We are asked to **determine the half-cell potential for the reaction given below:**

**No ^{2}^{+} (aq) + 2 e^{- }→ No (s) ε°_{red} = ???**

Recall:

$\overline{){\mathbf{}}{\mathbf{\Delta G}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}{\mathbf{nF\epsilon}}{\mathbf{\xb0}}}$

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 No^{3+} to a lower oxidation state were performed at 25 °C, and the results are given below.

No^{3+} (aq) + 1 e^{- }→ No^{2+} (aq) ε°_{red} = +1.45 V

No^{3+} (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.

No^{2}^{+} (aq) + 2 e^{- }→ No (s) ε°_{red} = ???

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the The Nernst Equation concept. You can view video lessons to learn The Nernst Equation. Or if you need more The Nernst Equation practice, you can also practice The Nernst Equation practice problems.