Asked by .
Q: Calculate ℰ° values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations. Standard reduction potentials are found in the Standard Reduction Potentials table. (Use the lowest possible whole number coefficients. Include states-of-matter under the given conditions in your answer.)
A:
When writing the balanced eq from:
MnO4−(aq) + I −(aq) <=> I2(s) + MnO2(s)
I get:
2MnO4-(aq) + 8H+(g) + 4I-(aq) <=> 3MnO2(g)+6H2O(l)+3I2(s)
I probably wrote a half rxn wrong idk?
A:
When writing the balanced eq from:
MnO4−(aq) + I −(aq) <=> I2(s) + MnO2(s)
I get:
2MnO4-(aq) + 8H+(g) + 4I-(aq) <=> 3MnO2(g)+6H2O(l)+3I2(s)
I probably wrote a half rxn wrong idk?
Answers
Answered by
DrBob222
You should know your equation isn't balanced. You can see there are 2 Mn on the left and 3 on the right. There are 4 I on the left and 6 on the right. Why do you not KNOW it isn't balanced?
Answered by
Doc48
2MnO₄ˉ + 4H₂O + 6eˉ => 2MnO₂ + 8OHˉ ….. E⁰ = + 0.83 volts
……………..……..6Iˉ => 3I₂ + 6eˉ …….…… E⁰ = + 0.54 volts
…………………………………………………………………………
2MnO₄ˉ + 6Iˉ + 4H₂O => 3I₂ + 2MnO₂ + 8OHˉ
E⁰(net) = E⁰(Reduction) - E⁰(Oxidation) = (+0.83 volts) – (+0.54 volts) = 0.29 volts
……………..……..6Iˉ => 3I₂ + 6eˉ …….…… E⁰ = + 0.54 volts
…………………………………………………………………………
2MnO₄ˉ + 6Iˉ + 4H₂O => 3I₂ + 2MnO₂ + 8OHˉ
E⁰(net) = E⁰(Reduction) - E⁰(Oxidation) = (+0.83 volts) – (+0.54 volts) = 0.29 volts
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.