Chemistry: The ksp of copper(ii) ferrocyanide (cu2[fe(cn)6]) is 1.3 × 10−16 at 25°c. determine the potential of a concentration cell in which one half-cell consists of a copper electrode in 1.00 m copper(ii) nitrate, and the other consists of a copper electrode in a saturated solution of cu2[fe(cn)6]. ferrocyanide, ([fe(cn)6]4−), is a complex ion.

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The ksp of copper(ii) ferrocyanide (cu2[fe(cn)6])

Ответ:

briibearr

12.01.2022

Chemistry

Explanation:

Expression for $K_{sp}$ of the given reaction is as follows.

$K_{sp} = [Cu^{2+}]^{2}[Fe(CN)_{6}]$

Let us assume that the concentration of given species is "s". As the value of $K_{sp}$ is given as $1.3 \times 10^{-16}$ .

$K_{sp} = [Cu^{2+}]^{2}[Fe(CN)_{6}]$

$1.3 \times 10^{-16} = s^{2} \times s$

$s^{3} = 1.3 \times 10^{-16}$

s = $3.19 \times 10^{-6}$

Therefore, concentration of $Cu^{2+}$ will be calculated as follows.

$Cu^{2+}$ = 2s

= $2 \times 3.19 \times 10^{-6}$

= $6.38 \times 10^{-6}$ M

Now, we will calculate the value of $E_{cell}$ as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0591}{2} \times log \frac{6.38 \times 10^{-6}}{1}$

= $0 - \frac{0.0591}{2} \times log \frac{6.38 \times 10^{-6}}{1}$

= 0.1535 V

Thus, we can conclude that the potential of given cell is 0.1535 V.

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