Electrolysis sends the current through the solution to force a redox reaction that would not happen on its own — cations go to the cathode and get reduced, anions (or water) go to the anode and get oxidized. Which species actually reacts is decided by ease of reduction/oxidation, not by which ion "belongs" to the salt.Two ideas run through all four parts:
1. At the cathode, whichever species is
easiest to reduce (the more positive standard reduction potential, \(\displaystyle E^\circ\)) is discharged first. Relevant values:
\[\text{Ag}^+ + e^- \rightarrow \text{Ag}(s), \quad E^\circ = +0.80\ \text{V}
\]
\[\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s), \quad E^\circ = +0.34\ \text{V}
\]
\[2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g), \quad E^\circ = 0.00\ \text{V}
\]
Since \(\displaystyle \text{Ag}^+\) and \(\displaystyle \text{Cu}^{2+}\) are both easier to reduce than \(\displaystyle \text{H}^+\), they are deposited as metal
before any \(\displaystyle \text{H}_2\) forms — a metal below hydrogen in the electrochemical series always wins at the cathode.
2. At the anode, whichever species is
easiest to oxidize is discharged first — normally the one with the
lowest \(\displaystyle E^\circ\) — but there is a catch: gas evolution (especially \(\displaystyle \text{O}_2\)) needs extra voltage in practice ("overvoltage") beyond what the tables predict, so a poorly-oxidizable ion like \(\displaystyle \text{Cl}^-\) can still beat water to the anode. Also, if the electrode metal itself is easily oxidized (a reactive, "active" electrode such as Ag or Cu, as opposed to an inert one like Pt), the electrode dissolves instead of the solution reacting.
A step people skip: \(\displaystyle \text{NO}_3^-\) and \(\displaystyle \text{SO}_4^{2-}\) are never discharged at an inert anode in dilute aqueous solution — nitrogen and sulfur are already in their highest common oxidation states (+$\displaystyle 5$ and +$\displaystyle 6$), so these ions have nothing further to give up. Water is oxidized in their place.
(i) Aqueous AgNO₃ with silver electrodesHere the electrodes are
active — made of the same metal as the cation in solution. \(\displaystyle \text{Ag}(s)\) is oxidized far more easily (\(\displaystyle E^\circ = +0.80\ \text{V}\)) than water (\(\displaystyle E^\circ = +1.23\ \text{V}\) for \(\displaystyle \text{O}_2/\text{H}_2\text{O}\)), so the silver anode itself dissolves instead of water being oxidized:
Anode (oxidation): \(\displaystyle \text{Ag}(s) \rightarrow \text{Ag}^+(aq) + e^-\)
Cathode (reduction): \(\displaystyle \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)\)
Each equation already balances one silver atom and one electron on each side. Silver metal dissolves off the anode as \(\displaystyle \text{Ag}^+\) ions and an equal amount of silver metal plates back out at the cathode — the \(\displaystyle \text{AgNO}_3\) concentration in solution stays essentially unchanged. This is exactly the principle used to electrorefine silver.
(ii) Aqueous AgNO₃ with platinum electrodesPlatinum is inert, so it cannot dissolve. At the cathode, \(\displaystyle \text{Ag}^+\) is still easier to reduce than \(\displaystyle \text{H}^+\), so silver still plates out:
Cathode: \(\displaystyle \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)\)
At the anode, \(\displaystyle \text{NO}_3^-\) cannot be oxidized further (N is already +$\displaystyle 5$), so water is oxidized instead, releasing oxygen gas:
Anode: \(\displaystyle 2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-\)
To combine them, the cathode step must supply the same $\displaystyle 4$ electrons the anode step releases, so multiply it by $\displaystyle 4$:
\[4\text{Ag}^+(aq) + 4e^- \rightarrow 4\text{Ag}(s)
\]
\[2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-
\]
Adding and cancelling the \(\displaystyle 4e^-\):
\[4\text{Ag}^+(aq) + 2\text{H}_2\text{O}(l) \rightarrow 4\text{Ag}(s) + \text{O}_2(g) + 4\text{H}^+(aq)
\]
Check the balance: $\displaystyle 4$ Ag, $\displaystyle 4$ H, $\displaystyle 2$ O on each side, and charge \(\displaystyle +4\) on both sides. Silver metal deposits at the cathode, oxygen gas is evolved at the anode, and the solution is left with the \(\displaystyle \text{H}^+\) produced (it slowly turns into dilute \(\displaystyle \text{HNO}_3\) as the \(\displaystyle \text{Ag}^+\) is used up).
(iii) Dilute H₂SO₄ with platinum electrodes\(\displaystyle \text{SO}_4^{2-}\) cannot be oxidized (sulfur is already +$\displaystyle 6$), so, just as with \(\displaystyle \text{NO}_3^-\) above, water is oxidized at the anode. At the cathode, the only cation present besides water is \(\displaystyle \text{H}^+\), so hydrogen gas is evolved:
Cathode: \(\displaystyle 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)\)
Anode: \(\displaystyle 2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-\)
Multiply the cathode step by $\displaystyle 2$ so both steps carry $\displaystyle 4$ electrons, then add:
\[4\text{H}^+(aq) + 4e^- \rightarrow 2\text{H}_2(g)
\]
\[2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-
\]
The \(\displaystyle 4\text{H}^+\) and \(\displaystyle 4e^-\) cancel exactly, leaving
\[2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g)
\]
This is simply the electrolysis of water — the \(\displaystyle \text{H}_2\text{SO}_4\) is regenerated, not consumed, and only serves to carry the current. Hydrogen gas collects at the cathode and oxygen gas at the anode, in the $\displaystyle 2$ : $\displaystyle 1$ mole ratio that matches water's own composition.
(iv) Aqueous CuCl₂ with platinum electrodesAt the cathode, \(\displaystyle \text{Cu}^{2+}\) (\(\displaystyle E^\circ = +0.34\ \text{V}\)) is easier to reduce than \(\displaystyle \text{H}^+\) (\(\displaystyle E^\circ = 0.00\ \text{V}\)), so copper metal deposits:
Cathode: \(\displaystyle \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\)
At the anode there is a genuine competition between \(\displaystyle \text{Cl}^-\) and water. By the tabulated potentials, water should be
easier to oxidize (\(\displaystyle E^\circ = +1.23\ \text{V}\) for \(\displaystyle \text{O}_2/\text{H}_2\text{O}\), against \(\displaystyle +1.36\ \text{V}\) for \(\displaystyle \text{Cl}_2/\text{Cl}^-\)) — this is the step where it's easy to guess "oxygen" and be wrong. In practice, oxygen gas evolution on an electrode surface carries a large
overvoltage (an extra energy cost beyond the thermodynamic value), so with an appreciable concentration of \(\displaystyle \text{Cl}^-\) present, chlorine gas is discharged instead:
Anode: \(\displaystyle 2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-\)
The cathode and anode steps both involve exactly $\displaystyle 2$ electrons, so they add directly:
\[\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)
\]
\[2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-
\]
\[\text{Cu}^{2+}(aq) + 2\text{Cl}^-(aq) \rightarrow \text{Cu}(s) + \text{Cl}_2(g)
\]
i.e. \(\displaystyle \text{CuCl}_2(aq) \rightarrow \text{Cu}(s) + \text{Cl}_2(g)\) — one copper and two chlorines on each side, charge zero on each side. Copper metal deposits at the cathode and chlorine gas is liberated at the anode.
Answer: (i) With Ag electrodes, the silver anode dissolves (\(\displaystyle \text{Ag} \rightarrow \text{Ag}^+ + e^-\)) and an equal amount of silver metal deposits at the cathode — the electrode metal itself reacts, and the solution composition is unchanged. (ii) With Pt electrodes, silver metal deposits at the cathode while oxygen gas is evolved at the anode (\(\displaystyle 4\text{Ag}^+ + 2\text{H}_2\text{O} \rightarrow 4\text{Ag} + \text{O}_2 + 4\text{H}^+\)). (iii) Dilute \(\displaystyle \text{H}_2\text{SO}_4\) with Pt electrodes simply electrolyses water: hydrogen gas at the cathode and oxygen gas at the anode, in a $\displaystyle 2$ : $\displaystyle 1$ mole ratio (\(\displaystyle 2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2\)). (iv) With CuCl₂ and Pt electrodes, copper metal deposits at the cathode and chlorine gas is liberated at the anode (\(\displaystyle \text{CuCl}_2 \rightarrow \text{Cu} + \text{Cl}_2\)), chlorine winning out over oxygen because of oxygen's overvoltage.