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  • An element A burns with golden flame in air. It reacts with another element B, atomic number 17 to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. Identify A, B, C and D. Also write down the eq

An element A burns with golden flame in air. It reacts with another element B, atomic number 17 to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. Identify A, B, C and D. Also write down the equations for the reactions involved.

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An element A burns with golden flame in air. So, A is sodium (Na) because it burns with golden flame in air.
Atomic number of B is given as 17 so B is chlorine (Cl).
When Sodium reacts with Chlorine, sodium chloride (NaCl) is formed. Hence, C is sodium chloride (NaCl)

2Na\left ( s \right )+Cl_{2}\left ( g \right )\rightarrow 2NaCl\left ( s \right )
Now it is given that an aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen
Aqueous solution of NaCl, on electrolysis gives sodium hydroxide.
Electrolysis Reaction:
The reaction at the cathode is:
H_{2}O\left ( l \right )+2e^{-}\rightarrow H_{2}\left ( g \right )+2OH^{-}
The reaction at the anode is:
Cl^{-}\rightarrow \frac{1}{2}Cl_{2}\left ( g \right )+1e^{-}
The overall reaction is:
2NaCl\left ( aq \right )+2H_{2}O\left ( l \right )\rightarrow 2NaOH\left ( aq \right )+Cl_{2}\left ( g \right )+H_{2}\left ( g \right )
Thus, D is sodium hydroxide (NaOH)

 

 

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