Question

A galvanic cell Crs)Cr2(aq) || Pb2*(a)|Pb( is constructed using a completely immersed Cr electrode that weighs 25.0 g and a Pb electrode immersed in 632 mL of 1.00 M Pb*(aq) solution. A steady current of 0.0719 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? Enter symbol (b) How long does it take for the cell to be completely discharged? (c) How much mass has the Pb clectrode gained when the cell is completely discharged? (d) What is the concentration of the Pb2*(aq) when the cell is completely discharged? (Assume that the limiting reactant is 100% reacted.)

Answer 1

The cell reaction of the galvanic cell is:

Cr(s) + Pb²⁺(aq) = Cr²⁺(aq) + Pb(s)

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

Now, number of moles of Cr(s) present in the cell = (25.0/52.0) mol = 0.481 mol [ MW of Cr = 52.0 g mol^-1]

Again, the number of moles of Pb²⁺ present in the solution = (1.00*632/1000) mol = 0.632 mol

We know the species present in less amount than the others is the limiting reagent.

(a) Therefore, Cr is the limiting reagent.

(b) Now 1 mole of Cr can produce 2 Faraday= 96500*2 C = 193000 C

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

At the rate of 0.0719 A current, the time required for discharged = 92833/0.0719 s = 1291140.47 s

(c) Clearly, 0.481 mole of Pb will be produced when the cell is completely discharged.

Mass of that amount of Pb = (0.481*207.2) g = 99.6632 g

(d) The excess mole number of Pb²⁺ in the solution when the cell is completely discharged = (0.632-0.481) mol = 0.151 mol

Therefore then the conc. of Pb²⁺(aq) will be (0.151*1000/632) M = 0.24 M