The color of the Universal pH indicator at various pH is shown in Figure 2.ROYG BIV[H+] = 104[H+]/[OH-] = 10-7[OH-] = 10-4Figure 2. Colors of the Universal Indicator based on pHUsing the information from Figure 1 and Figure 2, answer the following questions.oxidation, 1)oxidation, 2)reduction, 1)reduction, 2)The positive electrode produces the half reaction, which corresponds to equation We will now consider the reaction of our experiment.The two half reactions for electrolysis of water are:_-)2H₂O(l) + 2e¯¯ →>> H₂(g) + OH¯(aq)ndC)H₂O(l) → 2H+ (aq) + 1/2O₂(g) + 2e¯The experiment is set up such that two connected funnels contain 50 mL of 0.5 M K₂SO4 and Universal Indicator.An electrode is then placed in each funnel, and an electric current is applied (Figure 1).(+)(-)

The two half-cell reactions given for the electrolysis of water are as follows:2 H2O (l) + 2 e- H2…

Using the information from Figure 1 and Figure 2, answer the following questions. The positive electrode produces the oxidation, 1) oxidation, 2) reduction, 1) reduction, 2) half reaction, which corresponds to equation

Using the information from Figure 1 and Figure 2, answer the following questions. The positive electrode produces the oxidation, 1) oxidation, 2) reduction, 1) reduction, 2) half reaction, which corresponds to equation

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

The color of the Universal pH indicator at various pH is shown in Figure 2.
ROYG BIV
[H+] = 104
[H+]/[OH-] = 10-7
[OH-] = 10-4
Figure 2. Colors of the Universal Indicator based on pH
Using the information from Figure 1 and Figure 2, answer the following questions.
oxidation, 1)
oxidation, 2)
reduction, 1)
reduction, 2)
The positive electrode produces the half reaction, which corresponds to equation

We will now consider the reaction of our experiment.
The two half reactions for electrolysis of water are:
_-)
2H₂O(l) + 2e¯¯ →>> H₂(g) + OH¯(aq)
nd
C)
H₂O(l) → 2H+ (aq) + 1/2O₂(g) + 2e¯
The experiment is set up such that two connected funnels contain 50 mL of 0.5 M K₂SO4 and Universal Indicator.
An electrode is then placed in each funnel, and an electric current is applied (Figure 1).
(+)
(-)

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