Question

$M+NS{O}_4\to MS{O}_4+N$
$N+PC{I}_2\to NC{l}_2+P$
Compare the electrode potentials of M, N and P.

• A. $P>N>M$
• B. $M>N>P$
• C. $N>P>M$
• D. $M>P>N$

Answer 1

Answer: (B)

$M>N>P$

$M$ displaces $N$ from $NS{O}_4$. $M$ is more active metal than $N$. This means the oxidation potential of $M$ is greater than $N$. In the second reaction, N displaces $P$ showing that $N$ has greater oxidation potential than $P$. Hence, the oxidation electrode potentials are as follows:

$M>N>P$

The reduction potential order would be $P>N>M$.