For the reaction 2Fe(s) + 3Cu 2+ (aq, 0.050 molal) à 2Fe 3+ (aq, 0.100 molal) +

Question

For the reaction 2Fe(s) + 3Cu2+ (aq, 0.050 molal) à 2Fe3+ (aq, 0.100 molal) + 3Cu(s), determine the expected voltage using the Debye –Hückel limiting law. The reaction occurs at 25oC. The value of B at this T is 2.32 X 109 m-1-molal-½; A is 1.171 molal-½. Assume that the molal concentrations are close enough to molar concentrations that they can be used directly. Also assume that the anion is NO3-, that is, we are in reality considering 0.050 molal Cu(NO3)2 and 0.1000-molal Fe(NO3)3 solutions. Also use the fact that the ionic diameters for Fe3+ and Cu2+ are 9.0 angstroms and 6.0 angstroms, respectively.

Explanation / Answer

For the reaction 2Fe(s) + 3Cu 2+ (aq, 0.050 molal) à 2Fe 3+ (aq, 0.100 molal) +

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.