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BigorU [14]
3 years ago
5

Which in bigger 43 mg or 5 g

Chemistry
2 answers:
AleksandrR [38]3 years ago
6 0
5 g is bigger than 43 mg.

I hope this is right, I apologize if I'm wrong.
strojnjashka [21]3 years ago
5 0
5g is bigger. 43 mg is only 0.043g.
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Solution :

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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}

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

n = number of electrons = 2

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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

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Explanation:

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