Question

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

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