🤓 Based on our data, we think this question is relevant for Professor Egolf's class at Penn State University Park.
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
Cr2O72- + 14H+ + 6e- → 2Cr 3+ + 7H2O
[H2O2 + 2H+ + 2e- → 2H2O]×3
Exercise 46. Give the balanced cell equation and determine E° for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table 17‑1.
a. Cr2O72- + 14H+ + 6e- → 2Cr 3+ + 7H2O
H2O2 + 2H+ + 2e- → 2H2O
b. 2H + + 2e- → H2
Al 3+ + 3e- → Al
Calculate the maximum amount of work that can be obtained from the galvanic cells at standard conditions in Exercise 46.
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Our tutors have indicated that to solve this problem you will need to apply the Cell Potential concept. If you need more Cell Potential practice, you can also practice Cell Potential practice problems.
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Based on our data, we think this problem is relevant for Professor Egolf's class at Penn State University Park.
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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.