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Math  /  Algebra

Question1. What is hydrodynamic voltammetry? State two (2) applications of voltammetry.
2. What is the potential of a cell consisting of a silver electrode dipped in silver nitrate solution with [Ag+]=0.01M[\mathrm{Ag}+]=0.01 \mathrm{M} and a standard calomel reference electrode (Eref=+0.242\left(\mathrm{E}_{\mathrm{ref}}=+0.242\right. V). Neglect liquid-junction potential.
3. Briefly describe a simple electrochemical cell used in potentiometric analysis. Write a cell diagram for this cell.
4. Calculate the ionic strengths of the following solutions A) 0.20 M NaCl and 0.30MNa2SO40.30 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4} B) 0.1MK2Cr2O70.1 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}

Studdy Solution

STEP 1

1. The problem involves multiple parts related to electrochemistry and ionic strength calculations.
2. We need to explain hydrodynamic voltammetry and its applications.
3. We need to calculate the cell potential using the Nernst equation.
4. We need to describe a simple electrochemical cell and write its cell diagram.
5. We need to calculate the ionic strengths of given solutions.

STEP 2

1. Explain hydrodynamic voltammetry and state its applications.
2. Calculate the potential of the given electrochemical cell.
3. Describe a simple electrochemical cell and write its cell diagram.
4. Calculate the ionic strengths of the given solutions.

STEP 3

Explain hydrodynamic voltammetry and state two applications.
Hydrodynamic voltammetry is a technique in electrochemistry where the working electrode is rotated or the solution is stirred to enhance mass transport to the electrode surface. This technique is used to study the kinetics of electrochemical reactions and to determine the concentration of analytes in a solution.
Two applications of voltammetry are:
1. Determination of trace metal concentrations in environmental samples.
2. Analysis of organic compounds in pharmaceuticals.

STEP 4

Calculate the potential of the given electrochemical cell.
The potential of the cell can be calculated using the Nernst equation:
Ecell=Ecathode−Eanode E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}
Given: - The standard reduction potential for the silver electrode is EAg+/Ag∘=+0.799 V E^{\circ}_{\text{Ag}^+/\text{Ag}} = +0.799 \, \text{V} . - The concentration of [Ag+][\text{Ag}^+] is 0.01 M0.01 \, \text{M}. - The reference electrode potential is Eref=+0.242 V E_{\text{ref}} = +0.242 \, \text{V} .
Using the Nernst equation for the silver electrode:
EAg=EAg+/Ag∘−0.05911log⁡(1[Ag+]) E_{\text{Ag}} = E^{\circ}_{\text{Ag}^+/\text{Ag}} - \frac{0.0591}{1} \log \left( \frac{1}{[\text{Ag}^+]} \right)
EAg=0.799−0.05911log⁡(10.01) E_{\text{Ag}} = 0.799 - \frac{0.0591}{1} \log \left( \frac{1}{0.01} \right)
EAg=0.799−0.0591×2 E_{\text{Ag}} = 0.799 - 0.0591 \times 2
EAg=0.799−0.1182 E_{\text{Ag}} = 0.799 - 0.1182
EAg=0.6808 V E_{\text{Ag}} = 0.6808 \, \text{V}
Now, calculate the cell potential:
Ecell=EAg−Eref E_{\text{cell}} = E_{\text{Ag}} - E_{\text{ref}}
Ecell=0.6808−0.242 E_{\text{cell}} = 0.6808 - 0.242
Ecell=0.4388 V E_{\text{cell}} = 0.4388 \, \text{V}

STEP 5

Describe a simple electrochemical cell and write its cell diagram.
A simple electrochemical cell consists of two electrodes immersed in electrolyte solutions, connected by a salt bridge or a porous membrane. One electrode acts as the anode, where oxidation occurs, and the other acts as the cathode, where reduction takes place. The cell generates an electric current as a result of the redox reactions.
Cell diagram for a simple electrochemical cell:
Anode ∣ Anode solution ∣∣ Cathode solution ∣ Cathode \text{Anode} \, | \, \text{Anode solution} \, || \, \text{Cathode solution} \, | \, \text{Cathode}
Example cell diagram:
Zn (s) ∣ Zn2+(aq) ∣∣ Cu2+(aq) ∣ Cu (s) \text{Zn (s)} \, | \, \text{Zn}^{2+} (\text{aq}) \, || \, \text{Cu}^{2+} (\text{aq}) \, | \, \text{Cu (s)}

STEP 6

Calculate the ionic strengths of the given solutions.
A) For 0.20 M NaCl0.20 \, \text{M} \, \text{NaCl} and 0.30 M Na2SO40.30 \, \text{M} \, \text{Na}_2\text{SO}_4:
The ionic strength I I is given by:
I=12∑cizi2 I = \frac{1}{2} \sum c_i z_i^2
For NaCl: - cNa+=0.20 M c_{\text{Na}^+} = 0.20 \, \text{M} , zNa+=+1 z_{\text{Na}^+} = +1 - cCl−=0.20 M c_{\text{Cl}^-} = 0.20 \, \text{M} , zCl−=−1 z_{\text{Cl}^-} = -1
INaCl=12(0.20×12+0.20×12) I_{\text{NaCl}} = \frac{1}{2} \left( 0.20 \times 1^2 + 0.20 \times 1^2 \right)
INaCl=12×0.40=0.20 M I_{\text{NaCl}} = \frac{1}{2} \times 0.40 = 0.20 \, \text{M}
For Na2SO4: - cNa+=2×0.30=0.60 M c_{\text{Na}^+} = 2 \times 0.30 = 0.60 \, \text{M} , zNa+=+1 z_{\text{Na}^+} = +1 - cSO42−=0.30 M c_{\text{SO}_4^{2-}} = 0.30 \, \text{M} , zSO42−=−2 z_{\text{SO}_4^{2-}} = -2
INa2SO4=12(0.60×12+0.30×22) I_{\text{Na}_2\text{SO}_4} = \frac{1}{2} \left( 0.60 \times 1^2 + 0.30 \times 2^2 \right)
INa2SO4=12(0.60+1.20) I_{\text{Na}_2\text{SO}_4} = \frac{1}{2} \left( 0.60 + 1.20 \right)
INa2SO4=12×1.80=0.90 M I_{\text{Na}_2\text{SO}_4} = \frac{1}{2} \times 1.80 = 0.90 \, \text{M}
Total ionic strength for the mixture:
Itotal=INaCl+INa2SO4=0.20+0.90=1.10 M I_{\text{total}} = I_{\text{NaCl}} + I_{\text{Na}_2\text{SO}_4} = 0.20 + 0.90 = 1.10 \, \text{M}
B) For 0.1 M K2Cr2O70.1 \, \text{M} \, \text{K}_2\text{Cr}_2\text{O}_7:
- cK+=2×0.1=0.2 M c_{\text{K}^+} = 2 \times 0.1 = 0.2 \, \text{M} , zK+=+1 z_{\text{K}^+} = +1 - cCr2O72−=0.1 M c_{\text{Cr}_2\text{O}_7^{2-}} = 0.1 \, \text{M} , zCr2O72−=−2 z_{\text{Cr}_2\text{O}_7^{2-}} = -2
I=12(0.2×12+0.1×22) I = \frac{1}{2} \left( 0.2 \times 1^2 + 0.1 \times 2^2 \right)
I=12(0.2+0.4) I = \frac{1}{2} \left( 0.2 + 0.4 \right)
I=12×0.6=0.3 M I = \frac{1}{2} \times 0.6 = 0.3 \, \text{M}

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