Monday

October 20, 2014

October 20, 2014

Posted by **bik** on Wednesday, May 16, 2007 at 1:11pm.

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

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