I. FeCr_2O_4 + NaOH + text{air} rightarrow (A | vedclass

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(A) $I$. $4FeCr_2O_4 + 16NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8H_2O$. Here,$(A)$ is $Na_2CrO_4$.
$II$. $2Na_2CrO_4 + 2H^+ \rightarrow Na_2

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$Al$

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(A) $I$. $4FeCr_2O_4 + 16NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8H_2O$. Here,$(A)$ is $Na_2CrO_4$.
$II$. $2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$. Here,$(B)$ is $H^+$.
$III$. $Na_2Cr_2O_7 + 2C \rightarrow Cr_2O_3 + Na_2CO_3 + CO$. Here,$X$ is $C$.
$IV$. $Cr_2O_3 + 2Al \rightarrow 2Cr + Al_2O_3$. Here,$Y$ is $Al$.
For highly reactive metals like $Al$ (Aluminum),which are at the top of the electrochemical series,electrolytic reduction is required at high temperatures (Hall-Heroult process) because they cannot be reduced by common reducing agents like $C$ or $CO$.

$I$. $FeCr_2O_4 + NaOH + \text{air} \rightarrow (A) + Fe_2O_3$
$II$. $(A) + (B) \rightarrow Na_2Cr_2O_7$
$III$. $Na_2Cr_2O_7 + X \xrightarrow{\Delta} Cr_2O_3$
$IV$. $Cr_2O_3 + Y \xrightarrow{\Delta} Cr$
High temperature $(> 1000 \ ^\circ C)$ electrolytic reduction is necessary for isolating:

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$I$. $FeCr_2O_4 + NaOH + \text{air} \rightarrow (A) + Fe_2O_3$ $II$. $(A) + (B) \rightarrow Na_2Cr_2O_7$ $III$. $Na_2Cr_2O_7 + X \xrightarrow{\Delta} Cr_2O_3$ $IV$. $Cr_2O_3 + Y \xrightarrow{\Delta} Cr$ High temperature $(> 1000 \ ^\circ C)$ electrolytic reduction is necessary for isolating: