Δ-nFE=-nFEo+RT ln [C]c [D]d/[A]a [B]b

E = Eo – RT/nF ln [C]c [D]d/[A]a [B]b

Equation above is called Nerst equation as basic calculation of chemical electric and implement to cation Cu2+ that occur reduction become Cu like this reaction.

Cu2+ + 2e ⟷ CuE Cu2+/Cu = Eo

Cu2+/Cu – RT/2F ln [Cu]/[Cu2+]This equation will implement at potential measurement from some concentration of Cu2+ solution and determining concentration Cu2+ in the sample.

III. TOOLS AND MATERIALS- CuSO4 0,1 M- Aquades - Cu2+ solution - pH meter- beaker glass

IV. PROCEDURE

0,1 M CuSO4

Diluted with aquades

Made standart solution(10-3, 2x10-3, 4x10-3, 8x10-3, 16x10-3)

Prepared Cu2+ for sample

Used pH meter

Set E-scale

Measured potential for each standart solutionPotential (V)

Make graph

Calculate concentration of

sampleSample (M)

V. OBSERVATION DATA

Procedure Result Hypothesis Conclusion

[Cu2+] V ( mV)A 1x10-3 38,3B 2x10-3 46,3C 4x10-3 55,9D 8x10-3 65,9E 16x10-3 72,9sample 55,6

Cu2+ + 2e ⟷ CuSo thatECu2+/Cu = Eo

Cu2+/Cu – RT/2F ln [Cu]/[Cu2+]

Potential of Cu2+ concentration in solution A,B,C,D,E is increasing. And sample solution has concentartion 55,6 mV

[Cu2+] =

0,1 M CuSO4

Diluted with aquades

Made standart solution(10-3, 2x10-3, 4x10-3, 8x10-3, 16x10-3)

Prepared Cu2+ for sample

Used pH meter

Set E-scale

Measured potential for each standart solutionPotential (V)

Make graph

Calculate concentration of

sampleSample (M)

VI. ANALYSIS AND DISCUSSION

Potentiometric method has several kinds are distinguished by the type of electrode used. Two components of the electrodes used are indicator electrode and a reference electrode. Reference electrode is an electrode comparator used to compare the potential value is read by the indicator electrode.

The requirements that must be completed to be used as a reference electrode is:a. Comply Nersnt equation is reversibleb. Have a constant electrode potential by the timec. Right back to initial potential value when small current flowingd. An electrode that is ideally nonpolarization

Indicator electrode is a second electrode used. This electrode is used to analyze or read potential generated by the sample and the value will indicate how large the sample concentration. Indicator electrode is generally divided into 2 parts:

a. metal indicator electrode

Metal indicator electrode is an electrode which is made using metal plates or wire dipped in an electrolyte solution. Metal electrodes can be grouped into the first type electrode (first kind), the second type electrode (second kind), the third type of electrode (third kind), redox electrode. Of various kinds of metal electrodes, in this experiment using the first type of electrode, the electrode directly continuous with the cations from the metal

b. membrane indicator electrode

The indicator electrode is usually sensitive to only one type of ion. Voltage generated depends on the number of ions in the solution on the surface. It can be seen from the amount or concentration of ions in solution, but this practice is not to use the membrane indicator electrode.

After copper wire was inserted and filled with CuSO4 solution, Cu metal as an indicator electrode.

Standard solution used in this experiment is CuSO4 solution with a concentration of 1x10-3 M, 2x10-3 M, 4x10-3 M, 8x10-3 M, 16x10-3 M and 0.1 M CuSO4 solution as its main solution. After all the samples were made then measured the samples potential using the electrode has been created. And results such as those set forth in the table of observations. From these results, made the graph that relates between the concentration with the potential difference. Where the higher the concentration the higher the potential difference and could mean that the concentration is directly proportional to the potential difference. This is because at higher concentrations, more activity of ions so resulting in a higher current and also showed a large potential difference. Below is a graph of potential difference and concentration.

Picture 1:Graph showing the relationship between the concentration of the potential differencethe experiments we did, the results obtained Cu 2 + concentration is very much different as it should which is about 0.003 M. This is due to several factors, namely less accurate in reading the potential value of each standard solution. because of the value that appears in the tool changing and difficult to remain constant

VII. CONCLUSION

Based on the result of experiment, it may be concluded that:- The higher the concentration the higher the potential difference and could

mean that the concentration is directly proportional to the potential difference.

- The principle of potentiometric method is measurement of the potential difference when no current is flowing.

VIII. REFERRENCES

Day, R. A and Underwood, A. L. 1991. Quantitative Analysis Sixth Eddition. New Jersey: Prentice Hall. Inc.

Poedjiastoeti, Sri, dkk. 2011. Panduan Kimia Analitik II: Dasar-Dasar Pemisahan Kimia. Surabaya: Unesa press.

Soebagio, dkk. 2013. Common Text Book Kimia Analitik II. Malang: UNM

ANSWER & QUESTION

1. Write the equation of linear regression from the relation between X variable as

concentration and Y as potential!

Answer:

Y=12,811x + 126,6R2=0,9972. How to determine the amount of electron that occurs in reaction?

Answer:

The tendency of electrons to flow from one chemical to another is known as electrochemistry. This is what occurs in a concentration cell. The electrons flow from the left side (or left beaker) to the right side (or right beaker). Because the left side is losing electrons and the right is gaining them, the left side is called the oxidation side and the right side is the reduction side. Although you could switch the two to be on the opposite sides, this is the general way in which the set up is done. The oxidation side is called the anode and the reduction side is the cathode. It is the flow of the electrons that cause one side to be oxidized and the other to be reduced.Cu2+ + 2e- ↔ Cu Zn2+ + 2e- ↔ ZnOxidation + amount of electron ↔ reduction

Y=ax + ba=tg =slopeαb=

3. Can the concentration of Cu2+ from the pH? Explain!

Answer:

We can determine the concentration of ion from the pH if we use Standard

Hydrogen Electrode (SHE)

2H+ + 2e → H2E H⁰ +/H2=0,00 V

E H+/H2= E H⁰ +/H2 + RT/2F ln [H+]2

E H+/H2= E H⁰ +/H2 + RT/F ln [H+]

E H+/H2= E H⁰ +/H2 – 2,303 RT/nF pH

4. Determine the equilibrium constant from Cu2+ becomes Cu!

Answer:

Cu2+ + 2e- ↔ Cu Eo = 0.337

Zn(s) ↔ Zn2+ + 2e- Eo= 0.763

Cu2+ + Zn(s) ↔ Zn2+ + Cu Eo = 1.10 volts

E = Eo - (0.0592/n) (log Q)

which at equilibrium, where the cell is dead and E = zero, becomes:

0 = Eo - (0.0592/n) (log K)

0 = 1.10 - (0.0592/2) (log K)

-1.10 = - 0.0296 logK

log K = 37.162

K = 1.45x1037

ATTACHMENT

CALCULATION

Y=12,881x + 126,6R2=0,997E Cu/Cu2+ = -E Cu/Cu2+ -

55,6 = 126,6 – 12,881

55,6 = 126,6 + 12,881

12,881 = - 126,6

= -126,6/12,881

= 9,828

[Cu2+] = 18,03 M