Predict the products of electrolysis in each of the following :

An aqueous solution of $A g N O_{3}$ with silver electrodes

An aqueous solution of $A g N O_{3}$ with platinum electrodes

A dilute solution of $H_{2} S O_{4}$ with platinum electrodes.

An aqueous solution of $C u C l_{2}$ with platinum electrodes.

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Solution

An aqueous solution of $A g N O_{3}$ with silver electrodes.

In aqueous solution, ionisation of $A g N O_{3}$ and $H_{2} O$ takes place.

$A g N O_{3} (s) (a q) ⇌ A g^{+} (a q) + N O_{3}^{-} (a q)$
$H_{2} O (l) ⇌ H^{+} (a q) + O H^{-} (a q)$
At cathode $A g^{+}$ ions has less discharge potential than $H^{+}$ ions so silver will be deposited at cathode.
$A g^{+} (a q) + e^{-} \to A g (s)$
at anode An equivalent amount of silver will be oxidised to Ag^+ ions by releasing electrons
$A g (s) \to A g^{-} (a q) + e^{-}$
Ag anode is attaked by $N O_{3}^{-}$ ions, so it will also produce $A g^{+}$ in the solution

An aqueous solution $A g N O_{3}$ with platinum electrodes. In aqueous solution ionisation of $A g N O_{3}$ and $H_{2} O$ occurs

$A g N O_{3} (s) (A q) ⇌ A g^{+} (a q) + N O_{3}^{-} (a q) H_{2} O (l) ⇌ H^{+} (a q) + O H^{-} (a q)$
As platinum electrodes are non-attackable electrodes, they will not be reacted upon by $N O_{3}^{-}$ ions.
At cathode Ag will be deposited at cathode.
$A g^{+} (a q) + e^{-} \to A g (s)$
At anode Out of $N O_{3}^{-}$ and $O H^{-}$ ions, only $O H^{-}$ ions will be oxidised (due to less discharge potential) preferentially and $N O_{3}^{-}$ ions will remain in the solution.
$O H^{-} (a q) \to O H + e^{-} 4 O H \to 2 H_{2} O (l) + O_{2} (g)$
So, oxygen gas is produced at anode. The solution remains acidic due to the presence of $H N O_{3} .$
$H^{+} (a q) + N O_{3}^{-} (a q) ⇌ H N O_{3} (a q)$

A dilute solution of $H_{2} S O_{4}$ with platinum elections.

Both $H_{2} S O_{4}$ and water ionise in the solution

$H_{2} S O_{4} (a q) ⇌ 2 H^{+} (a q) + S O_{4}^{2 -} (a q) H_{2} O (l) ⇌ 2 H^{+} (a q) + O H^{-} (a q)$
At cathode $H^{+}$ ioins will be reduced and hydrogen gas is produced at cathode.
$H^{+} (a q) + e^{-} \to H (g) H (g) + H (g) \to H_{2} (g)$
At anode $O H^{-}$ ions will be released preferentially and not $S O_{4}^{-}$ ions due to less discharge potential.
$O H^{-} (a q) \to O H + e^{-} 4 O H \to 2 H_{2} O (l) + O_{2} (g)$
Oxygen gas is produced at anode.
Oxygen gas is produced at anode.
Solution will be acidic and will contain $H_{2} S O_{4}$

An aqueous solution of $C u C l_{2}$ with platinum electrodes. Both $C u C l_{2}$ and water ionise as usual.

$C u C l_{2} (a q) ⇌ C u^{2 +} (a q) + 2 C l^{-} (a q) H_{2} O (l) ⇌ 2 H^{+} (a q) + O H^{-} (a q)$
At cathode $C u^{2 +}$ ions will be reduced preferentially due to less discharge potential than $H^{+}$ ions.
$C u^{2 + (a q)} + 2 e^{-} \to C u (s)$
Copper metal is deposited at cathode.
At anode $C l^{-}$ ions will be discharged in preference to $O H^{-}$ ions and chlorine gas is produced at anode.
$C l^{-} (a q) \to C l (g) + e^{-} C l (g) + C l (g) \to C l_{2} (g)$

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