When gold ores are leached with solutions, gold forms a complex ion,$\mathbf{\text{Au(CN}}{{\mathbf{\text{)}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{-}}}$ , (a) Find ${\mathbf{\text{E}}}_{\mathbf{\text{cell}}}$for the oxidation in air$\left({\text{P}}_{{\text{O}}_{\text{2}}}\text{= 0.2l}\right)$ of${\mathbf{\text{Au}}}^{}$ torole="math" localid="1663334129218" ${\mathbf{\text{Au}}}^{\mathbf{\text{+}}}$ in basic $\mathbf{\text{(pH13.55)}}$solution with$\left[{\text{Au}}^{\text{+}}\right]\mathbf{\text{= 0.50M}}$ Is the reaction${\mathbf{\text{Au}}}^{\mathbf{\text{+}}}{\mathbf{\text{(aq) +e}}}^{\mathbf{\text{-}}}\mathbf{\to }\mathbf{\text{Au(s)}}$ ${\mathit{E}}^{\mathbf{°}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{68}\mathit{V}$spontaneous? (b) How does formation of the complex ion change so that the oxidation can be accomplished?

Oxidation is the reaction triggered by contact between compounds and oxygen molecules. Such substances may be metals or non-metals, such as living tissues. In a more technical sense, oxidation is defined as the loss of one electron during the interaction of two or more elements.

(a)

For the oxidation of gold in air to gold ion, the cell potential is computed as

$$\begin{gathered} pH=13.55 \\ pOH=14.00-pH \\ =0.45 \\ \left[O{H}^{-}\right]={10}^{-0.45} \\ =0.3548M \\ {E}^{°}={E^{°}}_{O_2}-{E_{Au}}^{°} \\ =0.40V-1.68\text{V} \\ E=E^{°}-\frac{0.0592}{n}\log Q \\ =-1.28\text{V}-\frac{0.0592}{n}\log \frac{{[0.50]}^4{[0.3458]}^4}{[0.21]} \\ =1.25 \end{gathered}$$

Because the voltage is negative, this reaction is not spontaneous.

Therefore, the oxidation of gold in air to gold ion, the cell potential is$E=1.25$ .

(b)

Qualitatively, the addition of cyanide ion will lower the concentration of free gold ion by forming the goldcyanide complex ion. If the concentration is decreased, the second term offsets the first and the cell potential becomes favorable.

Hence, If the concentration is decreased, the second term offsets the first and the cell potential becomes favorable.