Answer : The cell potential for this cell 0.434 V
Solution :
The balanced cell reaction will be,
Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
![E^o_{[Cu^{2+}/Cu]}=0.34V](https://tex.z-dn.net/?f=E%5Eo_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D0.34V)
![E^o_{[Ag^{+}/Ag]}=0.80V](https://tex.z-dn.net/?f=E%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D0.80V)
![E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}](https://tex.z-dn.net/?f=E%5Eo%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D)
Now we have to calculate the concentration of cell potential for this cell.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCu%5E%7B2%2B%7D%5D%5BAg%5D%5E2%7D%7B%5BCu%5D%5BAg%5E%2B%5D%5E2%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this cell 0.434 V
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An ion is created by the transfer of electrons. The metals give away the elections and become positively charged. The non - metals take on electrons.
Balance.
So an ion is any atom that either gives away or takes on electrons.
Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
T₂ = 283.15k
205 = 2.0 × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.
The height of the object will be -5.19 cm
A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.
Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall
So let,
v = Image distance from the mirror = -33.5 cm
u = object distance from the mirror (concave) = 24 cm
hi = Image height = 7.25 cm
h = height of the object = ?
Using below formula to find height of the object
-v/u = hi/h
Putting all value in the formula we get
-(-33.5)/(-24) = 7.25/h
h = -5.19 cm
Therefore the height of the object will be -5.19 cm
Learn more about Concave mirror here:
brainly.com/question/3727024
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