We are asked to determine the half-cell potential for the reaction given below:
No2+ (aq) + 2 e- → No (s) ε°red = ???
Recall:
where:
n = number of moles electrons
F = Faraday’s constant, 96485 C/mol e-
ε° = standard reduction potential
Briefly answer the following questions.
Nobelium (No) is a radioactive metal element that does not occur naturally. Several experiments measuring the half-cell potentials for the reduction of No3+ to a lower oxidation state were performed at 25 °C, and the results are given below.
No3+ (aq) + 1 e- → No2+ (aq) ε°red = +1.45 V
No3+ (aq) + 3 e- → No (s) ε°red = -1.20 V
Using the data given above, determine the half-cell potential for the reaction given below. Please circle your answer.
No2+ (aq) + 2 e- → No (s) ε°red = ???
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