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Whitepunk [10]
3 years ago
8

Metals posses varying reduction potentials. Iron can reduce Cu+2 (aq) to copper metal. Silver can reduce Au+ (aq) to gold metal.

Sodium can reduce Fe3+(aq) to iron metal and copper can reduce Ag+(aq) to silver metal. Which of the following statements is true?copper can reduce Na+ (aq) to sodium metalIron can reduce Au+(aq) to gold metalsilver can reduce Fe3+(aq) to ironnone of the above statements is true
Chemistry
1 answer:
natta225 [31]3 years ago
7 0

<u>Answer:</u> The true statement is iron can reduce Au^+(aq) to gold metal

<u>Explanation:</u>

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

A+BC\rightarrow AC+B

Metal A is more reactive than metal B.

We are given:

Iron can reduce copper, silver can reduce gold, sodium can reduce iron and copper can reduce silver metal.

The increasing order of reactivity thus follows:

Au

where, sodium is most reactive and gold is least reactive

For the given options:

<u>Option 1:</u>  Copper cannot easily reduce sodium ion to sodium metal because it is less reactive.

Cu(s)+Na^+(aq.)\rightarrow \text{ No reaction}

<u>Option 2:</u>  Iron cant easily reduce gold ion to gold metal because it is more reactive.

Fe(s)+3Au^+(aq.)\rightarrow Fe^{3+}(aq.)+3Au(s)

<u>Option 3:</u>  Silver cannot easily reduce iron ion to iron metal because it is less reactive.

Ag(s)+Fe^{3+}(aq.)\rightarrow \text{ No reaction}

Hence, the true statement is iron can reduce Au^+(aq) to gold metal

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weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

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Answer:

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Molecular\ weight = \frac{Mass\ in\ g}{Moles}

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