Write the net chemical equation for below equations
H2O2 + 2I- + 2H+ -----> 2H2O + I2
2 S2O32- + I2 -----> 2I- + S4O62-
How do you solve for this problem
To write the net chemical equation, we need to balance the individual equations and then combine them in such a way that the reactants common in both equations cancel out.
1. Balancing the first equation:
H2O2 + 2I- + 2H+ → 2H2O + I2
There are 2 iodine (I) atoms on the reactant side, so let's balance the iodine by multiplying the second equation by 2:
2 × (2 S2O32- + I2 → 2I- + S4O62-)
4 S2O32- + 2 I2 → 4 I- + 2 S4O62-
Now, we have 4 iodine (I) atoms on each side.
2. Combining the equations:
H2O2 + 2I- + 2H+ → 2H2O + I2
4 S2O32- + 2 I2 → 4 I- + 2 S4O62-
Since 2 I- on the left side reacts with 2 I- on the right side, they cancel each other out. Similarly, the water (H2O) formed on the right side can be canceled.
The final net chemical equation is:
H2O2 + 2H+ + 4 S2O32- → 2 H2O + 2 S4O62-