H2O+S2-+MnO4-->S+MnS+OH-

This equation is a redox reaction .
Reaction is in basic medium because OH- are the reaction.
This is unbalanced equation

H₂O+S^2-+MnO₄^-->S+MnS+OH^-
Use oxidation number method and follows the step - In the equation S^2---> S is oxidation with a loss of 2 electrons but S^2- to MnS is no change in Oxidation number of S^2-.
MnO₄^- to MnS is gain of  5 electrons.Now ignoring  water and  OH^- we will get
H₂O+7 S^2-+2 MnO₄^-->5 S+2 MnS+OH^-
Now follow other steps as discussed in the class and your final equation is
8 H₂O+7 S^2-+2 MnO₄^-->5 S+2 MnS+16 OH^-

H2O+7 S2-+2 MnO4-->5 S+2 MnS+OH-
Ma'am., how do we get 7 S2?
i got H2O+5 S2-+2 MnO4-->5 S+2 MnS+OH-, since the factor is 5/2.

Quote from: Ram_Manav_Soni on November 23, 2021, 08:41:50 AM
H2O+7 S2-+2 MnO4-->5 S+2 MnS+OH-
Ma'am., how do we get 7 S2?
i got H2O+5 S2-+2 MnO4-->5 S+2 MnS+OH-, since the factor is 5/2.

On the right side you have 7 S so you need 7 S on the left side.

6 H2O + 2 MnO4- + 3 (S) 2- → 3 S + 8 OH- + 2 MnO (OH) 2
This is an oxidation-reduction (redox) reaction:
2 Mn VII + 6 e- → 2 Mn IV (reduction)
3 S - II - 6 e- → 3 S0 (oxidation)
MnO4- is an oxidizing agent, S2- is a reducing agent.
Reactants for the reaction
For H2O ? Water, oxidane
Other names: Water (H2O), Hydrogen hydroxide (HH or HOH), Hydrogen oxide Dihydrogen monoxide (DHMO) (systematic name), Hydrogen monoxide, Dihydrogen oxide, Hydric acid, Hydrohydroxic acid, Hydroxic acid, Hydrol, Μ-oxide dihydrogen, Κ1 - hydroxyl hydrogen (0)
Appearance: White crystalline solid, almost colorless liquid with a hint of a blue, colorless gas
MnO4-
S2-
Products of the reaction
For S
Names: Sulfur, S, Sulphur element 16, Flowers of sulfur, Flour sulfur, Brimstone
Appearance: Yellow solid in various forms
For OH- ? Hydroxide
Other names: A hydroxide ion
, and MnO(OH)2

Quote from: Ram_Manav_Soni on November 23, 2021, 08:41:50 AM
H2O+7 S2-+2 MnO4-->5 S+2 MnS+OH-
Ma'am., how do we get 7 S2?
i got H2O+5 S2-+2 MnO4-->5 S+2 MnS+OH-, since the factor is 5/2.

Two S2- are not undergoing oxidation /reduction so they need to be balanced just by checking S on the right side and left side. Out of 7 S2- ,only 5 are undergoing oxidation and rest two are as such with no change in oxidation number.