Question

A few elements in order of decreasing reactivity are: $\mathrm{K}>\mathrm{Ca}>\mathrm{Mg}>\mathrm{Fe}>\mathrm{H}>\mathrm{Sb}>\mathrm{Au}$. Identify the incorrect statement.

• A. $\mathrm{Au}$ is the weakest reducing agent.
• B. $\mathrm{K}$ is the strongest oxidising agent.
• C. The oxide of $\mathrm{Sb}$ can be reduced with hydrogen.
• D. $\mathrm{Ca}$ displaces hydrogen from water more easily than $\mathrm{Fe}$.

Answer 1

The correct option is B

$\mathrm{K}$ is the strongest oxidising agent.

Option B

Potassium $\mathrm{K}$ is the strongest oxidising agent.

Explanation of correct option:

• Oxidation means the loss of electrons in this oxidation state increases.
• Reduction means the gain of electrons in this oxidation state decreases.
• The one who gets oxidised acts as a reducing agent and the one who gets reduced acts as an oxidising agent.

$\mathrm{K}>\mathrm{Ca}>\mathrm{Mg}>\mathrm{Fe}>\mathrm{H}>\mathrm{Sb}>\mathrm{Au}$

• Here potassium is the most reactive metal.
• Higher reactive metal donates electron easily and oxidises so they are strong reducing agents.

So this statement is false. $\mathrm{K}$ is strongest reducing agent not strongest oxidising agent.

Explanation of incorrect options:

Option A

Silver $\mathrm{Au}$ is the weakest reducing agent.

• Reactivity series

$\mathrm{K}>\mathrm{Na}>\mathrm{Li}>\mathrm{Ba}>\mathrm{Sr}>\mathrm{Ca}>\mathrm{Mg}>\mathrm{Al}>\mathrm{Mn}>\mathrm{Zn}>\mathrm{Cr}>\mathrm{Fe}>\mathrm{Cd}>\mathrm{Co}>\mathrm{Ni}>\mathrm{Sn}>\mathrm{Pb}>\mathrm{H}>\mathrm{Sb}>\mathrm{Bi}>\mathrm{Cu}>\mathrm{Hg}>\mathrm{Ag}>\mathrm{Au}>\mathrm{Pt}$

• Here potassium is the most reactive metal and platinum is the least reactive metal in this series.
• Higher reactive metal donates electrons easily and oxidises they are strong reducing agents.

$\mathrm{K}>\mathrm{Ca}>\mathrm{Mg}>\mathrm{Fe}>\mathrm{H}>\mathrm{Sb}>\mathrm{Au}$

• Since $\mathrm{Au}$ reactivity is less so it will be the weakest reducing agent.
• So this statement is correct.

Option C

The oxide of antimony $\mathrm{Sb}$ can be reduced with hydrogen.

Here $\mathrm{H}$ is more reactive than $\mathrm{Sb}$ so it can displace oxide of $\mathrm{Sb}$.

• Reaction:

${\mathrm{Sb}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2\mathrm{Sb}\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Antimony(III) oxide Hydrogen Antimony Water

• Here oxidation state of $\mathrm{Sb}$ in ${\mathrm{Sb}}_2{\mathrm{O}}_3$ is $+3$ and that of $\mathrm{H}$ in ${\mathrm{H}}_2$ is $0$ at the reactant side but the oxidation state of $\mathrm{Sb}$ in $\mathrm{Sb}$ is $0$ and that of $\mathrm{H}$ in ${\mathrm{H}}_2\mathrm{O}$ is $+1$ at the product side.
• So, we can say that ${\mathrm{H}}_2$ is getting oxidised since its oxidation number is increasing from $0$ to $+1$ and ${\mathrm{Sb}}_2{\mathrm{O}}_3$ is getting reduced since its oxidation number is decreasing from $+3$ to $0$.
• The one who gets oxidised acts as a reducing agent and the one who gets reduced acts as an oxidising agent.
• So here ${\mathrm{H}}_2$ is a reducing agent and ${\mathrm{Sb}}_2{\mathrm{O}}_3$is an oxidising agent.

So this statement is correct.

Option D

calcium $\mathrm{Ca}$ displaces hydrogen from water more easily than$\mathrm{Fe}$.

$\mathrm{K}>\mathrm{Ca}>\mathrm{Mg}>\mathrm{Fe}>\mathrm{H}>\mathrm{Sb}>\mathrm{Au}$

• Here potassium is the most reactive metal.
• Higher reactive metal can displace less reactive metal.
• Here reactivity of $\mathrm{Ca}$ is greater than $\mathrm{H}$ as well as reactivity of $\mathrm{Fe}$ is also greater than $\mathrm{H}$ so both these metal can displace hydrogen from water.
• But reactivity of $\mathrm{Ca}$ is greater than that of $\mathrm{Fe}$ so $\mathrm{Ca}$ can displace $\mathrm{H}$ more easily is comparison to $\mathrm{Fe}$.

So this statement is correct.

The correct answer is Option B $\mathrm{K}$ is the strongest oxidising agent. This is incorrect statement.