How does temperature differ from heat?

What is the calculated value of the cell potential at 298K for an

Tpy6a [65]

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?

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

Answer: 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

4 0

2 years ago

The characteristic temperature at which a pure solid changes to a liquid is its ______ point.

Tasya [4]

It is known as the melting point

5 0

3 years ago

The melting point of your product is 10 degrees lower than the expected one. What conclusion can you make about the purity of yo

DanielleElmas [232]

Answer:

The product is significantly impure

Explanation:

In order to test for the purity of a specific sample that was synthesized, the melting point of a compound is measured. Basically speaking, the melting point identifies how pure a compound is. There are several cases that are worth noting:

if the measured melting point is significantly lower than theoretical, e. g., lower by 3 or more degrees, we conclude that our compound contains a substantial amount of impurities;
wide range in the melting point indicates impurities, unless it agrees with the theoretical range.

Since our compound is even 10 degrees Celsius lower than expected, it indicates that the compound is significantly impure.

5 0

3 years ago

(40 Points) Complete, balance, compute the amounts of the products assuming 100% yield. 10g Na + 10g Oxygen.

Lilit [14]

13.5g

Explanation:

Given parameters:

Mass of Na = 10g

Mass of O₂ = 10g

Unknown:

Mass of products formed = ?

Balanced equation = ?

Solution:

The balanced chemical equation is shown below:

4Na + O₂ ⇒ 2Na₂O

In any reaction, the specie in short supply determines the extent of the reaction.

This reaction is not an exclusion. We need to first determine the specie in short supply and use it to estimate the amount of product since we have a 100% yield which signifies that all was used up.

let us convert to moles;

Number of moles of Na = $\frac{mass }{molar mass} = \frac{10}{23}$ = 0.435mole