Question

If
x
g of silver are deposited in cell 1, what volume of oxygen, in dm3 at STP, is given off in cell 2?

Ar(Ag) = 108; Molar volume of an ideal gas at STP = 22.7 dm3 mol−1

A.
x
108
×
1
4
×
22.7
B.
x
108
×
4
×
22.7
C.
x
108
×
1
2
×
22.7
D.
x
108
×
2
×
22.7

Answer 1

Answer:

Explanation:

Oops

Answer 2

Hey Dear,

◆ Answer -

x×22.4 / 4×108 dm^3 O2

● Explaination -

Half-cell reaction in cell 1 (cathode),

Ag+ + e- --> Ag

Half-cell reaction in cell 2 (anode),

2O-- --> O2 + 4e-

Overall balanced cell reaction -

4Ag+ + 2O-- --> 4Ag + O2

That is

4 mol Ag => 1 mol O2

4×108 g Ag => 22.4 dm^3 O2

1 g Ag => 22.4/(4×108) dm^3 O2

x g Ag => x×22.4/4×108 dm^3 O2

Therefore, x g Ag in cell 1 gives off (x×22.4)/(4×108) dm^3 O2 in cell 2.

Hope this helps you.