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.
(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 agoFluorine reacts with ice and results in the change:
\( H_{ 2 }O_{ (s) } \; + \; F_{ 2 \; (g) } \; \rightarrow \; HF_{ (g) } \; + \; HOF_{ (g) } \)
Justify that this reaction is a redox reaction
Depict the galvanic cell in which the reaction is:
\( Zn_{\left ( s \right )} \;+ \;2 \;Ag_{\left ( aq \right )}^{+} \;\rightarrow \;Zn^{2+}_{\left ( aq \right )} \;+ \;2 \;Ag_{\left ( s \right )} \)
Further show:
(i) which of the electrode is negatively charged?
(ii) the carriers of the current in the cell.
(iii) individual reaction at each electrode.
Given the standard electrode potentials,
\( \dfrac{ K^{+}}{K}\)= –2.93V
\( \dfrac{ Ag^{+}}{Ag}\) = 0.80V
\( \dfrac{Hg^{2+}}{Hg}\) = 0.79V
\(\dfrac{ Mg^{2+}}{Mg}\) = –2.37V
\(\dfrac{Cr^{3+}}{Cr}\) = –0.74V
Arrange these metals in their increasing order of reducing power.
Arrange the given metals in the order in which they displace each other from the solution of their salts.
Al, Fe, Cu, Zn, Mg
Using the standard electrode potentials given, predict if the reaction between the following is feasible:
(a) \( Fe^{ 3+ }_{ (aq) }+ and\; I^{ – }_{ (aq) }\)
(b) \( Ag^{ + }_{ (aq) }+ and\; Cu^{ }_{ (s) }\)
(c) \( Fe^{ 3+ }_{ (aq) }+and\; Cu^{ }_{ (s) }\)
(d) \( Ag^{ }_{ (s) } and \;Fe^{ 3+ }_{ (aq) }\)
(e) \( Br_{ 2 \; (aq) }and\; Fe^{ 2+ }_{ (aq) }\)