Can u help me in balancing the following eqn. through oxidation no. method or half reaction method.
Also please show and explain the steps.
HNO3(aq.) + KI(aq.) -----> NO(g) + I2(s)+ H2O(l) + KNO3(aq.)
Hoping for a quicker response.
I goes from -1 to 0
N goes from +5 to +2
There must be three I for each N that is changed.
so... 2NO and 3I2 would be a starting point.
check my thinking.
Hey mr. Bobpursley, I didn't get ur answer clearly.Can u please explain me in proper steps.i tried to balance from ur hint but coudnt do that.
I showed you the change in oxidation steps, and that there must be three changes of I for each change of N. Then I put the final coefficents in those producets to balance. The rest is for you.
I showed you the change in oxidation steps, and that there must be three changes of I for each change of N. Then I put the final coefficents in those producets to balance. The rest is for you.
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.
Sure, I can help you with balancing the equation using the oxidation number method. Here are the steps:
Step 1: Assign oxidation numbers to each element in the equation.
HNO3(aq.) + KI(aq.) → NO(g) + I2(s) + H2O(l) + KNO3(aq.)
H: +1
N: +5
O: -2
K: +1
I: -1
Step 2: Identify the elements undergoing oxidation and reduction.
In this equation, nitrogen (N) is being reduced, and iodine (I) is being oxidized.
Step 3: Determine the change in oxidation number for each element.
Nitrogen (N) goes from +5 to +2, a decrease of 3.
Iodine (I) goes from -1 to 0, an increase of 1.
Step 4: Balance the equation by adjusting coefficients.
To balance the change in oxidation number, we need three iodines (I) for each nitrogen (N) that changes. Therefore, we start with 2NO and 3I2.
2NO + 3I2 + H2O + KNO3 → NO(g) + I2(s) + H2O(l) + KNO3(aq.)
Next, balance the other elements by adding coefficients where needed. In this case, you already have equal numbers of potassium (K) atoms on both sides, and the hydrogen (H) and oxygen (O) atoms are balanced due to the water (H2O) already being present.
The final balanced equation is:
2HNO3(aq.) + 5KI(aq.) → 2NO(g) + 3I2(s) + H2O(l) + 5KNO3(aq.)
This is the balanced equation using the oxidation number method.