Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficients (integers) into the cells before each substance below the equation. ___ AsO2−(aq) + 3 Mn2+(aq) + ___ H2O(l) \rightarrow→ ___ As(s) + ___ MnO4−(aq) + ___ H+(aq)

The balanced redox reaction is

5AsO⁻₂(aq) + 3Mn²⁺(aq) + 2H₂O(l) → 5As(s) + 3MnO₄−(aq) + 4H+(aq)

The unbalanced redox reaction is :

AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

The above equation can be balanced  by ensuring the atom of the elements

on the left hand side is equal to those on the right hand side.

We have 3 atoms of Mn on the left side of the equation and 1 atom on the

right hand side.This will be balanced by putting 3 in front of MnO₄− as shown

below:

AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

We have 3 atoms of O on the left hand side and 12 atoms of O on the right

hand side. This is  balanced by putting 5 in front of AsO₂− and 2 in front of

H₂O as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)

We have 4 atoms of H on the left hand side and 1 atom of H on the right

hand side.We can balance by putting 4 in front of H⁺ as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + 4H⁺(aq)

We have 5 atoms of As on the left hand side and 1 atom of As on the right

hand side. We can balance by putting 5 in front of As as shown below:

5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → 5As(s) + MnO₄−(aq) + 4H⁺(aq)

The equation is therefore now balanced as the number of atoms of the

element on the left hand side are equal with those on the right hand side.

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_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.