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Kaplan Faraday's problems.

Started by GRod18
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GRod18

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Faraday's constant problem from Kaplan Topicals
Can anyone explain this problem a little better than what Kaplan gave?
It says to use Q=It, but I got pretty lost on the dimensional analysis for this problem.

Two electrolytic cells are set up in series and
attached to a power source as shown in Figure 1. The
power source produces 2.0 A at 0.5 V. Electrons are
flowing as indicated by the arrows. Data from a table of
reduction potentials are shown in Table 1.

Answer is A

If the current is allowed to flow for 10 minutes, how
much copper will be deposited on Electrode 1?
(1F = 96,500 C/mole)
A . 0.39 grams
B . 23.7 grams
C . 47.4 grams
D . 63.5 grams

Half-reaction E°(V)
2H+ + 2e– → H2 0.00
Cu2+ + 2e– → Cu 0.34
Na+ + e– → Na –2.71
O2 + 2H2O + 4e– → 4OH– 0.40
SO4
2– + 4H+ + 2e– → H2SO3 + H2O 0.20
 
If a power source produces 2 amps (i.e. 2 C/sec) of current, that means in 10 minutes, it will produce 2 amps x 10 minutes x 60 seconds = 1200 coloumbs of charge.

1200 C x 1 mole of e/96500 = 0.012 moles of electrons (estimating)

So 0.012 moles of electrons were moved around the circuit.

According to the half reaction, for every mole of copper produced, 2 moles of electrons are used up. So 0.006 moles of copper is deposited. MM of Cu is ~60 g/mole. That's about 0.36 grams. A comes the closest.