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Use the following cell to answer the questions: Cr / Cr3+ (1.00 M) // Ni2+ (1.00...

Use the following cell to answer the questions:

Cr / Cr3+ (1.00 M) // Ni2+ (1.00 M) / Ni.

a) How many milligrams of Ni would be plated on the cathode if a current of 0.36 A is drawn for 3.5 hrs?

b) What is [Cr3+] when [Ni2+] has dropped to 10-4 M?

[Cr3+] =

c) What is the cell potential at the conditions described in Part b?

E =

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Homework Answers

Answer #1

a) Mass of nickel plated on cathode (W) = Current X time X Eqt. Wt/96500

W = 0.36 X12600 X 29.35/96500 = 1.38g = 1380mg

b) The cell reaction is

2Cr(s) + 3Ni2+(aq) <------> 3Ni(s) + 2Cr3+(aq)

For every three moles of Ni2+ consumed 2 moles of Cr3+ are produced.

Moles of Ni2+ consumed = 1.0 - 10-4 = 0.9999

moles of Cr3+ formed = 2/3 X 0.9999 = 0.6666

therefore concentration of Cr3+ = 1.0 + 0.66666 = 1.6666

C) cell voltage at this concentration (E) = Eo - 0.0591log[Cr3+]/[Ni2+]

Eo = Eo cathode - Eo anode = -0.28 - (-0.74) = 0.46V

E = Eo - 0.0591log[1.6666]/[0.9999]

E = 0.46 - 0.0591log[1.6666]/[0.9999]

E = 0.448V

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