Question

In: Chemistry

A galvanic cell consists of a chromium anode immersed in a CrSO4 solution and a lead...

A galvanic cell consists of a chromium anode immersed in a CrSO4 solution and a lead cathode immersed in a PbSO4 solution. A salt bridge connects the two half-cells.

(a) Write a balanced equation for the cell reaction.

-------- (aq) or (s) or (l) or (g) + -------- (aq) or (s) or (l) or (g) = --------- (aq) or (s) or (l) or (g) + ------- (aq) or (s) or (l) or (g)



(b) A current of 1.16 A is observed to flow for a period of 1.68 hours. How much charge passes through the circuit during this time? How many moles of electrons is this charge equivalent to?

------- C
------- mol



(c) Calculate the change in mass of the chromium electrode. ---------- g is _______ (lost or gained)

(d) Calculate the change in mass of the lead electrode. ---------- g is _______ (lost or gained)

Solutions

Expert Solution

a)

CrSO4 + PbSO4; therefore, we have Cr+2 and Pb+2 ions

Cr2+(aq) + 2e– → Cr(s) –0.91

Pb2+(aq) + 2e– → Pb(s) –0.13

therefore, the Pb wil form Pb(s) since it has a higher reduction potential, the solid Cr(s) from the anode will form Cr+2

The reaction that occurs:

Pb+2(aq) + Cr(s) = Cr+2(aQ) + Pb(s)

B)

I = 1.16 A = 1.16 C/s

t = 1.68 h = 1.68*3600 = 6048 s

now, calcualte total charge

C = I*t = 1.16*6048 = 7,015.68 C

now, we also require moles of e-

so, according to faraday,

1 mol of e- = 96500 C approx

x mol of e- = 7,015.68 C

x = 7,015.68/96500 = 0.0727 moles of e-

C)

change in mass of Cr:

0.0727 mol of e- = 0.0727/2 = 0.03635 mol of Cr will react

mass of Cr = 0.03635*51.99610 = 1.8900 g

this is LOST

for the lead:

now, similar for Pb, Pb(s)

mass of Pb = 0.03635*207.2 = 7.53172 g of Pb

now, we know this is produced so

mass of Pb INCREASES/GAINS


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