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2 years ago

For the following reactions, write a balanced equation using half-reactions and calculate the voltage to be expected.

Chemistry

1 answer:

Iteru [2.4K]2 years ago

8 0

a) $2Na(s) +2H_2O(I)$ → $2Na^+(aq) + OH^-(aq)+ H_2(g)$

b) $2Ag (s) +2H^(aq)$ → $2 Ag^+ (aq) +H_2(g)$

<h3>What are half-reactions?</h3>

The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.

$2Na(s) +2H_2O(I)$ → $2Na^+(aq) + OH^-(aq)+ H_2(g)$

$E^0 cell = E^0 (reduction) - E^0 (oxidation)$

= $E^0(\frac{H_2O}{H_2}, OH^-) -E^0(Na^+/Na)$

= -0.83 - (-2.71) =1.88V

$2Ag (s) +2H^(aq)$ → $2 Ag^+ (aq) +H_2(g)$

$E^0cell$ = $E^0(H^+/H_2) -E^0(Ag^+/Ag)$

$E^0cell$ =-0. - (0.8) =-0.8V

Learn more about the half-reactions here:

https://brainly.in/question/18053421

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3 years ago

Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI, and the rate constant is 9.7×10−6M−

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Answer : The molarity after a reaction time of 5.00 days is, 0.109 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $9.7\times 10^{-6}M^{-1}s^{-1}$

t = time taken = 5.00 days

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$[A]_o$ = Initial concentration = 0.110 M

Now put all the given values in above equation, we get:

$9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)$

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3 years ago

The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and

Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

<em>...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).</em>

<em />

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

<em>Moles KSCN = Moles Ag⁺ in the filtrate:</em>

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

<em>Total moles Ag⁺:</em>

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

<em>Moles of Ag⁺ in the precipitate:</em>

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

<em>Moles AsO₄ = Moles As:</em>

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

<em>Moles As₂O₃:</em>

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

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