Chemistry, 01.08.2019 12:00 mcclendoncassandra

Based on your measured potential for this cell and the literature value for the standard reduction potential for the ag/agcl reference electrode, what would you expect the overall potential to be for the spontaneous reaction between your cu2+/cu electrode and a standard hydrogen eletrode? type your calculation for the expected standard reduction potential vs the she as well as the % error between this value and the literature value.

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Answer from: 1937

I have attached an image that shows the measured cell potentials, and this question specifically refers to cell #4 with the following reaction:

Cu²⁺(aq) + 2Ag(s) → Cu(s) + 2Ag⁺(aq) Ecell = 0.137 V

We can separate this reaction into its two half reactions to identify the anode and cathode:

Cathode: Cu²⁺ + 2e⁻ → Cu E° = 0.34 V
Anode: Ag → Ag⁺ + 1e⁻ E° = 0.22 V

The formula for cell potential is:

Ecell = Ecathode - Eanode

We are asked to solve for the cell potential of the Cu²⁺/Cu electrode using the standard Ag/AgCl electrode potential and the measured potential:

0.137 V = Ecathode - 0.22 V
Ecathode = 0.137 + 0.22
Ecathode = 0.357 V

Now we can compare this value for the Cu²⁺/Cu electrode with the standard hydrogen electrode:

Ecell = 0.357 V - 0 V
Ecell = 0.357 V

We found a value of 0.357 V, however, the literature value is 0.34 V and now we can solve for the % error in our experimental value:

% error = ([experimental - theoretical]/[theoretical]) x 100%

% error = ((0.357 - 0.34)/0.34) x 100%

% error = 5 %

The experimental electrode potential for the Cu²⁺/Cu electrode was 0.357 V which has a 5% error compared to the literature value.
Based on your measured potential for this cell and the literature value for the standard reduction p

Based on your measured potential for this cell and the literature value for the standard reduction p

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Answer from: Quest

Ireally don’t know bruh

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