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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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Answer:

The value of the equilibrium constant KC is 1.244

Explanation:

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.550 M, [B] = 1.40 M, and [C] = 0.600 M. The following reaction occurs and equilibrium is established: A+2B<->C

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Step 1: The balanced equation

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Step 3: The concentraions at equilibrium

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