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Answer: <span>A-Ce is oxidized because it is losing electrons and Cu is reduced because it is gaining electrons</span><span>.
</span>There are two reactions in the equation, oxidation and reduction. A molecule that oxidized will lose electrons while the molecule that reduced will gain electrons. In this case, Cu2+ changed into Cu which means its oxidation number reduced from +2 into 0. Ce oxidation number increased from 0 into +3
The answer would be Rocks, metals, hydrogen compounds, hydrogen and helium, all in gaseous form.
Answer:
- <em>The mystery substance is</em> <u>C. Bromine (Br) </u>
Explanation:
<em>Argon (Ar) </em>is a noble gas. Whose freezing point is -189 °C (very low), thus it cannot be the frozen substance. Also, it is not reactive, thus is would have not reacted with iron. Hence, argon is not the mystery substance.
<em>Scandium (Sc) </em>is a metal from group 3 of the periodic table, thus is will not react with iron. Thus, scandium is not the mystery substance.
Both <em>bromine</em> and <em>iodine</em> are halogens (group 17 of the periodic table).
The freezing point of bromine is −7.2 °C, and the freezing point of iodine is 113.7 °C. Thus, both could be solids (frozen) in the lab.
The reactivity of the halogens decrease from top to bottom inside the group. Bromine is above iodine. Then bromine is more reactive than iodine.
Bromine is reactive enough to react with iron. Iodine is not reactive enough to react with iron.
You can find in the internet that bromine vapour over hot iron reacts producing iron(III) bromide. Also, that bromine vapors are red-brown.
Therefore, <em>the mystery substance is bromine (Br).</em>
Answer: 0.18 V
Explanation:-
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
=-0.40V[/tex]
=-0.24V[/tex]
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials.
![E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BNi%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BCd%5E%7B2%2B%7D%2FCd%5D%7D)
Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCd%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential = 0.16 V
![E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D0.16-%5Cfrac%7B0.0592%7D%7B2%7D%5Clog%20%5Cfrac%7B%5B0.10%5D%7D%7B%5B0.5%5D%7D)
Thus the potential of the following electrochemical cell is 0.18 V.
