I assume you’re looking for a balanced equation.
SiCl4 + 2H2O = SiO2 + 4HCl
Answer:
All are having different valent cation and anion like mono,di and trivalent polyatomic ions .
A. RbNO3
B. K2S
C. NaHS
D. Mg3(PO4)2 formed by divalent Mg+2 and trivalent PO43-
E. CaHPO4
F. PbCO3 , lead is in Pb+2 form
G. SnF2
H. (NH4)2SO4
I. AgClO4
J. BCl3
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
173.1f is the answer I believe, please let me know if I'm wrong then I would try to make up for it
That number has 6 significant numbers. It's precision would be 602.
Hope it helped.