In order to oxidize a mixture of 1 text{ mole | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) In an acidic medium, $MnO_4^-$ acts as an oxidizing agent and is reduced to $Mn^{2+}$, gaining $5$ electrons per mole $(MnO_4^- + 8H^+ + 5e^- \rig

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) In an acidic medium, $MnO_4^-$ acts as an oxidizing agent and is reduced to $Mn^{2+}$, gaining $5$ electrons per mole $(MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O)$.We calculate the moles of electrons lost by each component:$1$) $FeC_2O_4 ightarrow Fe^{3+} + 2CO_2 + 3e^-$ ($1 	ext{ mole}$ loses $3 	ext{ moles of } e^-$).$2$) $Fe_2(C_2O_4)_3 ightarrow 2Fe^{3+} + 6CO_2 + 6e^-$ ($1 	ext{ mole}$ loses $6 	ext{ moles of } e^-$).$3$) $FeSO_4 ightarrow Fe^{3+} + SO_4^{2-} + 1e^-$ ($1 	ext{ mole}$ loses $1 	ext{ mole of } e^-$).$4$) $Fe_2(SO_4)_3$: Iron is already in the $+3$ oxidation state and sulfate is in its highest oxidation state, so no further oxidation occurs $(0 	ext{ moles of } e^-)$.Total moles of electrons lost = $3 + 6 + 1 + 0 = 10 	ext{ moles}$.Since $1 	ext{ mole}$ of $KMnO_4$ accepts $5 	ext{ moles}$ of electrons, the moles of $KMnO_4$ required = $\frac{10}{5} = 2 	ext{ moles}$.

In order to oxidize a mixture of $1 \text{ mole}$ each of $FeC_2O_4$, $Fe_2(C_2O_4)_3$, $FeSO_4$, and $Fe_2(SO_4)_3$ in an acidic medium, the number of moles of $KMnO_4$ required is:

Similar Questions

The number of moles of $KMnO_4$ reduced by $1 \ mol$ of $KI$ in basic medium is .............

Assign $A, B, C, D$ from the given type of reaction.
$Ba(OH)_2 + CO_2 \longrightarrow BaCO_3 \downarrow + H_2O$

If salt $Q$ undergoes a redox reaction with $H_2S$ in an acidic medium,then which of the following species cannot be a possible product?

Which of the following reactions is not feasible?

Vedclass Test Series

Exam Paper Generator

Online Exam Module