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skelet666 [1.2K]
3 years ago
12

Consider four different samples: aqueous LiBr , molten LiBr , aqueous AgBr , and molten AgBr . Current run through each sample p

roduces one of the following products at the cathode: solid lithium, solid silver, or hydrogen gas. Match each sample to its cathodic product.Math following to Cathoid Products-Solid Lithium, Solid Silver, and Hydrogen GasSamples1. Aqueous LiBr2. Molten LiBr3. Aqueous AgBr4. Molten AgBr
Chemistry
1 answer:
Charra [1.4K]3 years ago
6 0

Answer:

a) Aqueous LiBr = Hydrogen Gas

b) Aqueous AgBr = solid Ag

c) Molten LiBr = solid Li

c) Molten AgBr = Solid Ag

Explanation:

a) Aqueous LiBr

This sample produces Hydrogen gas, because the H+ (conteined in the water) has a reduction potential higher than the Li+ from the salt. Therefore the hydrogen cation will reduce instead of the lithium one and form the gas.

b) Aqueous AgBr

This sample produces Solid Ag, because the Ag+ has a reduction potential higher than the H+ from the water. Therefore the silver cation will reduce instead of the hydrogen one and form the solid.

c) Molten LiBr

In a molten binary salt like LiBr there is only one cation present in the cathod. In this case the Li+, so it will reduce and form solid Li.  

c) Molten AgBr

The same as the item above: there is only one cation present in the cathod. In this case the Ag+, so it will reduce and form solid Ag.  

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<u>Answer:</u> The value of \Delta G^o of the reaction is 28.38 kJ/mol

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For the given chemical reaction:

SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)

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\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]

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\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]

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\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}

where,

\Delta H^o_{rxn} = standard enthalpy change of the reaction =-67200 J/mol

\Delta S^o_{rxn} = standard entropy change of the reaction =-159.3 J/Kmol

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Putting values in above equation, we get:

\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol

Hence, the value of \Delta G^o of the reaction is 28.38 kJ/mol

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