Question

A galvanic cell Fe(s)|Fe²⁺(aq) || Pb²⁺(aq)|Pb(s) is constructed using a completely immersed Fe electrode that weighs 35.6 g and a Pbelectrode immersed in 510 mL of 1.00 M Pb²⁺(aq) solution. A steady current of 0.0672 A is drawn from the cell as the electrons move from the Fe electrode to the Pb electrode.

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

The given cell is

Fe(s)|Fe²⁺(aq) || Pb²⁺(aq)| Pb(s)

The half cell reactions in the cell will be

At Anode (oxidation)

Fe → Fe²⁺ +2e^-

At Cathode (reduction)

Pb²⁺ +2e^-→ Pb

Overall cell reaction

Fe + Pb²⁺ → Fe²⁺ + Pb

(a) From the overall cell reaction, we see that 1 mole of Pb²⁺ions are reduced by 1 mole of Fe.

We will calculate the number of moles of Fe and Pb²⁺ using given information to find the limiting reagent.

Mass of Fe electrode = 35.6 g

Molar mass of Fe=55.845 g /mol

No. of moles of Fe in Fe electrode = 35.6 g / 55.845 g mol^-1

=0.6375 moles

Concentration of Pb²⁺ = 1.00 M

This means 1.00 mole of Pb²⁺ ions are present in 1L (1000 mL ) of solution.

No. of moles of Pb²⁺ ions present in 510 mL of 1.00 M solution =( 1 mole/ 1000 mL) * 510 mL

= 0.510 moles

We know that 1 mole of Fe reacts with 1 mole of Pb²⁺.

Therefore, no. of moles of Pb²⁺ required to react with 0.6375 moles of Fe = 0.6375 moles

No. of moles of Pb²⁺ available =0.510 moles

As the no. of moles of Pb²⁺ are lesser than the required no. of moles to completely react with available Fe, the limiting reagent is Pb²⁺.

(b) As Pb²⁺ is the limiting reagent, it will get completely consumed during the reaction.

We know that 1 mole of Fe reacts with 1 mole of Pb²⁺.

The number of moles of Fe that will undergo oxidation will be equal to number of moles of Pb²⁺ reduced, that is  0.510 moles.

Oxidation reaction of Fe is

Fe → Fe²⁺ +2e^-

1 mole of Fe produce 2 moles of electrons.

Therefore, No. of moles of electrons produced by 0.510 moles of Fe=2* 0.510 moles = 1.02 moles

Charge on 1 moles of electrons = 96485 C mol^-1

Charge on 1.02 moles of electrons = 1.02 moles *96485 C mol^-1

= 98,414.7 C

Therefore, the charge produced by galvanic cell, Q = 98,414.57 C

Current drawn from the cell, I = 0.0672 A

Using , Q= It

t= Q/I

= 98,414.57 C / 0.0672 A

=1,464,504.464 s

Time required for cell to get completely discharged = 1,464,504.464 s

(c) From the reduction reaction of Pb²⁺ , we know that each mole of Pb²⁺  deposits 1 mole of Pb.

Pb²⁺  +2e^-→ Pb

Therefore, 0.510 moles of Pb²⁺  will deposit 0.510 moles of Pb.

Molar mass of Pb = 207.2 g mol^-1

Mass of 0.510 moles of Pb =0.510 moles * 207.2 g mol^-1

= 105.672 g

Mass gained by Pb electrode when cell is completely discharged =105.672 g

(d) As Pb²⁺ is the limiting reagent, it will get completely consumed as the cell gets completely discharged.

Therefore, the concentration of the Pb²⁺ (aq) when the cell is completely discharged = 0 M