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love history [14]
3 years ago
9

A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M 1.0 M nickel(II) ion solution and another elect

rode composed of silver in a 1.0 M 1.0 M silver ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C . 25 °C.
Chemistry
1 answer:
Katarina [22]3 years ago
3 0

Answer:

<h3>So the standard electrode potential of the cell is 1.05 V</h3>

Explanation:

electrode composed of nickel in a 1.0 M

1.0 M nickel(II) ion solution and another electrode composed of silver in a 1.0 M 1.0 M silver ion solution,

connected by a salt bridge. Calculate the standard potential for this cell at 25 °C .

2Ag+ + Ni -----> 2 Ag + Ni +2

Eocell = Eo cathode - Eo anode

= Eo Ag+/Ag - Eo Ni+2/Ni

=0.80 - (-0.25)

=0.80 + 0.25

= 1.05

<h3>So the standard electrode potential of the cell is 1.05 V</h3>
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20 protons, 20 electrons, and 21 neutrons

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20 protons, 20 electrons, and 21 neutrons

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A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h
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Explanation:

First, calculate the moles of CO_{2} using ideal gas equation as follows.

                PV = nRT

or,          n = \frac{PV}{RT}

                = \frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = \frac{mass}{volume}

Hence, mass of water will be as follows.

                Density = \frac{mass}{volume}

             0.998 g/ml = \frac{mass}{3.26 ml}    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles = \frac{mass}{\text{molar mass}}

                              =  \frac{3.25 g}{18.02 g/mol}            

                              = 0.180 mol

Moles of hydrogen = 0.180 \times 2 = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

          0.183 mol =  \frac{mass}{12 g/mol}            

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                               =  \frac{0.95 g}{16 g/mol}            

                              = 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is C_{3}H_{6}O.

Thus, we can conclude that empirical formula of the given compound is C_{3}H_{6}O.            

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