Sure, I can help you with that. Let's balance the equation step by step using the oxidation number method.
Step 1: Identify the oxidation numbers for each element in the equation.
HNO3(aq) + KI(aq) -----> NO(g) + I2(s) + H2O(l) + KNO3(aq)
H in HNO3: +1
N in HNO3: +5
O in HNO3: -2
K in KI: +1
I in KI: -1
N in NO: +2
O in NO: -2
I in I2: 0
H in H2O: +1
O in H2O: -2
K in KNO3: +1
N in KNO3: +5
O in KNO3: -2
Step 2: Identify the changes in oxidation numbers for the elements that undergo oxidation or reduction.
From the oxidation numbers, we can observe that:
- N in HNO3 is reduced from +5 to +2.
- I in KI is oxidized from -1 to 0.
Step 3: Write the half-reactions for each oxidation or reduction process.
For the oxidation:
2I- (from KI) -> I2
For the reduction:
HNO3 -> NO + H2O
Step 4: Balance the atoms in the half-reactions.
For the oxidation:
2I- (from KI) -> I2
For the reduction:
2HNO3 -> 2NO + 3H2O
Step 5: Balance the charges in the half-reactions.
For the oxidation:
2I- (from KI) -> I2 + 2e-
For the reduction:
6HNO3 + 4e- -> 6NO + 6H2O
Step 6: Multiply the half-reactions by appropriate coefficients to make the number of electrons transferred equal.
For the oxidation:
2I- (from KI) -> I2 + 2e-
For the reduction:
12HNO3 + 8e- -> 12NO + 12H2O
Step 7: Add the balanced half-reactions together.
2I- + 12HNO3 + I2 -> 12NO + 12H2O
Step 8: Balance the remaining atoms in the equation.
2HNO3 + 10KI + 5H2SO4 -> 2NO + I2 + 4H2O + 10KNO3 + 5H2SO4
So, the balanced equation is:
2HNO3(aq) + 10KI(aq) + 5H2SO4(aq) -> 2NO(g) + I2(s) + 4H2O(l) + 10KNO3(aq) + 5H2SO4(aq)
I hope this helps! Let me know if you have any further questions.