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# cell emf and free energy

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### #1HeatherL

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Posted 31 May 2009 - 10:07 AM

Hey!!
Can someone explain to me why you dont multiply the answer for standard free energy change for the cell by six in this example?

A lead-acid storage cell has a standard cell potential of 1.924 V. The anode reaction is:

Pb(s) Pb2+(aq) + 2e-

Six such cells are connected in series in a car battery.

Also, in the G=-nFE equation is n = the number of electrons released on the oxidation of the substance at the anode?

Thanks,
Heather

### #2Marcus

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Posted 31 May 2009 - 05:14 PM

Because the free energy change you calculate is for one mole of Pb being oxidised, it is irrelevant where, and in what quantities this oxidation takes place.

Even if one mole reacts in each cell, then if you sum the free energies, you would get the free energy for six moles, but you only want one so you would still have to divide by six leaving you right back where you started...if you get my drift

post back if you still don't understand and I'll try and explain this better :S
=-=-=Marcus=-=-=

### #3ginneswatson

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Posted 01 June 2009 - 09:28 AM

QUOTE (HeatherL @ May 31 2009, 11:07 AM) <{POST_SNAPBACK}>
Hey!!
Can someone explain to me why you dont multiply the answer for standard free energy change for the cell by six in this example?

A lead-acid storage cell has a standard cell potential of 1.924 V. The anode reaction is:

Pb(s) Pb2+(aq) + 2e-

Six such cells are connected in series in a car battery.

Also, in the G=-nFE equation is n = the number of electrons released on the oxidation of the substance at the anode?

Thanks,
Heather

"Also, in the G=-nFE equation is n = the number of electrons released on the oxidation of the substance at the anode?"

Yes, if the oxidation reaction is the only one you care about, n=2 and your G is in terms of one mole of Pb.

Confusion can result when one reactant loses/gains 2 electrons while the other reactant maybe only loses/gains 1 electron. Normally you'd balance the equations by multiplying, eg

Pb to Pb2+ + 2e-

and 2Ag+ to 2Ag + 2e-

Unusually, Eo values are one of the few values that are not mole dependent so doubling the Ag equation makes no difference to the overall Eo of the reaction and, hence, the G value calculated.

However, this is where things can get confusing. The G calculated will be in kJ mol-1.

But per mole of what? In this case it can only be per mole of Pb.

To express the G in terms of per mole of Ag, you would need to half the G value.

To spare you this problem, the SQA exam will 'conveniently' stick to reactions where both reactants are losing/gaining the same number of electrons.

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