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 0.39 A is drawn for 2.0 hrs?

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

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

Table of standard reduction potentials: http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

Answer 1

a) faradays first law

w = zit

w = (E/F)it

E = equivalent weight of Ni = 58.7/2 = 29.35 g/equiv

    = (29.35/96500)*0.39*2*60

   Weight of Ni diposited = 0.0142 g

b) 3Ni^2+(aq) + 2Cr(s) -----> 3Ni(s) + 2Cr^3+(aq)

E0cell = E0cathode - E0anode

        = -0.25-(-0.74)

            = 0.49 V

b)  

decrease in concentration of Ni^2+ = 1 - (10^-4) = 0.9999 M

   mass of Ni deposited during this = M*V*Mwt

                         = 0.9999*1*58.7
                      
                         = 58.7 g

   no of faradys of electricity produced = mwt / ewt = 58.7/29.35 = 2 F

    3 faradays = 1 mole Cr

    2 Faradays = 2/3 = 0.67 mole

spontaneously concentration of Cr^3+ increased = 0.67/1 = 0.67 M

final concentration of Cr^3+ increased = 1+0.67 = 1.67 M

C) Ecell = E0cell - 0.0591/nlog((Cr^3+)^2/(Ni^2+)^3)

        = 0.49 - (0.0591/6)log(1.67^2/(10^-4)^3)

        = 0.367 V