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?
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
Well, don't worry, I'm here to help you clarify things!
Let's break it down step by step:
First, we'll write the half-reactions for each component:
1. Reduction half-reaction: MnO4^-(aq) + 3e^- → MnO2(s)
(This is the reduction of MnO4^- to MnO2)
2. Oxidation half-reaction: 2I^-(aq) → I2(s) + 2e^-
(This is the oxidation of I^- to I2)
Next, we need to balance the half-reactions by adding coefficients to make the number of electrons equal in both reactions:
1. Reduction half-reaction: 2MnO4^-(aq) + 16H^+(aq) + 10e^- → 2MnO2(s) + 8H2O(l)
2. Oxidation half-reaction: 10I^-(aq) → 5I2(s) + 10e^-
Now, we can add the two balanced half-reactions together to get the overall balanced equation:
2MnO4^-(aq) + 16H^+(aq) + 10I^-(aq) → 2MnO2(s) + 8H2O(l) + 5I2(s)
To determine if the reaction is spontaneous as written, we need to check the standard cell potential (ℰ°). The overall ℰ° is calculated by subtracting the ℰ° of the oxidation half-reaction from the ℰ° of the reduction half-reaction.
Since the reduction half-reaction involves the reduction of MnO4^-, we will look up the ℰ° value for MnO4^-(aq) to MnO2(s) in the Standard Reduction Potentials table. Similarly, we will look up the ℰ° value for I^-(aq) to I2(s).
After finding the ℰ° values, we subtract the oxidation ℰ° from the reduction ℰ° to get the overall ℰ° value. If the overall ℰ° is positive, the reaction is spontaneous. If it's negative, the reaction is non-spontaneous.
I hope this clears things up! Feel free to ask if you have any more questions.
To balance the equation correctly, you need to make sure that the number of atoms of each element is balanced on both sides.
First, let's balance the atoms other than oxygen and hydrogen:
MnO4^-(aq) + I^-(aq) -> MnO2(s) + I2(s)
To balance the equation, you can start by balancing the iodine (I) atoms. Since there are four iodine atoms on the left side and only two on the right side, you can multiply the I2(s) on the right side by 2:
MnO4^-(aq) + 2I^-(aq) -> MnO2(s) + I2(s)
Next, balance the oxygen (O) atoms. There are four O atoms on the left side, and only two on the right side. To balance the O atoms, you can add two H2O(l) molecules on the right side:
MnO4^-(aq) + 2I^-(aq) -> MnO2(s) + I2(s) + 2H2O(l)
Finally, balance the charge by adding electrons (e^-) to the appropriate side. On the left side, there are eight negative charges (four from MnO4^- and four from the two I^- ions) and no charges on the right side. Therefore, you need to add eight electrons (e^-) to the left side:
MnO4^-(aq) + 2I^-(aq) + 8e^- -> MnO2(s) + I2(s) + 2H2O(l)
Now the equation is balanced correctly.
To calculate the E° value, you can use the standard reduction potentials table. The E° value for the oxidation half-reaction of MnO4^-(aq) is +1.51 V, and the reduction half-reaction for I^-(aq) is -0.54 V.
The overall E° value for the cell can be calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode:
E° = E°(cathode) - E°(anode)
= (-0.54 V) - (+1.51 V)
= -2.05 V
Since the E° value is negative, the reaction is not spontaneous under standard conditions.
To calculate the ℰ° values for the given cells, you need to use the standard reduction potentials from the Standard Reduction Potentials table. The ℰ° value is the standard potential for a half-reaction, which indicates the tendency of a species to be reduced or oxidized.
For the given cell:
MnO4−(aq) + I−(aq) ⇌ I2(s) + MnO2(s)
First, let's write the half-reactions for the reduction and oxidation processes individually:
Oxidation: MnO4−(aq) → MnO2(s)
Reduction: I−(aq) → I2(s)
We can look up the standard reduction potentials for these half-reactions in the Standard Reduction Potentials table.
From the table:
ℰ°(MnO4−/MnO2) = 1.51 V
ℰ°(I−/I2) = 0.54 V
To determine the ℰ° value for the overall cell reaction, you need to subtract the ℰ° value of the oxidation half-reaction from the ℰ° value of the reduction half-reaction:
ℰ°(cell) = ℰ°(reduction) - ℰ°(oxidation)
ℰ°(cell) = 0.54 V - 1.51 V
ℰ°(cell) = -0.97 V
The negative value indicates that the reaction is not spontaneous under standard conditions.
To balance the equation, we can start with the reduction half-reaction since it has a lower ℰ° value:
I−(aq) → I2(s)
To balance the number of iodide ions (I−) and iodine (I2) atoms, we need to add 2 iodide ions:
2I−(aq) → I2(s)
Next, we balance the charge by adding 2 electrons (e−) to the left side:
2I−(aq) + 2e− → I2(s)
Now, for the oxidation half-reaction:
MnO4−(aq) → MnO2(s)
To balance the number of oxygen (O) atoms, we need to add 2 water (H2O) molecules to the left side:
MnO4−(aq) + 2H2O(l) → MnO2(s)
Finally, to balance the hydrogen (H) atoms, we add 8 hydrogen ions (H+) to the left side:
MnO4−(aq) + 8H+(aq) + 2H2O(l) → MnO2(s)
Therefore, the balanced equation is:
2MnO4−(aq) + 8H+(aq) + 4I−(aq) → 2MnO2(s) + 4H2O(l) + 2I2(s)
After balancing the equation and calculating the ℰ° value, we can see that the reaction is not spontaneous as written under standard conditions.