Chemistry, 25.04.2020 22:49 kactus

What is the calculated value of the cell potential at 298K for an
electrochemical cell with the following reaction, when the H2
pressure is 6.56x10- atm, the H concentration is 1.39M, and
the Sn2+ concentration is 9.35x10-4M?
2H+ (aq) + Sn(s)—_H2(g) + Sn2+(aq)

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Answer from: justiceharris8

The question is incomplete, here is the complete question:

What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the H₂ pressure is 6.56 x 10⁻² atm, the H⁺ concentration is 1.39 M, and the Sn²⁺ concentration is 9.35 x 10⁻⁴ M?

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

The cell potential of the given electrochemical cell is 0.273 V

Explanation:

For the given chemical equation:

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

The half cell reactions for the given equation follows:

Oxidation half reaction: $Sn(s)\rightarrow Sn^{2+}(aq)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V$

Reduction half reaction: $H_2+2e^-\rightarrow H_2(g);E^o_{2H^{+}/H_2}=0.0V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.0-(-0.14)=0.14V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]\times p_{H_2}}{[H^+]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ?

$E^o_{cell}$ = standard electrode potential of the cell = +0.14 V

n = number of electrons exchanged = 2

$[H^{+}]=1.39M$

$[Sn^{2+}]=9.35\times 10^{-4}M$

$p_{H_2}=6.56\times 10^{-2}atm$

Putting values in above equation, we get:

$E_{cell}=0.14-\frac{0.059}{2}\times \log(\frac{(9.35\times 10^{-4})\times (6.56\times 10^{-2})}{(1.39)^2})\\\\E_{cell}=0.273V$

Hence, the cell potential of the given electrochemical cell is 0.273 V

Answer from: Quest

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

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