Question

What happens when: (i) $Cl_{2}$ is passed through a solution of $F^{-}$ and $I^{-}$ ions. (ii) Iron nails are added to $AgNO_{3}$ solution. (iii) A dilute solution of $AgNO_{3}$ is electrolysed using silver electrodes.

Answer 1

(i) ${ Cl }_{ 2 }$ is passed through a solution of ${ I }^{ - }$ ions. When ${ Cl }_{ 2 }$ is passed through the KI solution then chlorine replace the I from KI and make KCl and ${ I }_{ 2 }$ gas is liberated. This is an example for displacement reaction.

(ii)  Iron nails are added to $Ag{ NO }_{ 3 }$ solution the following reaction takes place. Iron replaces silver in this reaction. Iron is more reactive than silver.

                 $Fe(s) + 2Ag{ NO }_{ 3 }(aq) \rightarrow  Fe{ \left( { NO }_{ 3 } \right)  }_{ 2 }(aq) + 2Ag(s)$

(iii)  A dilute solution of $Ag{ NO }_{ 3 }$ is electrolysed using silver electrodes. The Ag or silver metal gets oxidized at the anode part.

(iv)  $Ag{ NO }_{ 3 }$ ionizes in aqueous solutions to form ${ Ag }^{ + }$ and ${ NO }_{ 3 }^{ - }$ ions. On electrolysis, either ${ Ag }^{ + }$ ions or ${ H }_{ 2 }O$ molecules can be reduced at the cathode. But the reduction potential of ${ Ag }^{ + }$ ions is higher than that of ${ H }_{ 2 }O.$

                  ${ Ag }_{ (aq) }^{ + } + { e }^{ - } \rightarrow  Ag(s);$

                  E° = +0.08 V

                  $2{ H }_{ 2 }O(l) + 2{ e }^{ - } \rightarrow { H }_{ 2 }(g) + 2{ OH }^{ - }(aq);$

                  E° = -0.83 V

Hence, ${ Ag }^{ + }$ ions are reduced at the cathode. Similarly, Ag metal or ${ H }_{ 2 }O$ molecules can be oxidised at the anode. But the oxidation potential of Ag is higher than that of { H }_{ 2 }O molecules.

                  $Ag(s) \rightarrow { Ag }^{ + }(aq) + { e }^{ - };$

                  E° = -0.80 V

                  $2{ H }_{ 2 }O(l) \rightarrow { O }_{ 2 }(g) + 4{ H }^{ + }(aq) + 4{ e }^{ - };$

                  E° = -1.23 V

Therefore, Ag metal gets oxidized at the anode.

Answer 2

HEYA MATE.....

Ag metal gets OXIDISED at the Anode.......