So (Because I'm so nice) I'm going to let somebody pick a random Redox Question and show me how to work through it step by step.

You ask why? I'll tell you! ; Because I suck at Redox Questions.

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# Redox

Started by Nancie, Apr 04 2008 02:47 PM

3 replies to this topic

### #1

Posted 04 April 2008 - 02:47 PM

### #2

Posted 07 April 2008 - 05:24 PM

Hey. This is one i picked randomly from my Higher notes

For the following reaction combine the oxidation(loss of electrons) and reduction(gain of electrons) to give a balanced redox equation:

Al-> Al(3+) + 3e

2H(+)+2e-> H2

Step 1: Work out which reaction is oxidation and which is reduction:

Al-> Al(3+)+3e - Oxidation

2H(+)+2e-> H2 - Reduction

Step 2: To form a balanced redox equation you add the equations together by cancelling the electrons in each ion-electron equation. If the two equations have the same number of electrons in each equation simply cancel the electrons and add the equations. If the number of electrons are different multiply one or both the ion-electron equations to give an equal number of electrons:

2Al-> 2Al(3+) + 6e(multiplied by 2)

6H(+) + 6e-> 3H2(multiplied by 3)

Cancel the electrons as they are now equal:

2Al-> 2Al(3+)

6H(+)-> 3H2

Step 3: Add the two equations. In the equation the reactants and products from the ion-electron equations should stay on the same side:

2Al+6H(+)-> 2Al(3+) + 3H2

This gives the balanced redox equation for this reaction

Hope this helps. If you are still stuck i'll try and help again.

For the following reaction combine the oxidation(loss of electrons) and reduction(gain of electrons) to give a balanced redox equation:

Al-> Al(3+) + 3e

2H(+)+2e-> H2

Step 1: Work out which reaction is oxidation and which is reduction:

Al-> Al(3+)+3e - Oxidation

2H(+)+2e-> H2 - Reduction

Step 2: To form a balanced redox equation you add the equations together by cancelling the electrons in each ion-electron equation. If the two equations have the same number of electrons in each equation simply cancel the electrons and add the equations. If the number of electrons are different multiply one or both the ion-electron equations to give an equal number of electrons:

2Al-> 2Al(3+) + 6e(multiplied by 2)

6H(+) + 6e-> 3H2(multiplied by 3)

Cancel the electrons as they are now equal:

2Al-> 2Al(3+)

6H(+)-> 3H2

Step 3: Add the two equations. In the equation the reactants and products from the ion-electron equations should stay on the same side:

2Al+6H(+)-> 2Al(3+) + 3H2

This gives the balanced redox equation for this reaction

Hope this helps. If you are still stuck i'll try and help again.

### #3

Posted 12 April 2008 - 02:19 PM

Thanks very much. It really helps. Just redox confuses me! :L

But I'm understanding it now.

Thanks

But I'm understanding it now.

Thanks

### #4

Posted 12 April 2008 - 09:00 PM

No problem

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