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Natalija [7]
3 years ago
7

When determining the standard reduction potential of a substance by using a standard hydrogen electrode as a reference, the stan

dard reduction potential will always be equal to:______.
ta. he reduction potential for the standard hydrogen electrode.
b. one-half the cell potential.
c. the overall cell potential.
d. impossible to predict.
Chemistry
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

the overall cell potential

Explanation:

We must bear in mind that the standard hydrogen electrode is a reference electrode whose electrode potential has been arbitrarily set at 0 V.

The standard hydrogen electrode consists of hydrogen ion solution and hydrogen gas together with a platinum electrode.

The overall cell potential is the reduction potential of the substance being determined using the standard hydrogen electrode as a reference electrode since its electrode potential is set at zero volts.

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3 years ago
Which statement best describes the reaction pathway graph for an exothermic reaction but not an endothermic reaction? It has a h
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Answer: "The reactants are higher in energy than the products"

Explanation:

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Calculate the molar solubility of CaF2 in a 0.25 m solution of NaF(aq).
Kipish [7]

Answer:

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

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