# Use half-cell potentials to calculate K_(eq) for the reaction below?

## $2 C u$ (s)$+ 2 {H}^{+}$ (aq)$\setminus \to {H}_{2}$ (g) E°_{Cu\toCu^(2+)}=-0.34 V My work: Jul 29, 2018

${K}_{e q} = 2.16 \times {10}^{-} 12$

#### Explanation:

$C {u}^{o} \left(s\right) + 2 {H}^{+} \left(a q\right) \implies C {u}^{+ 2} \left(a q\right) + {H}_{2} \left(g\right)$

Using ${E}^{o} = \left(\frac{0.0592}{n}\right) \log {K}_{e q}$ => $\log {K}_{e q} = \left(\frac{n \cdot {E}^{o}}{0.0592}\right)$

For above reaction:
n = 2 and
${E}^{o} \left(c e l l\right) = {E}^{o} \left(\text{Reduction") - E^o("Oxidation}\right)$ = (0.00 volt) - (+0.34 volt) = -0.34 volt

$\log {K}_{e q} = \frac{2 \left(- 0.34\right)}{0.0592} = - 11.5$

${K}_{e q} = {10}^{- 11.5} = 3.16 \times {10}^{-} 12$

Jul 30, 2018

Well, for this nonspontaneous reaction... $K = 3.16 \times {10}^{- 12}$.

First of all, let's re-balance the reaction, which should be:

${\text{Cu"(s) -> "Cu}}^{2 +} \left(a q\right) + 2 {e}^{-}$, ${E}_{red}^{\circ} = + \text{0.34 V}$
$\underline{2 {\text{H"^(+)(aq) + 2e^(-) -> "H}}_{2} \left(g\right)}$, ${E}_{red}^{\circ} = \text{0.00 V}$
$2 {\text{Cu"(s) + 2"H"^(+)(aq) -> 2"Cu"^(2+)(aq) + "H}}_{2} \left(g\right)$

The reduction potential of copper cation is more positive, so copper would rather be reduced.

Hence, this reaction is nonspontaneous as-written if we try our hardest and certainly fail to oxidize copper using $\text{HCl}$.

The (nonspontaneous) cell potential is:

${E}_{c e l l}^{\circ}$

$= {\overbrace{{E}_{red}^{\circ}}}^{\text{reduction" +overbrace(E_(o x)^@)^"oxidation}}$, OR overbrace(E_("cathode")^@)^"reduction" - overbrace(E_"anode"^@)^"reduction"

= "0.00 V" + (-"0.34 V"), OR "0.00 V" - (+"0.34 V")

$= - \text{0.34 V}$

From this we obtain the positive $\Delta {G}_{r x n}^{\circ}$:

$\Delta {G}^{\circ} = - n F {E}_{c e l l}^{\circ}$

= -(2 cancel("mol e"^(-)))/("1 mol Cu") cdot "96485 C/"cancel("mol e"^(-)) cdot (-"0.34 V")

$=$ $\text{65610 J/mol}$

As a result, since we are at equilibrium,

${\cancel{\Delta G}}^{0} = \Delta {G}^{\circ} + R T \ln {\cancel{Q}}^{K}$

and thus, we get the physically impractical equilibrium constant:

$\textcolor{b l u e}{K} = {e}^{- \Delta {G}^{\circ} / R T}$

= e^(-(+"65610 J/mol")//("8.314 J/mol"cdot"K" cdot "298 K")

$= {e}^{- 26.5}$

$= \textcolor{b l u e}{3.16 \times {10}^{- 12}}$