A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2+(aq) and Mg(s), producing Cd(s) and Mg2+(aq). Wr
2 answers:
half-reactions
cathode : Cd²⁺ (aq) + 2e⁻ ---> Cd (s)
anode : Mg (s) → Mg²⁺ (aq) + 2e−
a balanced cell reaction
Cd²⁺(aq) + Mg(s)→ Cd(s) + Mg²⁺ (aq)
<h3>Further explanation </h3>Cell potential (E °) is the potential difference between the two electrodes in an electrochemical cell.
Electric current moves from a high potential pole to a low potential, so the cell potential is the difference between an electrode that has a high electrode potential (cathode) and an electrode that has a low electrode potential (anode)
or:
E ° cell = E ° reduction-E ° oxidation
(At the cathode the reduction reaction occurs, the anode oxidation reaction occurs)
The value of E cells uses a reference electrode which is used as a comparison called the Standard Electrode which is the hydrogen-platinum electrode
In reaction:
Cd²⁺ + Mg → Cd + Mg²⁺
half-reactions
- at the cathode (reduction reaction) Cd²⁺ (aq) + 2e⁻ ---> Cd (s)
- at the anode (oxidation reaction) Mg (s) → Mg²⁺ (aq) + 2e−
a balanced cell reaction
Cd²⁺(aq) + Mg(s)→ Cd(s) + Mg²⁺ (aq)
<h3>Learn more </h3>The standard cell potential
Answer : The balanced two-half reactions will be,
Oxidation half reaction (anode) :
Reduction half reaction (cathode) :
Thus the overall reaction will be,
Explanation :
Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.
The given redox reaction occurs between the magnesium and cadmium.
In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.
The balanced two-half reactions will be,
Oxidation half reaction (anode) :
Reduction half reaction (cathode) :
Thus the overall reaction will be,
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Answer:
The wavelength of the light decreases as it enters into the medium with the greater optical density.
The frequency of the light remains constant as it transitions between materials.
Explanation:
- When a ray of light crosses a boundary between two different materials, it undergoes refraction: the ray changes direction, and it also changes speed, according to the relationship:
where v is the speed of the ray of light in the material, c is the speed of light in a vacuum, n is the index of refraction of the material, which is larger for a medium with larger optical density. So, from the equation, we see that the larger the optical density, the smaller the speed of the wave.
- The frequency of the light does not depend on the properties of the medium, so it remains unchanged: therefore the statement
The frequency of the light remains constant as it transitions between materials.
is correct.
- Moreover, the wavelength of the ray of light is related to the speed and the frequency by the equation
where v is the speed and f the frequency. Since we have seen that v decreases and f remains constant, this means that the wavelength decreases as well, so the statement
The wavelength of the light decreases as it enters into the medium with the greater optical density.
is also correct.
Answer:
Thus, if field were sampled at same distance, the field due to short wire is greater than field due to long wire.
Explanation:
The magnetic field, B of long straight wire can be obtained by applying ampere's law
I is here current, and r's the distance from the wire to the field of measurement.
The magnetic field is obviously directly proportional to the current wire. From this expression.
As the resistance of the long cable is proportional to the cable length, the short cable becomes less resilient than the long cable, so going through the short cable (where filled with the same material) is a bigger amount of currents. If the field is measured at the same time, the field is therefore larger than the long wire because of the short wire.