What is the coefficient of H∗ after the following half-reaction is balanced? HSO4−(aq)⟶S2−(aq) Question 34 4 pts In a galvanic cell, the ion migration between the two half-cells occurs through the to complete the circuit and prevent charge built-up in the two half-cells. Cathode electrolyte solution Salt bridge Anode
Salt bridge
To balance the given half-reaction equation: HSO4−(aq) ⟶ S2−(aq), we need to balance the number of atoms on each side of the equation.
Step 1: Count the number of atoms on each side:
Left side (reactants): 1 hydrogen (H), 1 sulfur (S), 4 oxygen (O)
Right side (products): 1 sulfur (S)
Step 2: Balance the non-hydrogen and non-oxygen atoms first:
To balance the sulfur (S) atom, we need a coefficient of 1 on the left side.
HSO4−(aq) ⟶ S2−(aq)
Step 3: Balance the hydrogen (H) atoms:
On the left side, we have 1 hydrogen atom. To balance it, we need to add a coefficient of 8 in front of HSO4−.
8HSO4−(aq) ⟶ S2−(aq)
Step 4: Balance the oxygen (O) atoms:
On the left side, there are 8 hydrogen atoms, which means there are 32 oxygen atoms (8 x 4). On the right side, there is only 1 sulfur atom, which means there are no oxygen atoms. To balance the oxygen, we need to add water (H2O) molecules on the right side.
8HSO4−(aq) ⟶ S2−(aq) + 8H2O(l)
Therefore, after balancing the half-reaction, the coefficient of H∗ is 8.