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
Charge of electron = 1.6×10−¹⁹
(1.6×10−¹⁹)(1×10²) (2e)
= 3.2×10−¹⁷ J
Answer:
3ohms
Explanation:
From Ohm's Law
V = IR
V is that voltage = 3volts
I = current = 1amp
R = resistance in ohms
Putting those values into the above formula.
3volts = 1amp×R
Making R the subject
R = 3/1
R = 3ohms
The resistance of the light bulb is 3ohms.
30 + 6 = 36
36/12 = 3
So, it would take it 3 hours to go 12 kms downstream.
Note that we added 30 and 6 because it was going downstream. If it was going upstream, then we would have had to subtract.
Answer:
A step by step to walk
Explanation:
One- Make sure your shoes are tied so that you dont trip
Two- Make sure your way is a cleared path so you dont fall or even hurt yourself
three- use both set of lets to go in any direction you want.
Four- when walking make sure to try and keep a steedy pace so that both set of legs are going up and down but in harmony
