Question

9) A galvanic cell Cr(s)|Cr²⁺(aq) || Pb²⁺(aq)|Pb(s) is constructed using a completely immersed Cr electrode that weighs 26.5 g and a Pb electrode immersed in 665 mL of 1.00 M Pb²⁺(aq) solution. A steady current of 0.0598 A is drawn from the cell as the electrons move from the Cr electrode to the Pb electrode.

(a) Which reactant is the limiting reactant in this cell?
___

(b) How long does it take for the cell to be completely discharged? ___ s

(c) How much mass has the Pb electrode gained when the cell is completely discharged? ___ g

(d) What is the concentration of the Pb²⁺(aq) when the cell is completely discharged? (Assume that the limiting reactant is 100% reacted.) ___ M

Answer 1

Answer:

The cell response of the galvanic cell is:

Cr(s) + Pb2+(aq) = Cr2+(aq) + Pb(s)

So 1 mole of Cr(s) responds with 1 mole of Pb2+(aq).

Presently, number of moles of Cr(s) present in the cell = (26.5/52.0) mol

= 0.5096 mol [ MW of Cr = 52.0 g mol-1]

Once more, the quantity of moles of Pb2+ present in the arrangement = (1.00*665/1000) mol = 0.665 mol

We realize the species present in less sum than the others is the restricting reagent.

(a)

Hence, Cr is the limiting reagent.

(b)

Now 1 mole of Cr can create 2 Faraday = 96500*2 C = 193000 C

So 0.481 mol of Cr can create (193000*0.481) C = 92833 C

At the rate of 0.0598 A current, the time required for released = 92833/0.0598 s

= 1552391.304 s

(c)

Clearly, 0.481 mole of Pb will be created when the cell is totally released.

Mass of that measure of Pb = (0.481*207.2) g

= 99.6632 g

(d)

The overabundance mole number of Pb2+ in the arrangement when the cell is totally released = (0.665 - 0.481) mol

= 0.184 mol

In this way then the conc. of Pb2+(aq) will be (0.184*1000/665) M

= 0.28 M