Calculate the value of n in the given reactio | vedclass

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

Question

(C) In the given half-reaction,the manganese atom undergoes reduction.The oxidation state of $Mn$ in $MnO_4^{-}$ is calculated as: $x + 4(-2) = -1 \im

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) In the given half-reaction,the manganese atom undergoes reduction.The oxidation state of $Mn$ in $MnO_4^{-}$ is calculated as: $x + 4(-2) = -1 \implies x = +7$.The oxidation state of $Mn$ in $Mn^{+2}$ is $+2$.The change in oxidation state is $|7 - 2| = 5$.Since the reduction involves a gain of electrons,the number of electrons $n$ required to balance the charge and oxidation state is $5$.Therefore,$n = 5$.

Calculate the value of $n$ in the given reaction:
$MnO_4^{-} + 8H^{+} + n e^{-} \to Mn^{+2} + 4H_2O$

Similar Questions

Explain intermolecular redox reaction with an example.

Permanganate $(VII)$ ion,$MnO_4^-$,in basic solution oxidises iodide ion,$I^-$,to produce molecular iodine $(I_2)$ and manganese $(IV)$ oxide $(MnO_2)$. Write a balanced ionic equation to represent this redox reaction.

When thiosulphate ion is oxidised by iodine,which one of the following ions is produced?

How many moles of electrons are needed to convert one mole of nitrate ion to hydrazine?

In the reaction $2CuSO_4 + 4KI \rightarrow Cu_2I_2 + 2K_2SO_4 + I_2$,what is the ratio of the equivalent weight of $CuSO_4$ to its molecular weight?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Calculate the value of $n$ in the given reaction:$MnO_4^{-} + 8H^{+} + n e^{-} \to Mn^{+2} + 4H_2O$