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Mazyrski [523]
2 years ago
6

How is a standard reference electrode used to determine unknown E{half-cell)° values?

Chemistry
1 answer:
Andreas93 [3]2 years ago
3 0

A standard reference electrode can measure the potential difference between two different electrodes.

<h3>What is a standard reference electrode?</h3>

A standard reference electrode is an electrode whose potential is known.

It is physically impossible to measure the potential of a single electrode. Because only the difference between the two electrodes' potentials can be measured.

However, a way to measure the potential of different electrodes is to compare the standard cell potentials of two different galvanic cells that have one common electrode. This allows us to measure the potential difference between two different electrodes.

For example, the standard cell potential (E°) for the Zn/Cu is 1.10 V, and E° for the  Zn/Co is 0.51 V. So using these values we can determine the potential difference between the Co and Cu electrodes that is 1.10 V − 0.51 V = 0.59 V.

Hence, A standard reference electrode helps to measure the potential difference between two different electrodes.

Learn more about electrode:

brainly.com/question/4183610

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The given equation can be written as:
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2.92 A 50.0-g silver object and a 50.0-g gold object are both added
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Answer:

82.9 mL  

Explanation:

1. Volume of silver

\begin{array}{rcl}\text{Density}&=& \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho&=& \dfrac{m}{V}\\\\V &=& \dfrac{m}{\rho}\\\\& = & \dfrac{\text{50.0 g}}{\text{10.49 g$\cdot$mL}^{-1}}\\\\& = & \text{4.766 mL}\\\end{array}\\\text{The volume of the silver is $\large \boxed{\textbf{4.766 mL}}$}

2. Volume of gold

\begin{array}{rcl}V& = & \dfrac{\text{50.0 g}}{\text{19.30 g$\cdot$mL}^{-1}}\\\\& = & \text{2.591 mL}\\\end{array}\\\text{The volume of the gold is $\large \boxed{\textbf{2.591 mL}}$}

3. Total volume of silver and gold

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4 New reading of water level

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4 years ago
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The empirical formula is SCl_2.

The <em>empirical formula</em> (EF) is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the <em>molar ratio </em>of S to Cl.

Assume that you have 100 g of sample.

Then it contains 31.14 g S and 68.86 g Cl.

<em>Step</em> 1. Calculate the <em>moles of each element</em>

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<em>Step 2</em>. Calculate the <em>molar ratio</em> of each element

Divide each number by the smallest number of moles and round off to an integer

S:Cl = 0.971 30: 1.9425 = 1:1.9998 ≈ 1:2

<em>Step 3</em>: Write the <em>empirical formula</em>

EF = SCl_2

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