Does the entropy (△S) increase or decrease in each of the following processes? The wind blows a pile of leaves all over the yard
1 answer:
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Answer:
Explanation:
Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.
The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:
Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:
The net equation is then:
However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:
Actual anode half-equation:
Actual cathode half-equation:
Actual net reaction:
<u>Answer:</u> The balanced chemical equation is written below and for the reaction is -160.6 J/K
<u>Explanation:</u>
When calcium hydroxide reacts with sulfur dioxide, it leads to the formation of calcium sulfate and water molecule.
The chemical equation for the reaction of calcium hydroxide and sulfur dioxide follows:
To calculate the entropy change of the reaction, we use the equation:
For the given reaction:
Taking the standard entropy change values:
Putting values in above equation, we get:
Hence, the balanced chemical equation is written above and for the reaction is -160.6 J/K