Explain the similarities and differences between a voltaic andelectrolytic cell. Be sure to discuss how electrical energy and chemical energy areexchanged in a redox reaction. What results did this experiment end up with whether this lab was successful or not of the electrochemistry redox reaction (Oxidation Reduction) experiment? The results: Part 1: Percent Error Calculation for Voltaic Cells To calculate the percent error, use the formula: Percent Error=Theoretical Value∣Observed Value−Theoretical Value∣×100 Theoretical Voltages for Voltaic Cells To calculate the percent error, we first need the theoretical standard electrode potentials for the voltaic cells: Zn/Cu: EZn2+/Zn = −0.76 V ECu2+/Cu = +0.34 V Theoretical: Ecell =0.34−(−0.76) = 1.10 V Zn/Al: EAl3+/Al = −1.66 V Theoretical: Ecell = −1.66−(−0.76) = −0.90 V Zn/Ag: EAg+/Ag = +0.80 V Theoretical: Ecell = 0.80−(−0.76) = 1.56 V Al/Cu: Theoretical: Ecell = 0.34−(−1.66) = 2.00 V Ag/Cu: Theoretical: Ecell = 0.34−0.80 = −0.46 V Ag/Al: Theoretical: Ecell = 0.80−(−1.66) = 2.46 V Percent Error Calculation Zn/Cu Observed: 0.914 V Theoretical: 1.10 V Percent Error = (0.914−1.10)/1.10×100 = 0.186/1.10/×100 = 16.91% Zn/Al Observed: 0.210 V Theoretical: −0.90-0.90−0.90 V Percent Error = (0.210−(−0.90))/−0.90×100 = 1.11/0.90×100 = 123.33% Zn/Ag Observed: 1.330 V Theoretical: 1.56 V Percent Error = (1.330−1.56)/ 1.56 ×100 = 0.230/ 1.56×100 = 14.74% Al/Cu Observed: 0.672 V Theoretical: 2.00 V Percent Error = (0.672−2.00)/2.00 × 100 = 1.328/2.00×100 = 66.40% Ag/Cu Observed: 0.413 V Theoretical: −0.46-0.46−0.46 V Percent Error = (0.413−(−0.46))/−0.46×100 = 0.873/0.46×100 = 189.78% Ag/Al Observed: 1.000 V Theoretical: 2.46 V Percent Error = (1.000−2.46)/2.46×100 = 1.460/2.46×100 = 59.35% Calculating the mass of I2 produced from the pH change: Given: Initial pH: 5.22 Final pH: 10.74 The volume of solution: 100 mL (0.100 L) Electrolytic reaction: 2I−→I2+2e− Step 1: Calculate [OH−] from pH Initial pH = 5.22: pOH = 14−5.22 = 8.78 [OH−]initial=10−8.78 = 1.66×10−9 M Final pH = 10.74: pOH = 14−10.74 = 3.26 [OH−]final = 10−3.26 = 5.50×10−4 M Step 2: Calculate moles of OH− neutralized Δ[OH−] = 5.50×10−4−1.66×10−9 = 5.50×10−4 M Moles of OH−=Δ[OH−]×0.100 L = 5.50×10−5 mol Step 3: Relate moles of OH− to I2 For every mole of I2 formed, 2 moles of OH- are involved: Moles of I2 = (5.50×10−5)/2 = 2.75×10−5 mol Step 4: Calculate the mass of I2 Molar mass of I2 = 253.81 g/mol Mass of I2 = 2.75×10−5×253.81 = 0.00698 g

Explain the similarities and differences between a voltaic andelectrolytic cell. Be sure to discuss how electrical energy and chemical energy areexchanged in a redox reaction. What results did this experiment end up with whether this lab was successful or not of the electrochemistry redox reaction (Oxidation Reduction) experiment? The results: Part 1: Percent Error Calculation for Voltaic Cells To calculate the percent error, use the formula: Percent Error=Theoretical Value∣Observed Value−Theoretical Value∣×100 Theoretical Voltages for Voltaic Cells To calculate the percent error, we first need the theoretical standard electrode potentials for the voltaic cells: Zn/Cu: EZn2+/Zn = −0.76 V ECu2+/Cu = +0.34 V Theoretical: Ecell =0.34−(−0.76) = 1.10 V Zn/Al: EAl3+/Al = −1.66 V Theoretical: Ecell = −1.66−(−0.76) = −0.90 V Zn/Ag: EAg+/Ag = +0.80 V Theoretical: Ecell = 0.80−(−0.76) = 1.56 V Al/Cu: Theoretical: Ecell = 0.34−(−1.66) = 2.00 V Ag/Cu: Theoretical: Ecell = 0.34−0.80 = −0.46 V Ag/Al: Theoretical: Ecell = 0.80−(−1.66) = 2.46 V Percent Error Calculation Zn/Cu Observed: 0.914 V Theoretical: 1.10 V Percent Error = (0.914−1.10)/1.10×100 = 0.186/1.10/×100 = 16.91% Zn/Al Observed: 0.210 V Theoretical: −0.90-0.90−0.90 V Percent Error = (0.210−(−0.90))/−0.90×100 = 1.11/0.90×100 = 123.33% Zn/Ag Observed: 1.330 V Theoretical: 1.56 V Percent Error = (1.330−1.56)/ 1.56 ×100 = 0.230/ 1.56×100 = 14.74% Al/Cu Observed: 0.672 V Theoretical: 2.00 V Percent Error = (0.672−2.00)/2.00 × 100 = 1.328/2.00×100 = 66.40% Ag/Cu Observed: 0.413 V Theoretical: −0.46-0.46−0.46 V Percent Error = (0.413−(−0.46))/−0.46×100 = 0.873/0.46×100 = 189.78% Ag/Al Observed: 1.000 V Theoretical: 2.46 V Percent Error = (1.000−2.46)/2.46×100 = 1.460/2.46×100 = 59.35% Calculating the mass of I2 produced from the pH change: Given: Initial pH: 5.22 Final pH: 10.74 The volume of solution: 100 mL (0.100 L) Electrolytic reaction: 2I−→I2+2e− Step 1: Calculate [OH−] from pH Initial pH = 5.22: pOH = 14−5.22 = 8.78 [OH−]initial=10−8.78 = 1.66×10−9 M Final pH = 10.74: pOH = 14−10.74 = 3.26 [OH−]final = 10−3.26 = 5.50×10−4 M Step 2: Calculate moles of OH− neutralized Δ[OH−] = 5.50×10−4−1.66×10−9 = 5.50×10−4 M Moles of OH−=Δ[OH−]×0.100 L = 5.50×10−5 mol Step 3: Relate moles of OH− to I2 For every mole of I2 formed, 2 moles of OH- are involved: Moles of I2 = (5.50×10−5)/2 = 2.75×10−5 mol Step 4: Calculate the mass of I2 Molar mass of I2 = 253.81 g/mol Mass of I2 = 2.75×10−5×253.81 = 0.00698 g

Chemistry: Matter and Change

1st Edition

ISBN:9780078746376

Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom

Publisher:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom

Chapter20: Electrochemistry

Section20.1: Voltaic Cells

Problem 5PP

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Related questions

Question

Explain the similarities and differences between a voltaic and
electrolytic cell. Be sure to discuss how electrical energy and chemical energy are
exchanged in a redox reaction.

What results did this experiment end up with whether this lab was successful or not of the electrochemistry redox reaction (Oxidation Reduction) experiment?

The results:

Part 1: Percent Error Calculation for Voltaic Cells

To calculate the percent error, use the formula:

Percent Error=Theoretical Value∣Observed Value−Theoretical Value∣×100

Theoretical Voltages for Voltaic Cells

To calculate the percent error, we first need the theoretical standard electrode potentials for the voltaic cells:

Zn/Cu:

EZn2+/Zn = −0.76 V

ECu2+/Cu = +0.34 V

Theoretical: Ecell =0.34−(−0.76) = 1.10 V

Zn/Al:

EAl3+/Al = −1.66 V

Theoretical: Ecell = −1.66−(−0.76) = −0.90 V

Zn/Ag:

EAg+/Ag = +0.80 V

Theoretical: Ecell = 0.80−(−0.76) = 1.56 V

Al/Cu:

Theoretical: Ecell = 0.34−(−1.66) = 2.00 V

Ag/Cu:

Theoretical: Ecell = 0.34−0.80 = −0.46 V

Ag/Al:

Theoretical: Ecell = 0.80−(−1.66) = 2.46 V

Percent Error Calculation

Zn/Cu

Observed: 0.914 V

Theoretical: 1.10 V

Percent Error = (0.914−1.10)/1.10×100 = 0.186/1.10/×100 = 16.91%

Zn/Al

Observed: 0.210 V

Theoretical: −0.90-0.90−0.90 V

Percent Error = (0.210−(−0.90))/−0.90×100 = 1.11/0.90×100 = 123.33%

Zn/Ag

Observed: 1.330 V

Theoretical: 1.56 V

Percent Error = (1.330−1.56)/ 1.56 ×100 = 0.230/ 1.56×100 = 14.74%

Al/Cu

Observed: 0.672 V

Theoretical: 2.00 V

Percent Error = (0.672−2.00)/2.00 × 100 = 1.328/2.00×100 = 66.40%

Ag/Cu

Observed: 0.413 V

Theoretical: −0.46-0.46−0.46 V

Percent Error = (0.413−(−0.46))/−0.46×100 = 0.873/0.46×100 = 189.78%

Ag/Al

Observed: 1.000 V

Theoretical: 2.46 V

Percent Error = (1.000−2.46)/2.46×100 = 1.460/2.46×100 = 59.35%

Calculating the mass of I₂ produced from the pH change:

Given:

Initial pH: 5.22

Final pH: 10.74

The volume of solution: 100 mL (0.100 L)

Electrolytic reaction: 2I−→I2+2e−

Step 1: Calculate [OH−] from pH

Initial pH = 5.22:

pOH = 14−5.22 = 8.78

[OH−]initial=10^−8.78= 1.66×10⁻⁹M

Final pH = 10.74:

pOH = 14−10.74 = 3.26

[OH−]final = 10^−3.26= 5.50×10⁻⁴ M

Step 2: Calculate moles of OH− neutralized

Δ[OH⁻] = 5.50×10⁻⁴−1.66×10⁻⁹= 5.50×10⁻⁴M

Moles of OH−=Δ[OH−]×0.100 L = 5.50×10⁻⁵ mol

Step 3: Relate moles of OH− to I₂

For every mole of I₂ formed, 2 moles of OH^- are involved:

Moles of I₂= (5.50×10⁻⁵)/2 = 2.75×10⁻⁵ mol

Step 4: Calculate the mass of I₂

Molar mass of I₂= 253.81 g/mol

Mass of I₂= 2.75×10⁻⁵×253.81 = 0.00698 g

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