Hydrogen monofluoride

In an electrolytic cell, the magnitude of the standard cell potential is the:______

melamori03 [73]

Answer:

a. minimum voltage that must be supplied for a redox reaction to occur

c. always equal to Eanode - Ecathode

Explanation:

In an electrolytic cell; The electromotive force(the maximum standard potential difference) of the cell formed by the system is defined as the standard electrode potential of the right handed electrode minus the standard electrode potential of the left hand electrode. (i.e $\mathbf{E^{\theta}_{cell}=E_{anode} - E_{cathode}}$ )

As we all known that the process by which chemical energy is being converted to electrical energy is called the Electrochemical cell. It consists of two half cells , an oxidation half cell reaction and a reduction half-cell reaction.The overall redox reaction results in a flow of electrons in an electric current which is produced by a minimum voltage.

Therefore, option a and c are both correct.

6 0

3 years ago

What happens when an atom of sulfur combines with two atoms of chlorine to produce SCI2?

Aloiza [94]

Answer:..A.) Each chlorine atom shares a pair of electrons with the sulfur atom

3 0

2 years ago

What is the broadest, most general level of taxonomy?

nikdorinn [45]

Answer:The levels of classification are as follows (Broadest to

narrowest): Domain, Kingdom, Phylum, Class, Order, Family, Genus,

Species

Explanation:

8 0

3 years ago

Why are yeast cells used in making bread?

tatiyna

Answer:

Yeast is used for the leavening of bread. Yeast uses the sugars and oxygen in dough to produce more yeast cells and carbon dioxide gas. The carbon dioxide makes the dough rise which gives the bread a light and spongy texture. Yeast also works on the gluten network.

Explanation:

5 0

2 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago