Balance th following equation in acid Solutiong and determind if it is spontaneous under standard conditions.
NO3^-1 + Sn^+2 ---> NO + Sn^+4
balanced directed to first post.
You need to calculate the E value to determine if it is spontaneous or not. Please show any work and explain what you don't understand if you can't proceed.
i missed all the classes in this session.. am totally out of track...
I tried to find a GOOD site to explain this from th beginning but I didn't so I'll try to give you some steps to help you get started. Remember, however, that a number of methods exist and your prof may not use this method.
Step 1. Assign oxidation states to the atoms and determine which elements have changed. I assume you know how to do this. At any rate, N is +5 on the left and +2 on the right. Sn is +2 on the left and +4 on the right.
Step 2. Separate the equation into two half equations. In this case, we will have
NO3^- ==> NO
Sn^+2 ==> Sn^+4
Step 3. To one of the equation, add electrons to the appropriate side to balance the change in oxidation state. To change N from +5 to +2, we must add 3 electrons to the left side like this.
NO3^- + 3e ==> NO
Step 4. Count up the charge on each side and add H^+ (if in acid solution) or OH^- (if in basic solution) to balance the charge. The problem states this is an acid solution; therefore, we add H^+. I see a -4 charge on the left and zero on the right; therefore, we add 4H^+ to the left like this. Then check the charge to make sure it is balanced.
NO3^- + 3e + 4H^+ ==> NO
Step 5. Add water to other side to balance the H^+. This should automatically balance the O atoms. Like this.
NO3^- + 3e + 4H^+ ==> NO = 2H2O
Step 6. Do the same for the other half equation.
Sn+2 ==> Sn+4 + 2e
Step 7. Multiply the equations by a number such that the electrons are equal. (Note that the electrons in one half cell are on one side of the equation and on the other side in the other half equation.) In this case the first equation is multiplied by 2 and the second by 3.
2 x (NO3^- + 3e + 4H^+ ==> NO + 2H2O)
3 x (Sn^+2 ==> Sn^+4 + 2e)
Then Add the two. I will multiply and add at the same time to shorten this.
2NO3^- + 6e + 8H^+ + 3Sn^+2 ==> 2NO + 4H2O + 3Sn^+4 + 6e
Step 8. Cancel ions etc common to both sides. In this case, only 6e can be canceled. Now check atoms and charge to make sure it is balanced.
I shall be happy to answer any questions. There is a subpart I omitted because it isn't necessary in this equation; however, you need to add it as Step 2a. Make sure we have the same number of atoms and the total charge for those atoms that have changed. In this example, we have 1N on both sides and 1 Sn on both side so it is ok. If we had Cr2O7^-2 ==> Cr^+3 we would need to make the total charge on the left Cr as +12, add a 2 to Cr^+3 on the right to look like this. 2Cr^+3. So the total charge is +12 for the 2Cr on the left and +6 for the 2 Cr on the right and the final equation for dichromate going to chromium ion is
Cr2O7^-2 + 6e + 14H^+ ==> 2Cr^+3 + 7H2O
To balance the given equation in acid solution (NO3^-1 + Sn^+2 ---> NO + Sn^+4), you can follow these steps:
Step 1: Assign oxidation states to the atoms and determine which elements have changed. In this case, N is +5 on the left and +2 on the right. Sn is +2 on the left and +4 on the right.
Step 2: Separate the equation into two half equations:
NO3^-1 ==> NO
Sn^+2 ==> Sn^+4
Step 3: To balance the change in oxidation state, add electrons to one side of each half equation. In this case, to change N from +5 to +2, you need to add 3 electrons to the left side: NO3^-1 + 3e^- ==> NO. The other half equation does not require any electron addition.
Step 4: Count up the charge on each side and add H^+ ions (since it's an acid solution) to balance the charge. In this case, the left side has a -1 charge and the right side has 0 charge, so you need to add 1 H^+ ion to the left side: NO3^-1 + 3e^- + 1H^+ ==> NO.
Step 5: Add water molecules to balance the hydrogen atoms. In this case, add 2H2O to the right side: NO3^-1 + 3e^- + 1H^+ ==> NO + 2H2O.
Step 6: Repeat the above steps for the other half equation. In this case, Sn^+2 ==> Sn^+4 can be balanced as it is.
Step 7: Multiply the half equations by appropriate numbers to balance the electrons. In this case, multiply the first half equation by 2 and the second half equation by 3:
2NO3^-1 + 6e^- + 2H^+ ==> 2NO + 4H2O
3Sn^+2 ==> 3Sn^+4
Step 8: Add the two balanced half equations together to get the balanced overall equation:
2NO3^-1 + 6e^- + 2H^+ + 3Sn^+2 ==> 2NO + 4H2O + 3Sn^+4 + 6e^-
Step 9: Cancel out the electrons that appear on both sides of the equation:
2NO3^-1 + 2H^+ + 3Sn^+2 ==> 2NO + 4H2O + 3Sn^+4
Now the equation is balanced in acid solution.
To determine if the reaction is spontaneous under standard conditions, you need to calculate the standard cell potential (E°) using the standard reduction potentials of the half reactions involved. Once you have the E° values, you can use the equation ΔG° = -nFE°, where ΔG° is the change in Gibbs free energy, n is the number of moles of electrons transferred in the balanced equation, F is Faraday's constant (96485 C/mol), and E° is the standard cell potential.
If ΔG° is negative, the reaction is spontaneous, and if ΔG° is positive, the reaction is non-spontaneous.