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2 years ago

Why are the oxidation and reduction half-reactions separated in an electrochemical cell?

Chemistry

2 answers:

Mrac [35]2 years ago

7 0

Answer:

The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire.

Explanation:

ELEN [110]2 years ago

3 0

Answer:

Electrochemical cells typically consist of two half-cells. The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire.

Explanation:

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Electrons are to bonding as neutrons are to-

-BARSIC- [3]

I think it’s b sorry if it isn’t it’s been a while since I’ve done this but it should be positive charge

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2 years ago

The pH of a saturated solution of a metal hydroxide MOH is 10.15. Calculate the Ksp for this compound.

Airida [17]

Answer:

$Ksp=2.00x10^{-8}$

Explanation:

Hello!

In this case, since the pH of the given metal is 10.15, we can compute the pOH as shown below:

$pOH=14-pH=14-10.15=3.85$

Now, we compute the concentration of hydroxyl ions in solution:

$[OH^-]=10^{-pOH}=10^{-3.95}=1.41x10^{-4}M$

Now, since this hydroxide has the form MOH, we infer the concentration of OH- equals the concentration of M^+ at equilibrium, assuming the following ionization reaction:

$MOH(s)\rightarrow M^+(aq)+OH^-(aq)$

Whose equilibrium expression is:

$Ksp=[M^+][OH^-]$

Therefore, the Ksp for the saturated solution turns out:

$Ksp=1.41x10^{-4}*1.41x10^{-4}\\\\Ksp=2.00x10^{-8}$

Best regards!

4 0

2 years ago

Explain hydrophobic and hydrophylic

Alika [10]

Answer:

Something defined as hydrophilic is actually attracted to water, while something that is hydrophobic resists water.

Explanation:

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2 years ago

Read 2 more answers

The decomposition of hydrogen peroxide follows first order kinetics and has a rate constant of 2.54 x 10-4 s-1 at a certain temp

Eva8 [605]

Answer:

$[A]_0=0.400M$

Explanation:

Hello.

In this case, since the first-order reaction is said to be linearly related to the rate of reaction:

$r=-k[A]$

Whereas [A] is the concentration of hydrogen peroxide, when writing it as a differential equation we have:

$\frac{d[A]}{dt} =-k[A]$

Which integrated is:

$ln(\frac{[A]}{[A]_0} )=-kt$

And we can calculate the initial concentration of the hydrogen peroxide as follows:

$[A]_0=\frac{[A]}{exp(-kt)}$

Thus, for the given data, we obtain:

$[A]_0=\frac{0.321M}{exp(-2.54x10^{-4}s^{-1}*855s)}$

$[A]_0=0.400M$

Best regards!

3 0

3 years ago

In order to promote the common ion effect, the concentration of the common ion must first:_____.

Svet_ta [14]

We need to increase the concentration of common ion first, in order to promote the common ion effect

<h3>What is the Common ion effect?</h3>

It is an effect that suppresses the dissociation of salt due to the addition of another salt having common ions.

For example, a saturated solution of silver chloride in equilibrium has Ag⁺ and Cl⁻ . Sodium Chloride is added to the solution and has a common ion Cl⁻. As a result, the equilibrium shifts to the left to form more silver chloride. Thus, solubility of AgCl decreases.

The Equilibrium law states that if a process is in equilibrium and is subjected to a change

in temperature,
pressure,
the concentration of reactant or product,

then the equilibrium shifts in a particular direction, according to the condition.

Thus, an increase in the concentration of common ion promotes the common ion effect.

Learn more about common ion effect:

brainly.com/question/23684003

#SPJ4

3 0

1 year ago