🤓 Based on our data, we think this question is relevant for Professor Massa's class at HUNTER.
cathode → reduction → oxidation number decreases
anode → oxidation → oxidation number increases
For ions: charge of the ion = oxidation number
For neutral atoms/compound: oxidation number = 0
ClO2- → ClO2 (oxidation number decreases)
Cl2 → 2 Cl- (oxidation number decreases)
From the reduction potential table, we have:
ClO2 + e- → ClO2- E°cell = 0.954 V anode
Cl2 + 2 e- → 2 Cl- E°cell = 1.36 V. cathode
Chlorine dioxide (ClO2), which is produced by the reaction
2NaClO2 (aq) + Cl2 (g) → 2ClO2 (g) + 2NaCl (aq)
has been tested as a disinfectant for municipal water treatment. Using data from Table 17‑1, calculate E° and ΔG° at 25°C for the production of ClO2.
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Based on our data, we think this problem is relevant for Professor Massa's class at HUNTER.
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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl 2nd Edition practice problems.