Predict the products of electrolysis in each of the following:

(i) An aqueous solution of $$AgNO_{3}$$ with silver electrodes

(ii) An aqueous solution $$AgNO_{3}$$​ with platinum electrodes

(iii) A dilute solution of $$H_{2}SO_{4}$$​ with platinum electrodes

(iv) An aqueous solution of $$CuCl_{2}$$with platinum electrodes.

Asked by Pragya Singh | 1 year ago |  147

##### Solution :-

(i) $$AgNO_{3}$$​ ionizes in aqueous solution to form $$Ag^{+}$$ and $$NO^{-}_{3}$$​ ions.

On electrolysis, either $$Ag^{+}$$ ion or $$H_{2}O$$ molecule can be decreased at cathode. But the reduction potential of $$Ag^{+}$$ ions is higher than that of $$H_{2}O$$.

$$Ag^{+}_{(aq)} \; + \; e^{-} \; \rightarrow Ag(s)​; E^{\circ} \; =+0.80V$$

$$2\; H_{2}O_{\left ( l \right )} \; +\; 2 \;e^{-} \;→H_2(g)​+2OH_{\left (aq \right )}^{−}​;$$

$$E^{\circ} \; =−0.83V$$

Therefore, $$Ag^{+}$$ ions are decreased at cathode. Same way, Ag metal or $$H_{2}O$$ molecules can be oxidized at anode. But the oxidation potential of Ag is greater than that of $$H_{2}O$$ molecules.

$$Ag_{\left ( s \right )} \; \rightarrow \;Ag^{+}\;+ ​e^{−}; E^{\circ} \; \; =−0.80V$$

$$2\; H_{2}O_{\left ( l \right )}\; \rightarrow \;O_{2\;\left ( g \right )}\; +\; 4 \;H^{+}_{\left ( aq \right )}\; +\; 4\; e^{-}=−1.23V$$

Hence, Ag metal gets oxidized at anode.

(ii) Pt cannot be oxidized very easily. Therefore, at anode, oxidation of water occurs to liberate $$O_{2}$$​.

At the cathode, $$Ag^{+}$$ ions are decreased and get deposited.

(iii) $$H_{2}SO_{4}$$​ ionizes in aqueous solutions to give $$H^{+}$$ and $$SO_{4}^{2-}$$​ ions.

$$H_{2}SO_{4\;\left ( aq \right )} \;\rightarrow \; 2\; H^{+}_{\left ( aq \right )} \; +\; SO^{2-}_{4 \left ( aq \right )}$$

On electrolysis, either of $$H_{2}O$$ molecules or $$H^{+}$$ ions can get decreased at cathode. But the decreased potential of $$H^{+}$$ions is higher than that of $$H_{2}O$$​ molecules.

$$2 \;H^{+}_{\left ( aq \right )} \;+ \;2 \;e^{-} \rightarrow H_{2(g)}​; E^{\circ} \; \; =0.0V$$

$$2 \;H_{2}O_{\left ( aq \right )} \;+ \;2 \;e^{-} \;\rightarrow \;H_{2\;\left ( g \right )} \;+ \;2 \;OH^{-}_{\left ( aq \right )} E^{\circ} \; = −0.83V$$

Therefore, at cathode, $$H^{+}$$ ions are decreased to free $$H_{2}$$​ gas.

On the other hand, at anode, either of $$H_{2}O$$ molecules or $$SO_{4}^{2-}$$​ ions can be oxidized. But the oxidation of $$SO_{4}^{2-}$$​ involves breaking of more bonds than that of $$H_{2}O$$molecules. Therefore, $$SO_{4}^{2-}$$​ ions have lower oxidation potential than $$H_{2}O$$. Hence, $$H_{2}O$$ is oxidized at anode to free $$O_{2}$$​ molecules.

(iv)  In aqueous solutions, $$CuCl_{2}$$​ ionizes to give $$Cu^{2+}$$ and $$Cl^{-}$$ ions as:

$$CuCl_{2\; \left ( aq \right )} \;\rightarrow \;Cu^{2+}_{\left ( aq \right )} \;+ \;2 \;Cl_{\left ( aq \right )}^{-}$$

$$CuCl_{2\;\left ( aq \right )} \;\rightarrow \;Cu^{2+}_{\left ( aq \right )} \;+ \;2 \;Cl_{\left ( aq \right )}^{-}$$

On electrolysis, either of $$Cu^{2+}$$ ions or $$H_{2}O$$ molecules can get decreased at cathode. But the decreased potential of

$$Cu^{2+}$$ is more than that of $$H_{2}O$$ molecules.

$$Cu^{2+}_{\left ( aq \right )} \;+ \;2 \;e^{-} \;\rightarrow \;Cu_{\left ( aq \right )}E^{\circ} \; = \; +0.34V$$

$$H_{2}O_{\left ( l \right )} \;+ \;2 \;e^{-} \;\rightarrow \;H_{2\;\left ( g \right )} \;+ \; 2 \;OH^{-}$$

Therefore, $$Cu^{2+}$$ ions are decreased at cathode and get deposited. In the same way, at anode, either of $$Cl^{-}$$ or $$H_{2}O$$ is oxidized. The oxidation potential of
is higher than that of $$Cl^{-}.$$

$$2 \;Cl^{-}_{\left ( aq \right )} \;\rightarrow \;Cl_{2\;\left ( g \right )} \;+ \;2 \;e^{-}; E^{\circ} \; = +0.34V$$

$$2\; H_{2}O_{\left ( l \right )} \;\rightarrow \;O_{2\;\left ( g \right )} \;+ \;4 \;H^{+}_{\left ( aq \right )} \;+ \;4 \;e^{-}$$

But oxidation of $$H_{2}O$$ molecules occurs at a lower electrode potential compared to that of $$Cl^{-}$$ ions because of over-voltage (extra voltage required to liberate gas). As a result, $$Cl^{-}$$ ions are oxidized at the anode to liberate $$Cl_{2}$$​ gas.

Answered by Abhisek | 1 year ago

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