A galvanic cell consists of a lead cathode immersed in a PbSO4 solution and a co

Question

A galvanic cell consists of a lead cathode immersed in a PbSO4 solution and a cobalt anode immersed in a CoS04 solution. A salt bridge connects the two half-cells. (a) Write a balanced cquation for the cell reaction. (b) A current of 1.34 A is observed to flow for a period of 1.56 hours. How much charge passes through the circuit during this time? How many moles of electrons is this charge equivalent to? mol (c) Calculate the change in mass of the lead electrode. g is (d) Calculate the change in mass of the cobalt electrode. lost Submit Answer gained 5 question attempts remaining

Explanation / Answer

a.)PbSO4(aq) + Co(s) ---> Pb(s) + CoSO4(aq) It is because the cobalt is the anode where oxidation or corrosion takes place and lead is the cathode where reduction or deposition takes place.

b.)Columb of charge transfered= current flown in 1 second * time in time

charge = 1.34*1.56*60*60

charge = 7525.44 C

Moles of electron transferred = charge transfered/charge transfered by 1 mol of electron

= 7525.44/96500 =0.0779838342 moles

c.) mass of lead gained on cathode as reduction takes place in cathode.

amount deposited => 19300 C = 1 mol

=>7525.44C = 0.0389919171 moles

1 mol lead = 207.2 g

lead deposited = 0.0389919171 moles * 207.2 g moles-1 = 8.0791 g

d.)

mass of cobalt lost  as oxidation takes place in anode

amount lost => 19300 C = 1 mol

=>7525.44C = 0.0389919171 moles

1 mol Cobalt = 59 g

lead deposited = 0.0389919171 moles * 59 g moles-1 = 2.3005 g

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