In: Chemistry
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)
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