Question

A galvanic cell Cr(s)/Cr2+(aq) || Sna+(aq)|Sn(s) is constructed using a completely immersed Cr electrode that weighs 35.5 g and a Sn electrode immersed in 535 mL of 1.00 M Sn2+(aq) solution. A steady current of 0.0692 A is drawn from the cell as the electrons move from the Cr electrode to the Sn 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 Sn electrode gained when the cell is completely discharged? (d) What is the concentration of the Sn2+(aq) when the cell is completely discharged? (Assume that the limiting reactant is 100% reacted.) M

Answer 1

Cr(s)|Cr⁺²(aq) || Sn⁺²(aq)|Sn(s)

the half cell reaction in the cell will be

at Anode (oxidation)

Cr --> Cr⁺² +2e^-

at cathode (reduction)

Sn⁺² + 2e^- --> Sn

overall cell reaction

Cr + Sn⁺² --> Cr⁺² + Sn

(a)

From the overall cell reaction we see that 1 mole of Sn⁺² ions are reduced by 1 moles of Sn

we will calculate the number of moles of Cr and Sn⁺² using given information to find the limiting reagent.

Mass of Cr electrode = 35.5 g

Molar mass of Cr = 51.9961 g/mol

No of moles of Cr in Cr electrode = 35.5 / 51.9961 g.mol^-1

= 0.6827 moles

concentration of Sn⁺² = 1.00 M

this means 1.00 moles of Sn⁺² are present in 1000 mL of solution.

No of moles of Sn⁺² ions present in 535 mL of 1.00 M solution(1 mole / 1000 mL) * 535 moles

=0.535 moles

we know that 1 mole of Cr reacts with 1 mole of Sn⁺²

as the no. of moles of Sn⁺² are lesser than the required no. of moles to completely react with available Cr the Limiting reagent is Sn⁺²

(b)

As Sn⁺² is limiting reagent , the number of moles of Cr that will undergo oxidation will be 0.535 moles

(oxidation) reaction of Cr is

Cr --> Cr⁺² +2e^-

i mole of Cr produce 2 moles of electrons

therefore No. of moles of electrons produced by 0.535 moles of Cr = 2 * 0.535 = 1.07 moles

charge of 1 mole of electrons =96485 C mol^-1

charge of 1.07 moles of electrons = 1.07 * 96485 = 103238.95 C

therefore the charge produced by galvanic cell , Q = 103238.95 C

current drawn from the cell I = 0.0692 A

using Q = It

t= Q/I

= 103238.95 C / 0.0692 A

= 1491892.341 sec

time required for cell to get completely discharged = 1491892.341 sec

(c)

From the reduction reaction of Sn⁺² we know that each mole of Sn⁺² deposits 1 mole of Sn.

Sn⁺² + 2e^- ---> Sn

therefore 0.535 moles of Sn⁺² will deposit 0.535 moles of Sn

molar mass of Sn = 118.71 g.mol^-1

mass of 0.535 moles of Sn = 0.535 * 118.71

= 63.51 g

mass gained by Sn electrode when cell is completely discharged = 63.51 g

(d)

As Sn⁺² is the limiting reagent, it will get completely consumed as the cell gets completely discharged

therefore the concentration of the Sn⁺²(aq) when the cell is completely discharged = 0 M

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