Question

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 0.18 A is drawn for 2.9 hrs?

mg Ni

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

[Cr3+] =  M

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

E =  V

Homework Answers

Answer #1

(a) It is just an unit conversion problem.

Use current (amp) x time (s) = amount of electric charge (C) = 0.18 amp * 2.9 h * 3600 sec = 1879.2 amp-sec

Charge (C)/96,486 = Faradays =1879.2/96486 = 0.0195 Faradays

Now, 1 Faraday = 1 mole so 0.0195 Faradays= 0.0195 mole

1 mole Ni corresponds to 58mg Ni.

0.0195 mole Ni corresponds to 58mg*0.0195 Ni = 1.131mg

(b) Red-ox reaction :

2 Cr + 3 Ni2+ = 2 Cr3+ + 3Ni

Now if [Ni2+] has dropped to 10^-4 M then [Cr3+]= 1.5 * 10^-4 M

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