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Tamiku [17]
3 years ago
5

Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu

2+(aq) E°(Pb2+/Pb) = -0.13 V and E°(Cu2+/Cu) = +0.34 V -41 kJ mol-1 +91 kJ mol-1 -21 kJ mol-1 -0.47 kJ mol-1 +46 kJ mol-1
Chemistry
1 answer:
nydimaria [60]3 years ago
6 0

Answer : The correct option is, +91 kJ/mole

Solution :

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^0_{[Pb^{2+}/Pb]}=-0.13V

E^0_{[Cu^{2+}/Cu]}=+0.34V

E^0_{cell}=E^0_{cathode}-E^0_{anode}

E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}

E^0_{cell}=-0.13V-(0.34V)=-0.47V

Now we have to calculate the standard Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = -0.47 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole

Therefore, the standard Gibbs free energy is +91 kJ/mole

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Calculate the mass of excess reagent remaining at the end of the reaction in which 90.0 g of SO2 are mixed with 100.0 g of O2
Radda [10]
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this is not balanced though. we have 3 oxygen on right and 4 on left
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now it is same on both sides. we have to figure out which is limiting reagent with the given amounts of reagents. we do this by comparing the ratio between them in terms of moles. we see that so2 has a coefficient of 2 and o2 has none which implies 1 and so3 has 2. this means that for every 2 moles of so2 reacting with 1 mole of o2, we get 2 moles of so3.

lets convert the given values to moles. to do this we know that molecular weight is measured in grams per mole. we are given grams and need to cancel out the grams to get moles. so the molecular weight:
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convert this amount of moles of so2 to moles of o2. we have 2 moles of so2 to 1 of o2
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so we see under ideal conditions that 90g of so2 would react with .702g of o2. lets see how many we actually have with 100g of o2
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so we have a lot more o2 than needed. we are looking for how much is left in grams. we have to figure out how much was used. to do this convert our ideal moles of o2 into grams.
.702 moles o2 * 32g/mol = 22.5g o2

so what we startrd with (100g) minus what we needed (22.5g) is what we have left
100 - 22.5 = 77.5g o2
6 0
3 years ago
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