Answer:
it would increase to twice what it was
When E° cell is an electrochemical cell which comprises of two half cells.
So,
when we have the balanced equation of this half cell :
Al3+(aq) + 3e- → Al(s) and E°1 = -1.66 V
and we have also this balanced equation of this half cell :
Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V
so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)
when E° = E°2 - E°1
∴E° =0.8 - (-1.66)
= 2.46 V
∴ the correct answer is 2.46 V
<u>Answer:</u> The solubility product of silver (I) phosphate is 
<u>Explanation:</u>
We are given:
Solubility of silver (I) phosphate = 1.02 g/L
To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:
Molar mass of silver (I) phosphate = 418.6 g/mol

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of silver (I) phosphate follows:
3s s
The expression of
for above equation follows:

We are given:

Putting values in above expression, we get:

Hence, the solubility product of silver (I) phosphate is 
Answer is 13.668! Or in scientific notation it would be .13668 x 10^2