Write a one or two paragraphs of -what was study in identification of a lectrochemistry redox reaction (Oxidation Reduction) experiment. -Summarize what happened in these results shown below -what results did this experiment end up with whether this lab was successful or not 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
Write a one or two paragraphs of
-what was study in identification of a lectrochemistry redox reaction (Oxidation Reduction) experiment.
-Summarize what happened in these results shown below
-what results did this experiment end up with whether this lab was successful or not
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
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