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3 years ago

Please help me! I don't understand this at all. All the info is in the picture. Thank you so much!!!

Chemistry

1 answer:

kykrilka [37]3 years ago

3 0

Answer:

H₂S + Cl₂ —> S + 2HCl

Explanation:

? + Cl₂ —> S + 2HCl

To balance the equation above, we must recognise what atoms are present in the products.

The products contains S, H and Cl.

Thus, S, H and Cl must also be present in the reactants.

Considering the equation given above, we can see clearly that H and S is missing in the reactants.

H and S together as a compound is expressed as H₂S.

Now, we shall input H₂S into the equation to obtain the complete equation. This is illustrated below:

? + Cl₂ —> S + 2HCl

H₂S + Cl₂ —> S + 2HCl

Next, we shall verify to see if the equation is balanced.

There are 2 atoms of H on both sides of the equation.

There are 2 atoms of Cl on both sides of the equation.

1 atom of S exist on both sides of the equation.

Thus, the equation is balanced.

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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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