Question

3. If x electrons are needed to displace 108 g silver from a solution which contains Ag ions,

how many electrons are needed to displace 9 g of aluminium from a solution which contains
Al ions?

[Relative atomic mass Al, 27; Ag, 108]
A X
B 3x
C 4x
D 9x​

Answer 1

Choice A.
`x` electrons would be required for displacing
`9\; \rm g` of aluminum from a solution of
`\rm Al^(3+)` ions.

Assumption: by "
`\rm Ag` ions" the question meant
`\rm Ag^(+)` with a charge of
`+1` on each ion.

Step-by-step explanation:

The question states that the relative atomic mass of
`\rm Ag` is
`108`. In other words, each mole of

Therefore, that
`108\; \rm g\!` of silver that were formed would contain
`1\; \rm mol` of silver atoms.

Metallic silver would precipitate out of this
`\rm Ag^(+)` solution only after these ions are turned into
`\rm Ag` atoms.

One
`\rm Ag^(+)` ion carries one unit of positive electrical charge. On the other hand, each
`e^(-)` carries one unit of negative electrical charge.

Therefore, each
`\rm Ag^(+)\!` ion will need to gain one electron to form a neutral
`\rm Ag` atom.

`{\rm Ag^(+)}\; (aq) + e^(-) \to {\rm Ag}\; (s)`.

At least
`1\; \rm mol` of electrons would be required to turn
`1\; \rm mol\!` of
`\rm Ag^(+)` ions into that
`1\; \rm mol\!\!` of silver atoms (which have a mass of
`108\; \rm g\!`.)

Hence,
`x = 1\; \rm mol`.

Unlike
`\rm Ag^(+)` ions, each aluminum ion
`\rm Al^(3+)` carries three units of positive electrical charge. That is three times the amount of charge on one
`\rm Ag^(+)\!` ion. Therefore, three electrons will be required to turn one
`\rm Al^(3+)\!` ion to an
`\rm Al` atom.

`{\rm Al^(3+)}\; (aq) + 3\, e^(-) \to {\rm Al}\; (s)`

The question states that the relative atomic mass of
`\rm Al` is
`27`. Therefore, each mole of
`\rm Al\!` atoms would have a mass
`27\; \rm g`. There would be
`\displaystyle (9\; \rm g)/(27\; \rm g \cdot mol^(-1)) = (1)/(3) \; \rm mol` of atoms in that
`9\; \rm g` of
`\rm Al\!\!`.

It takes
`3\; \rm mol` of electrons to turn one mole of
`\rm Al^(3+)` ions to one mole of
`\rm Al` atoms. Hence,
`\displaystyle (1)/(3)* 3\; \rm mol = 1\; \rm mol` of electrons would be required to produce that
`\displaystyle (1)/(3)\; \rm mol` of
`\rm Al\!` atoms (which has a mass of
`9\; \rm g`) from
`\rm Al^(3+)\!` ions.

That corresponds to the first choice,
`x` electrons.