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
Yeah!! It's possible for an object by changing it's direction....
For this question we should apply
a = v^2 - u^2 by t
a = 69 - 0 by 4.5
a = 69 by 4.5
a = 15.33
a = 6.85 m/s^2
If the answer in option is near to answer then , you can mark it as correct.
.:. The acceleration is 6.9 m/s^2
Answer:
His average speed is 980 units