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Monica [59]
4 years ago
9

Based on your measured potential for this cell and the literature value for the standard reduction potential for the ag/agcl ref

erence 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.

Chemistry
1 answer:
Marta_Voda [28]4 years ago
4 0
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.

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Answer:

The weight percent of potassium carbonate is 50,8 wt% and of potassium bicarbonate 49,2 wt%

Explanation:

The reactions of potassium carbonate (K₂CO₃) and potassium bicarbonate (KHCO₃) with HCl produce:

K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O

KHCO₃ + HCl → KCl + CO₂ + H₂O

That means that you need 2 moles of HCl to titrate potassium carbonate and 1 mol to titrate potassium bicarbonate.

The moles of HCl to titrate the mixture are:

0,03416L×\frac{0,762mol}{1L} = <em>0,02603 mol of HCl</em>

If X is mass of K₂CO₃ and Y is mass of KHCO₃ in the mixture, the moles of HCl to titrate the mixture are equals to:

0,02603 mol = 2X×\frac{138,2058 g}{1mol} + Y×\frac{100,1154 g}{1mol} <em>(1)</em>

As the mass of the mixture is 2,122g:

2,122g = X + Y <em>(2)</em>

Replacing (2) in (1):

0,02603 mol = 0,01447 (2,122-Y) + 9,988x10⁻³Y

0,02603 mol = 0,0307 - 0,01447Y + 9,988x10⁻³Y

-4,6778x10⁻³ = -4,4827x10⁻³Y

1,044g = Y <em>-mass of potassium bicarbonate-</em>

Thus:

X = 1,078g <em>-mass of potassium carbonate-</em>

The weight percent of potassium carbonate is:

\frac{1,078g}{2,122g}×100 =<em> 50,8 wt%</em>

The weight percent of potassium bicarbonate is:

\frac{1,044g}{2,122g}×100 = <em>49,2 wt%</em>

<em></em>

I hope it helps!

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Answer:

Q=87924J

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Hello,

At first, we must consider that at the beginning, both copper and silver are solid, so we must melt both of them, thus, the following enthalpic routes are proposed in order to know each heat because coppers melting points as a value of 1356K and silver's melting point 1235K:

H_{Cu}=n_{Cu}*(Cp_{solid,Cu}(T_2-T_1)+H_{melt})\\H_{Cu}=1mol*(0.3768J/gK*63.546g/mol*(1356-298)K+13590J/mol) \\H_{Cu}=38922.89J

H_{Ag}=n_{Ag}*(Cp_{solid,Ag}(T_m-T_1)+H_{melt}+Cp_{liquid,Ag}(T_2-T_m))\\H_{Ag}=1mol*(0.235J/gK*107.868g/mol*(1235-298)K+11300J/mol+0.28J/gK*107.868g/mol*(1356-1235)K) \\H_{Ag}=38706.56J

Now, the released heat due to the mixing process turns out into:

H_{mix}=2mol*(20590x_{Cu}x_{Ag}J/mol)\\H_{mix}=2mol*20590(0.5)(0.5)J/mol\\H_{mix}=10295J

Finally, the total heat required is found by:

Q=H_{Cu}+H_{Ag}+H_{mix}\\Q=38922.89J+38706.56J+10295J\\Q=87924J

Best regards.

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