
Write the balanced equation for the following chemical reactions:
(A) $2Na(s)+2{ H }_{ 2 }O(l)\longrightarrow 2NaOH(aq)+{ H }_{ 2 }(g)$
(B) $Na(s)+{ H }_{ 2 }O(l)\longrightarrow NaOH(aq)+{ H }_{ 2 }(g)$
(C) $2Na(s)+2{ H }_{ 2 }O(l)\longrightarrow NaOH(aq)+{ H }_{ 2 }(g)$
(D) $Na(s)+{ H }_{ 2 }O(l)\longrightarrow 2NaOH(aq)+{ H }_{ 2 }(g)$
Answer
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Hint: Sodium is an alkali metal which is highly reactive in nature. It reacts readily with water or moisture to produce hydrogen gas. Therefore it is stored in labs away from moisture and kept in kerosene.
Complete step by step solution:
Sodium is soft metal and can be easily cut with a knife. Sodium metal reacts vigorously with water and produces sodium hydroxide and hydrogen gas. Since sodium hydroxide is alkaline in nature, the resulting solution is basic. This reaction is exothermic in nature i.e. it produces heat. Most of the time, the sodium metal becomes so hot during the reaction that it catches fire and displays a characteristic orange hue. This reaction is faster than that of lithium but slower than that of potassium.
So why does sodium react so violently with water? This is because it is more active than hydrogen which can be seen in the activity series. At room temperature, there exist protons and hydroxide ions though in a very small quantity in water due to self- ionization. Now, the standard reduction potential of protons is zero while the standard reduction potential of Na+ is highly negative which implies that sodium will act as a reducing agent and will reduce the water to produce hydrogen gas while itself getting oxidised to Na+ ions. For this reaction, the Gibbs free energy change is negative which implies that this reaction is spontaneous. In fact, during this reaction such a high amount of energy is released that the hydrogen gas released can burn.
When sodium reacts with the protons present in water it gives sodium hydroxide along with hydrogen gas. This is a redox reaction. In order to balance such an equation we need to find out the n-factor for the sodium metal and the n-factor for protons. Since a sodium atom oxidizes to give a sodium ion, therefore its n-factor is 1. Since a proton is being reduced to give molecular hydrogen gas, its n-factor will be 2. Now write the complete equation:
$ Na(s)+{ H }^{ + }(aq)\longrightarrow N{ a }^{ + }(aq)+{ H }_{ 2 }(g) $
Now, cross-multiply the simplest ratio of the n-factors (sodium: proton=1:2)
$ 2Na(s)+{ H }^{ + }(aq)\longrightarrow N{ a }^{ + }(aq)+{ H }_{ 2 }(g) $
Now balance the atoms on both sides except for the hydrogen and the oxygen atoms if any:
$ 2Na(s)+{ H }^{ + }(aq)\longrightarrow 2N{ a }^{ + }(aq)+{ H }_{ 2 }(g) $
Now balance the oxygen atoms if any followed by the hydrogen atoms on both sides:
$2Na(s)+{ 2H }^{ + }(aq)\longrightarrow 2N{ a }^{ + }(aq)+{ H }_{ 2 }(g)$ ……(1)
Since this is a solvolysis reaction and the protons are present in a very small amount we want to replace the protons in the above equation by the solvent molecules therefore we will use:
$ { H }_{ 2 }O(l)\longrightarrow { H }^{ + }+{ OH }^{ - } $
Multiply the above equation by 2 and add it to (1)
$ \begin{matrix} 2Na(s)+2{ H }^{ + }(aq)\longrightarrow 2{ Na }^{ + }(aq)+{ H }_{ 2 }(g) \\ 2{ H }_{ 2 }O(l)\longrightarrow 2{ H }^{ + }(aq)+2{ OH }^{ - }(aq) \end{matrix} $
The final equation is:
$2Na(s)+2{ H }_{ 2 }O(l)\longrightarrow 2NaOH(aq)+{ H }_{ 2 }(g)$
Hence the answer is (a) $2Na(s)+2{ H }_{ 2 }O(l)\longrightarrow 2NaOH(aq)+{ H }_{ 2 }(g)$
Note: The n-factor of a particular species is not fixed, It changes with the type of reaction, the products of the reaction, the medium of the reaction whether acidic medium or basic medium or neutral.
Complete step by step solution:
Sodium is soft metal and can be easily cut with a knife. Sodium metal reacts vigorously with water and produces sodium hydroxide and hydrogen gas. Since sodium hydroxide is alkaline in nature, the resulting solution is basic. This reaction is exothermic in nature i.e. it produces heat. Most of the time, the sodium metal becomes so hot during the reaction that it catches fire and displays a characteristic orange hue. This reaction is faster than that of lithium but slower than that of potassium.
So why does sodium react so violently with water? This is because it is more active than hydrogen which can be seen in the activity series. At room temperature, there exist protons and hydroxide ions though in a very small quantity in water due to self- ionization. Now, the standard reduction potential of protons is zero while the standard reduction potential of Na+ is highly negative which implies that sodium will act as a reducing agent and will reduce the water to produce hydrogen gas while itself getting oxidised to Na+ ions. For this reaction, the Gibbs free energy change is negative which implies that this reaction is spontaneous. In fact, during this reaction such a high amount of energy is released that the hydrogen gas released can burn.
When sodium reacts with the protons present in water it gives sodium hydroxide along with hydrogen gas. This is a redox reaction. In order to balance such an equation we need to find out the n-factor for the sodium metal and the n-factor for protons. Since a sodium atom oxidizes to give a sodium ion, therefore its n-factor is 1. Since a proton is being reduced to give molecular hydrogen gas, its n-factor will be 2. Now write the complete equation:
$ Na(s)+{ H }^{ + }(aq)\longrightarrow N{ a }^{ + }(aq)+{ H }_{ 2 }(g) $
Now, cross-multiply the simplest ratio of the n-factors (sodium: proton=1:2)
$ 2Na(s)+{ H }^{ + }(aq)\longrightarrow N{ a }^{ + }(aq)+{ H }_{ 2 }(g) $
Now balance the atoms on both sides except for the hydrogen and the oxygen atoms if any:
$ 2Na(s)+{ H }^{ + }(aq)\longrightarrow 2N{ a }^{ + }(aq)+{ H }_{ 2 }(g) $
Now balance the oxygen atoms if any followed by the hydrogen atoms on both sides:
$2Na(s)+{ 2H }^{ + }(aq)\longrightarrow 2N{ a }^{ + }(aq)+{ H }_{ 2 }(g)$ ……(1)
Since this is a solvolysis reaction and the protons are present in a very small amount we want to replace the protons in the above equation by the solvent molecules therefore we will use:
$ { H }_{ 2 }O(l)\longrightarrow { H }^{ + }+{ OH }^{ - } $
Multiply the above equation by 2 and add it to (1)
$ \begin{matrix} 2Na(s)+2{ H }^{ + }(aq)\longrightarrow 2{ Na }^{ + }(aq)+{ H }_{ 2 }(g) \\ 2{ H }_{ 2 }O(l)\longrightarrow 2{ H }^{ + }(aq)+2{ OH }^{ - }(aq) \end{matrix} $
The final equation is:
$2Na(s)+2{ H }_{ 2 }O(l)\longrightarrow 2NaOH(aq)+{ H }_{ 2 }(g)$
Hence the answer is (a) $2Na(s)+2{ H }_{ 2 }O(l)\longrightarrow 2NaOH(aq)+{ H }_{ 2 }(g)$
Note: The n-factor of a particular species is not fixed, It changes with the type of reaction, the products of the reaction, the medium of the reaction whether acidic medium or basic medium or neutral.
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