Question

2 MnO4-(aq) + 10 Br-(aq) + 16 H+(aq) → 2 Mn2+(aq) + 5 Br2(aq) + 8 H2O(l). How many electrons are transferred in the reaction represented by the balanced equation above?

Answer 1

Answer : The number of electrons transferred are, 10

Explanation :

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given balanced redox reaction is,

$2MnO_4^-(aq)+10Br^-{aq)+16H^+(aq)\rightarrow 2Mn^{2+}(aq)+5Br_2(aq)+8H_2O(l)$

Step 1: Separate the skeleton equation into two half-reactions.

Oxidation : $Br^-\rightarrow Br_2$

Reduction : $MnO_4^-\rightarrow Mn^{2+}$

Step 2: Balance all atoms other than H and O.

Oxidation : $2Br^-\rightarrow Br_2$

Reduction : $MnO_4^-\rightarrow Mn^{2+}$

Step 3: Balance O.

Oxidation : $2Br^-\rightarrow Br_2$

Reduction : $MnO_4^-\rightarrow Mn^{2+}+4H_2O$

Step 4: Balance H.

Oxidation : $2Br^-\rightarrow Br_2$

Reduction : $MnO_4^-+8H^+\rightarrow Mn^{2+}+4H_2O$

Step 5: Balance the charge.

Oxidation : $2Br^-\rightarrow Br_2+2e^-$

Reduction : $MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O$

Step 6: Equalize electrons transferred.

Oxidation : $2Br^-\rightarrow Br_2+2e^-$    × 5

Reduction : $MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O$   × 2

and,

Oxidation : $10Br^-\rightarrow 5Br_2+10e^-$

Reduction : $2MnO_4^-+16H^++10e^-\rightarrow 2Mn^{2+}+8H_2O$

Step 7: Add the two half-reactions.

$2MnO_4^-(aq)+16H^+(aq)+10Br^-(aq)\rightarrow 2Mn^{2+}(aq)+8H_2O(l)+5Br_2(aq)$

In this reaction, there are 10 number of electrons transferred.

Hence, the number of electrons transferred are, 10

Answer 2

The total number of electrons transferred in the reaction is 10.

The number of electrons transferred can be given by half reactions:

Oxidation reaction: $\rm Br^-\rightarrow\;Br_2$

Reduction reaction : $\rm MnO_4^-\;\rightarrow\;Mn^2^+$

The transfer of electrons can be balanced with the addition of a water molecule to the reaction. If the hydrogen atoms are not balanced on both sides then add hydrogen ion at that side where the less number of hydrogen is present. The electron transfer will be:

Oxidation reaction : $\rm 2\;Br^-\;\rightarrow\;Br_2\;+\;2\;e^-$

Reduction reaction : $\rm MnO_4^-\;+\;H^+\;+\;5\;e^-\;\rightarrow\;Mn^2^+\;+\;2\;H_2O$.

By balancing the equation and electron transfer:

$\rm 2\;MnO_4^-\;+\;16\;H^+\;10\;Br^-\;\rightarrow\;2\;Mn^2^+\;8\;H_2O\;+\;5\;Br_2$

The total number of electrons transferred in the reaction is 10.

For more information about the electron transfer, refer to the link:

https://brainly.com/question/1160226