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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