Cd/Cd2+//H+/H2/Pt
Write the oxidation and reductionhalf reactions. Sketch a possible deisgn cell.
I am unclear on the Pt portion and the charges ?
Cd ==> Cd^+2 + 2e
2H^+ + 2e ==> H2
Oxidation is loss of electrons.
Reduction is the gain of electrons.
Pt is an inert electrode and doesn't enter into the reaction. I would look in the text for how to draw the cell.
Really
The notation Cd/Cd2+//H+/H2/Pt represents a redox reaction that can occur in a cell. Let's break it down step-by-step:
1. Identify the oxidation half-reaction:
Cd -> Cd2+ + 2e-
In this reaction, the Cd atom is being oxidized to Cd2+ ions, and it loses two electrons in the process.
2. Identify the reduction half-reaction:
2H+ + 2e- -> H2
In this reaction, two hydrogen ions (H+) are being reduced to form hydrogen gas (H2), and they gain two electrons in the process.
3. Sketch a possible design for the cell:
Pt | H2(g) | H+ (aq) || Cd2+ (aq) | Cd(s)
The Pt represents a platinum electrode, which is often used as an inert material to allow electron transfer in the cell but does not participate in the reaction. In this case, it is placed on the side of the reduction half-reaction (H2/H+). On the left side of the Pt electrode, hydrogen gas (H2) is reduced to form hydrogen ions (H+). On the right side of the Pt electrode, Cadmium ions (Cd2+) are reduced to form solid cadmium (Cd).
The double vertical lines (||) represent the salt bridge or the barrier between the two sides of the cell. It allows the flow of ions to maintain charge neutrality during the redox reaction.
Note that the direction of electron flow is from the Cd electrode (anode) to the Pt electrode (cathode).
The notation "Cd/Cd2+//H+/H2/Pt" is a simplified representation of an electrochemical cell. Let's break it down:
1. Cd/Cd2+: This represents the Cd electrode that is in contact with a Cd2+ solution. The single slash ("/") indicates a phase boundary, separating the electrode from the solution.
2. H+/H2: This represents the presence of H+ ions (acidic solution) and H2 gas. The double slash ("//") indicates a salt bridge or porous barrier that allows ion flow between the two half-cells without any direct mixing.
3. Pt: This refers to a platinum electrode. Platinum is often used as an inert electrode, meaning it doesn't undergo any chemical reactions itself. It acts as a conductor, allowing electron flow in the cell.
Now let's determine the oxidation and reduction half-reactions:
Oxidation half-reaction: Cd(s) → Cd2+(aq) + 2e-
In this reaction, solid Cd metal loses two electrons and forms Cd2+ ions in the solution. This occurs at the Cd electrode.
Reduction half-reaction: 2H+(aq) + 2e- → H2(g)
In this reaction, H+ ions in the solution gain two electrons and form gaseous H2. This occurs at the Pt electrode.
To sketch a possible design of the cell, you can represent it as follows:
Cd(s) | Cd2+(aq) || H+(aq) | H2(g) | Pt(s)
The vertical lines represent the phase boundaries, and the double vertical line ("||") represents the salt bridge or porous barrier. The metal electrodes (Cd and Pt) are on the ends, and the solutions (Cd2+ and H+) are in the middle.
Remember, this is just one possible design of the electrochemical cell, and there can be variations depending on specific experimental conditions or cell configurations.