Balance the following redox reaction by filling in the blank with the appropriate coefficient __H3AsO4 + SnCl2 + 2HCl -> H3AsO3 + SnCl4 + H2O

Answer 1

__H3AsO4 + __SnCl2 + __HCl -> __H3AsO3 + __SnCl4 + __H2O

To balance the redox reaction, we need to ensure that the number of atoms and charge is balanced on both sides of the equation.

First, let's balance the As (arsenic) atoms. There is only one As atom on each side of the equation, so we start with:

H3AsO4 + __SnCl2 + __HCl -> H3AsO3 + __SnCl4 + __H2O

Next, let's balance the Cl (chlorine) atoms. There are four Cl atoms on the right side of the equation (2 in SnCl4 and 2 in HCl). So, we need to have four Cl atoms on the left side as well. This can be achieved by adding 4 HCl:

H3AsO4 + __SnCl2 + 4HCl -> H3AsO3 + __SnCl4 + __H2O

Now, let's balance the H (hydrogen) atoms. There are already six H atoms on the left side (3 in H3AsO4 and 3 in 4HCl), so we need to have six H atoms on the right side as well. This can be achieved by adding 3 H2O:

H3AsO4 + __SnCl2 + 4HCl -> H3AsO3 + __SnCl4 + 3H2O

Now, it's time to balance the Sn (tin) atoms. There is one Sn atom on the left side, so we need to have one Sn atom on the right side as well. This can be achieved by adding a coefficient of 1 in front of SnCl2:

H3AsO4 + 1SnCl2 + 4HCl -> H3AsO3 + 1SnCl4 + 3H2O

Finally, let's check the charge balance. On the left side, we have -3 from H3AsO4 and -1 from each HCl, which gives us a total charge of -7. On the right side, we have -3 from H3AsO3 and -1 from each HCl, which gives us a total charge of -7 as well. Therefore, the equation is balanced.

The balanced equation is:

H3AsO4 + SnCl2 + 4HCl -> H3AsO3 + SnCl4 + 3H2O