Answer : The correct option is, (b) +0.799 V
Solution :
The values of standard reduction electrode potential of the cell are:
![E^0_{[H^{+}/H_2]}=+0.00V](https://tex.z-dn.net/?f=E%5E0_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D%3D%2B0.00V)
![E^0_{[Ag^{+}/Ag]}=+0.799V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.799V)
From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) :
Reaction at cathode (reduction) :
The balanced cell reaction will be,

Now we have to calculate the standard electrode potential of the cell.

![E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D)

Therefore, the standard cell potential will be +0.799 V
Answer:
150ml
Explanation:
For this question,
NaOH completely dissociates. It is a strong base
HCl also completely dissociates. It is a strong acid
So we have this equation
m1v1 = m2v2 ----> equation 1
M2 = 2m
V1= ??
M2 = 6m
V2 = 50m
When we input these into equation 1, we have:
2m x v1 = 6m x 50ml
V1 = 6m x 50ml/2
V1 = 300/2
V1 = 150ml
Therefore NaOH that is required to neutralize the solution of hydrochloric acid is 150ml.
Thank you
Answer:
A
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yes
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Answer:
It’s b even check quizzes
Explanation: