Question
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
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?
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
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