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liberstina [14]
3 years ago
6

Cr + 3fe3+cr3+ + 3fe2+ in the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced,

the oxidizing agent and the reducing agent.
Chemistry
1 answer:
anastassius [24]3 years ago
8 0
Element Cr goes from Cr --> Cr³⁺
element Fe goes from Fe³⁺ --> Fe²⁺
oxidation reactions are when the species gives out electrons and its oxidation number increases.
reduction reactions are when the species take in electrons and its oxidation number reduces.
Cr : oxidation number of Cr - 0 and Cr³⁺ - +3
Theres an increase in oxidation number from 0 to +3. Therefore its an oxidation reaction. Cr gets oxidised. 
Fe : oxidation number of Fe³⁺ - +3 and Fe²⁺ - +2
There's a decrease in oxidation number of Fe from +3 to +2, therefore its a reduction reaction and Fe gets reduced.
         
Cr reduces Fe - Cr is the reducing agent 
Fe oxidises Cr - Fe is the oxidising agent
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Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and
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Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

Explanation :

The given balanced chemical reaction is,

C_6H_6(l)\rightarrow C_6H_6(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0_{(C_6H_6(g))} = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

\Delta H_f^0_{(C_6H_6(l))} = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]

\Delta H^o=33.89kJ/mol=33890J/mol

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0_{(C_6H_6(g))} = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

\Delta S^0_{(C_6H_6(l))} = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]

\Delta S^o=95.94J/K.mol

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 25^oC\text{ or }298K.

\Delta G^o=(33890J)-(298K\times 95.94J/K)

\Delta G^o=5299.88J/mol=5.299kJ/mol

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

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Remember, OIL RIG 
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Reduction Is Gain of electrons 

If something goes from + to - then it has gained electrons 
If something goes from neutral to - it has gained electrons 
If it goes from - to + is loses electrons 
If it goes from neutral to + it loses electrons 

If something is oxidised, it is a reducing agent 
If something is reduced, it is an oxidising agent 


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
6 0
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