An accurate value of the Faraday constant is $96485 \mathrm{C} {\mathrm{mol}}^{-1}$. An accurate value for the charge on one electron is $1.6022 \times {10}^{-19} \mathrm{C}$. Use these values to calculate a value of the Avogadro constant to $5$ significant figures.

Refer to the list of electrode potentials below to answer the question.

${\mathrm{Ag}}^{+} \left(\mathrm{aq}\right) +{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Ag}\left(\mathrm{s}\right)$          voltage $= +0.80 \mathrm{V}$

${\mathrm{Co}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Co}\left(\mathrm{s}\right)$        voltage $= -0.28 \mathrm{V}$

${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Cu}\left(\mathrm{s}\right)$    voltage $= +0.34 \mathrm{V}$

${\mathrm{Pb}}^{2+} \left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Pb}\left(\mathrm{s}\right)$   voltage $= -0.13 \mathrm{V}$

${\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Zn}\left(\mathrm{s}\right)$     voltage $= -0.76 \mathrm{V}$

Which metal in the list is the best reducing agent?

Refer to the list of electrode potentials below to answer the question.

${\mathrm{Ag}}^{+} \left(\mathrm{aq}\right) +{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Ag}\left(\mathrm{s}\right)$          voltage $= +0.80 \mathrm{V}$

${\mathrm{Co}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Co}\left(\mathrm{s}\right)$      voltage $= -0.28 \mathrm{V}$

${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Cu}\left(\mathrm{s}\right)$    voltage $= +0.34 \mathrm{V}$

${\mathrm{Pb}}^{2+} \left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Pb}\left(\mathrm{s}\right)$   voltage $= -0.13 \mathrm{V}$

${\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Zn}\left(\mathrm{s}\right)$     voltage $= -0.76 \mathrm{V}$

Which metal ion in the list is most difficult to reduce?

Refer to the list of electrode potentials below to answer the question.

${\mathrm{Ag}}^{+} \left(\mathrm{aq}\right) +{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Ag}\left(\mathrm{s}\right)$          voltage $= +0.80 \mathrm{V}$

${\mathrm{Co}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Co}\left(\mathrm{s}\right)$     voltage $= -0.28 \mathrm{V}$

${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Cu}\left(\mathrm{s}\right)$    voltage $= +0.34 \mathrm{V}$

${\mathrm{Pb}}^{2+} \left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Pb}\left(\mathrm{s}\right)$   voltage $= -0.13 \mathrm{V}$

${\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Zn}\left(\mathrm{s}\right)$     voltage $= -0.76 \mathrm{V}$

Which metal in the list is most reactive?

Refer to the list of electrode potentials below to answer the question.

${\mathrm{Ag}}^{+} \left(\mathrm{aq}\right) +{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Ag}\left(\mathrm{s}\right)$         voltage $= +0.80 \mathrm{V}$

${\mathrm{Co}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Co}\left(\mathrm{s}\right)$     voltage $= -0.28 \mathrm{V}$

${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Cu}\left(\mathrm{s}\right)$    voltage $= +0.34 \mathrm{V}$

${\mathrm{Pb}}^{2+} \left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Pb}\left(\mathrm{s}\right)$   voltage $= -0.13 \mathrm{V}$

${\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Zn}\left(\mathrm{s}\right)$     voltage $= -0.76 \mathrm{V}$

Which metal ion in the list is the easiest to reduce?

Suggest why aqueous silver nitrate is not used in a salt bridge when connecting a half-cell containing $\mathrm{Zn}$ and $1.00 \mathrm{mol} {\mathrm{dm}}^{-3} {\mathrm{ZnCl}}_2\left(\mathrm{aq}\right)$ to another half-cell.

Write half-equation for the reaction taking place in the half-cell below. Write the equation as a reduction (electrons on the left-hand side of the equation). 
${\mathrm{Cr}}^{2+} / {\mathrm{Cr}}^{3+}$

Write half-equation for the reaction taking place in the half-cell below. Write the equation as a reduction (electrons on the left hand side of the equation).
${\mathrm{Br}}_2 /2{\mathrm{Br}}^{-}$

Write half-equation for the reaction taking place in the half-cell below. Write the equation as a reduction (electrons on the left hand side of the equation).
${\mathrm{O}}_2 + {\mathrm{H}}_2\mathrm{O}/{\mathrm{OH}}^{-}$ (make sure that you balance the equation)