AP Chemistry 1.7 Periodic Trends- Exam Style questions - FRQs- New Syllabus

(a)
For the correct answer:
\( \mathrm{Ag(s)} \): \( 0 \)      \( \mathrm{Ag_2S(s)} \): \( +1 \)
An element in its standard form has oxidation number \( 0 \). In \( \mathrm{Ag_2S} \), sulfur is typically \( -2 \), so the two silver atoms together must total \( +2 \), giving each silver atom \( +1 \).

(b)(i)
For a valid explanation:
Sterling silver is better classified as a substitutional alloy because silver and copper have similar atomic radii, \( 165\ \mathrm{pm} \) and \( 145\ \mathrm{pm} \), so copper atoms can replace silver atoms in the metal lattice rather than fitting into small holes between them.

(b)(ii)
For a valid explanation:
Silver has more occupied electron shells \((n=5)\) than copper \((n=4)\), so its valence electrons are farther from the nucleus and experience weaker Coulombic attraction. Therefore, silver has the larger atomic radius.

(c)
For the correct calculated mass of \( \mathrm{Ag_2S} \):
\( 409.21\ \mathrm{g} – 398.94\ \mathrm{g} = 10.27\ \mathrm{g} \)

For the correct calculated moles of silver atoms:
\( 10.27\ \mathrm{g\ Ag_2S} \times \dfrac{1\ \mathrm{mol\ Ag_2S}}{247.80\ \mathrm{g\ Ag_2S}} \times \dfrac{2\ \mathrm{mol\ Ag}}{1\ \mathrm{mol\ Ag_2S}} = 0.08289\ \mathrm{mol\ Ag} \)

So, the number of moles of silver atoms removed is \( \boxed{0.0829\ \mathrm{mol}} \).

(d)(i)
For the correct balanced equation (state symbols not required):
\( \mathrm{4\,Rh^{3+}(aq) + 6\,H_2O(l) \rightarrow 4\,Rh(s) + 3\,O_2(g) + 12\,H^+(aq)} \)
The reduction half-reaction is multiplied by \( 4 \), and the oxidation of water is the reverse of the given oxygen reduction half-reaction, multiplied by \( 3 \), so that \( 12 \) electrons cancel.

(d)(ii)
For the correct calculated value, consistent with part \( \mathrm{(d)(i)} \):
\( E^\circ_{\mathrm{cell}} = 0.80\ \mathrm{V} – 1.23\ \mathrm{V} = -0.43\ \mathrm{V} \)

The plating reaction uses the \( \mathrm{Rh^{3+}/Rh} \) reduction and the oxidation of water.
Because the oxidation potential of water is \( -1.23\ \mathrm{V} \) when written from the reduction table, the total cell potential is negative.

(d)(iii)
For a correct explanation, consistent with part \( \mathrm{(d)(ii)} \):
Since \( E^\circ_{\mathrm{cell}} \) is negative, the reaction is not thermodynamically favorable as written. Therefore, an external power source is required to drive the electroplating process.

(e)
For the correct calculated value of moles of electrons:
\( 2.8\ \mathrm{g\ Rh} \times \dfrac{1\ \mathrm{mol\ Rh}}{102.9\ \mathrm{g\ Rh}} \times \dfrac{3\ \mathrm{mol\ e^-}}{1\ \mathrm{mol\ Rh}} = 0.0816\ \mathrm{mol\ e^-} \)

For the correct calculated value of time:
\( 0.0816\ \mathrm{mol\ e^-} \times \dfrac{96{,}485\ \mathrm{C}}{1\ \mathrm{mol\ e^-}} \times \dfrac{1\ \mathrm{s}}{2.0\ \mathrm{C}} = 3.94\times10^3\ \mathrm{s} \)

Therefore, the time required is approximately \( \boxed{3900\ \mathrm{s}} \).
The small difference between \( 3940\ \mathrm{s} \) and \( 3900\ \mathrm{s} \) is due to rounding.