Question

Given the partial equation: MnO4−+ SO32− → Mn2++ SO42−, balance the reaction in acidic solution using the half-reaction method and fill in the coefficients. The missing blanks represent H2O, H+, or OH-, as required to balance the reaction. Enter the coefficients as integers, using the lowest whole numbers. If the coefficient for something is "1", make sure to type that in and not leave it blank. Enter only the coefficients.

Answer 1

`5SO_3^(2-)+2MnO_4^(-)+6H^+ \rightarrow 5SO_4^(2-)+ 2Mn^(2+)+3H_2O`

Step-by-step explanation:

Hello,

In this case, given the reaction:

`MnO_4^(-)+ SO_3^(2-) \rightarrow Mn^(2+)+ SO_4^(2-)`

We first identify the oxidation state of both manganese and sulfur at each side:

`Mn^(7+)O_4^(-)+ S^(4+)O_3^(2-) \rightarrow Mn^(2+)+ S^(6+)O_4^(2-)`

So we have the oxidation and reduction half-reactions below, including the addition of water and hydronium as it is in acidic media:

`S^(4+)O_3^(2-)+H_2O \rightarrow S^(6+)O_4^(2-)+2H^++2e^-`

`Mn^(7+)O_4^(-)+8H^++5e^- \rightarrow Mn^(2+)+4H_2O`

Next, we exchange the transferred electrons:

`5*(S^(4+)O_3^(2-)+H_2O \rightarrow S^(6+)O_4^(2-)+2H^++2e^-)\\2*(Mn^(7+)O_4^(-)+8H^++5e^- \rightarrow Mn^(2+)+4H_2O)\\\\5S^(4+)O_3^(2-)+5H_2O \rightarrow 5S^(6+)O_4^(2-)+10H^++10e^-\\2Mn^(7+)O_4^(-)+16H^++10e^- \rightarrow 2Mn^(2+)+8H_2O`

Then we add the resulting half-reactions and simplify the transferred electrons:

`5S^(4+)O_3^(2-)+5H_2O+2Mn^(7+)O_4^(-)+16H^+ \rightarrow 5S^(6+)O_4^(2-)+10H^++ 2Mn^(2+)+8H_2O`

We rearrange the terms in order to simplify water and hydronium molecules:

`5S^(4+)O_3^(2-)+2Mn^(7+)O_4^(-)+16H^+-10H^+ \rightarrow 5S^(6+)O_4^(2-)+ 2Mn^(2+)+8H_2O-5H_2O\\\\5S^(4+)O_3^(2-)+2Mn^(7+)O_4^(-)+6H^+ \rightarrow 5S^(6+)O_4^(2-)+ 2Mn^(2+)+3H_2O`

Finally we write the balanced reaction in acidic media:

`5SO_3^(2-)+2MnO_4^(-)+6H^+ \rightarrow 5SO_4^(2-)+ 2Mn^(2+)+3H_2O`

Best regards.

Answer 2

Step-by-step explanation:

MnO4−+ SO32− → Mn2++ SO42−

Splitting into half equations;

MnO4− → Mn2+

SO32− → SO42−

Balancing the electrons;

2 MnO4− + 10 e- → 2Mn2+

5SO32− → 5SO42− + 10 e-

In an acidic medium, it becomes;

2 MnO4− + 8 H+ → 2 Mn2+ + 4 H2O

5 SO32− + H2O → 5 SO42− + 2 H+

Final equation is;

2 MnO4- + 5 SO32- + 6 H+ → 2 (Mn)2+ + 5 SO42- + 3 H2O

Coefficient of H+ = 6

Coefficient of H2O = 3

Coefficient of MnO4- = 2

Coefficient of SO32- = 5

Coefficient of (Mn)2+- = 2

Coefficient of SO42- = 5