A salt bridge is used to provide electrons a path to flow from the electrode of one half-cell to another. TRUE FALSE

1 answer:

8 0

It is false. The salt bridge is not a path for electrons, but a path for ions to flow from one half-cell to another. It help to balance the charge between the oxidation and reduction vessels.

You might be interested in

If a solution initially contains 0.260 M HC2H3O2, what is the equilibrium concentration of H3O+ at 25 ∘C? Express your answer in

arlik [135]

Answer:

2.16 × 10⁻³

Explanation:

Step 1: Given data

Concentration of the acid (Ca): 0.260 M

Acid dissociation constant (Ka): 1.80 × 10⁻⁵

Step 2: Write the acid dissociation equation

HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)

Step 3: Calculate the concentration of H₃O⁺ at equilibrium

We will use the following expression.

$[H_3O^{+} ]= \sqrt{Ka \times Ca } = \sqrt{1.80 \times 10^{-5} \times 0.260 } = 2.16 \times 10^{-3}$

8 0

3 years ago

A student places 1.38g of unknown metal at 99.6C into 60.50g of water at 22.1C. The entire system reaches a uniform temperature

stiv31 [10]

Explanation:

Heat energy lost by metal = Heat energy gained by water.

(0.00138kg) * c * (99.6-31.6) = (0.006050kg) * 4.148 * (31.6-22.1)

c = 2.541J/kg°C

8 0

3 years ago

Express in scientific notation. Choose the answer with the proper number of significant figures.

CaHeK987 [17]

answer 3

Explanation:

when you multiply you round to the lowest number of sig figs

7 0

3 years ago

A sparingly soluble metal fluoride with formula mf2, where m is an unknown metal, has a ksp = 6. 65 x 10-6. calculate the concen

OleMash [197]

A sparingly soluble metal fluoride with formula MF₂, where m is an unknown metal, has a Ksp = 6.65 x 10⁻⁶. The concentration of F⁻ in solution 2.36 x 10⁻³M

Ksp is called the Molar solubility product and S is the Molar solubility of an ion in a solution.

According to given formula, the dissociation of metal fluoride MF₂ occurs as follows in aqueous solution:

MF₂ ------> M⁺² + 2F⁻

S 2S

Ksp = [S] [2S]²

Ksp = 4S³

Given, Ksp = 6.65 x 10⁻⁶

On substituting,

6.65 x 10⁻⁶ = 4S³

S³ = 1.66 x 10⁻⁶

S = 1.18 x 10⁻³

So, Solubility of F⁻ ion = 2S