Question

If $E^{o}{ }_{M^{+}}^{+} / M=-1.2\, V,\, E^{o}{ }_{X} /{M}^{-}=-1.1\, V$ and $E^{o}{ }_{O_{2} / H_{2} O}=1.23\, V$, then on electrolysis of aqueous solution of salt $M X$, the products obtained are

• A. $ M,X_{2}$
• B. $H_{2},X_{2}$
• C. $H_{2},O_{2}$
• D. $ M,O_{2}$

Answer 1

Answer: (B)

Given, $E^{o}{ }_{M^{+} / M}=-1.2\, V$
$E^{o}{ }_{x_{2} / x^{-}}=-1.1\, V$
$E^{o}{ }_{O_{2} / H_{2} O}=1.23\, V$
On electrolysis of aqueous solution of salt $MX$.
The products obtained depends on the given
$E ^{ o }$ values i.e. cathodic reaction will be one with higher $E ^{ o }$ red value. Anodic reaction will be one with higher $E ^{ o }{ }_{\text {oxi }}$ value or lower $E ^{ o }{ }_{\text {red }}$ value.
Hence, the products are $H_{2}$ and $X_{2}$.