# Question ef585

Feb 12, 2016

(3). 10%

#### Explanation:

The relationship between the 1/2 cell potential and concentration for dilute solutions is given by:

$E = {E}^{\circ} + \frac{R T}{n F} \times 2.303 \log \left(\left[\left[\text{oxidised"]]/[["reduced}\right]\right]\right)$

Where $n$ is the number of moles of electrons transferred which, in this case = 1.

For the $\text{Ag"^+"/Ag}$ 1/2 cell this becomes:

$E = {E}^{\circ} + \frac{R T}{F} \times 2.303 \log \left[A {g}^{+}\right]$

Putting in the values $\Rightarrow$

$0.62 = \left(+ 0.8\right) + 0.06 \log \left[A {g}^{+}\right]$

$\therefore \log \left[A {g}^{+}\right] = - \frac{0.18}{0.06} = - 3.0$

From which:

$\left[A {g}^{+}\right] = 0.001 \text{mol/l}$

To find the number of moles $n$ we know that:

$c = \frac{n}{v}$

$\therefore n = c \times v = 0.001 \times \frac{100}{1000} = 0.0001$

To convert this to grams ${m}_{A g}$ we multiply by the mass of 1 mole of silver:

${A}_{r} \left[A g\right] = 108$

$\therefore {m}_{A g} = 0.0001 \times 108 = 0.0108 \text{g}$

So the percentage by mass of silver in the alloy is given by:

%"Ag"=(0.0108)/(0.108)xx100=10#

So the answer is $\left(3\right)$.