Question

For the electrochemical cell \(\operatorname{Pt}(s) \mid \begin{aligned}&\mathrm{H}_{2}(g)\left|\mathrm{H}^{+}(1 M) \| \mathrm{Cu}^{2+}(1 M)\right| \mathrm{Cu}(s), \text { which one of the } \\\&1 \text { atm }\end{aligned}\) following statements is true ? (a) \(\mathrm{H}_{2}\) liberated at anode and \(\mathrm{Cu}\) is deposite at cathode. (b) \(\mathrm{H}_{2}\) liberated at cathode and \(\mathrm{Cu}\) is deposite at anode. (c) Oxidation occurs at cathode. (d) Reduction occurs at anode.

Short answer

Statement (a) is correct: H2 is liberated at the anode and Cu is deposited at the cathode.

Step-by-step solution

Understand the composition of the electrochemical cell

The given electrochemical cell can be represented by the notation: Pt(s) | H2(g) | H+(1M) || Cu2+(1M) | Cu(s). This signifies that on the left side of the cell, hydrogen gas is in contact with a platinum electrode, and hydrogen ions are present in the solution. On the right side, copper ions are in solution with a copper electrode.

Recognize the half-reactions at each electrode

At the anode, oxidation occurs, which means the loss of electrons. The half-reaction here would likely involve hydrogen gas releasing electrons to form hydrogen ions:2H2(g) -> 4H+(aq) + 4e-.At the cathode, reduction occurs, which is the gain of electrons. The copper ions in the solution would accept electrons to form solid copper:Cu2+(aq) + 2e- -> Cu(s).

Determine the reactions at the anode and cathode

According to the half-reactions, hydrogen gas is released at the anode (oxidation), and copper is deposited at the cathode (reduction).

Analyze the given statements against the half-reactions

By comparing the determined half-reactions to the given statements, it is observed that statement (a) H2 is liberated at the anode and Cu is deposited at the cathode correctly matches the half-reactions that take place in the electrochemical cell.

Validate the other statements

Assessing the other statements: (b) is incorrect since H2 is not liberated at the cathode; (c) is incorrect because oxidation, not reduction, occurs at the anode; and (d) is incorrect because reduction, not oxidation, occurs at the cathode.

Key concepts

Oxidation and Reduction Reactions

Understanding oxidation and reduction is crucial in electrochemistry, and to keep it simple, we remember one main idea: oxidation is the loss of electrons, and reduction is the gain of electrons. An easy way to recollect this is with the mnemonic 'OIL RIG' - Oxidation Is Loss, Reduction Is Gain.

These reactions occur simultaneously in an electrochemical cell; when one species loses electrons (oxidation), another one gains them (reduction). In the given exercise, the hydrogen is oxidized as it loses electrons, forming hydrogen ions. On the other side, metals like copper undergo reduction as they gain electrons to form solid metal from their ion state. This interplay is what drives the flow of electrons and generates electric current in an electrochemical cell.

Half-Reactions in Electrochemistry

Half-reactions help us to see what happens at each electrode in an electrochemical cell separately. Breaking down the overall reaction into two halves, one for oxidation and one for reduction, makes it easier to analyze the process.

In our exercise, the oxidation half-reaction can be expressed as: \[2\text{H}_2(g) \rightarrow 4\text{H}^+(aq) + 4e^-\]. This signifies hydrogen gas giving up electrons. Conversely, the reduction half-reaction is: \[\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\]. Here, copper ions are gaining electrons to become solid copper. These half-reactions are fundamental to understanding how electrochemical cells operate and how the electrode processes determine the flow of electric current.

Electrode Processes in Chemistry

Electrodes are where the magic of electrochemistry happens – they're the sites of oxidation and reduction. Each electrode has a distinct role; anode is for oxidation, and cathode for reduction.

In the text of the solution provided, we learn that the anode is where hydrogen is oxidizing and releasing electrons, whereas the cathode is where copper ions are reducing to form copper metal. It's essential to connect the dots between the half-reactions and these electrode processes to fully grasp how electrochemical cells function. This involves exploring the movement and transfer of electrons and how they contribute to the cell’s electrical output. Understanding these processes underlies much of modern chemistry and is key in applications like batteries, corrosion prevention, and even the production of metals.