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In-s [12.5K]
3 years ago
10

Complete the following single replacement reaction. If they don’t react, just write “NR”

Chemistry
1 answer:
Kipish [7]3 years ago
4 0

Here we have to complete the given single replacement reactions.

The replacement reactions are-

1) Fe (s) + CuCl₂ (aq) → FeCl₂ (aq) + Cu (s)

2) Cu (s) + FeCl₂ (aq) → NA

3) K (s) + NiBr₂ (aq) → NA

4) Ni (s) + KBr (aq) → NiBr₂ (aq) + K (s)

5) Zn (s) + Ca(NO₃)₂ (aq) → Zn(NO₃)₂ (aq)  + Ca (s)

6) Ca (s) + Zn(NO₃)₂ (aq) → NA

The replacement reactions can be explained in light of the redox potential.

The standard reduction potential of the half cells involved in these reactions are:

Fe²⁺ + 2e → Fe (E° = -0.441V); Cu²⁺ + 2e → Cu (E° = 0.674V)

Ni²⁺ + 2e → Ni (E° = -0.23V); Zn²⁺ + 2e → Zn (E° = -0.763V)

We know the half cell reactions in which the standard reduction potentials are positive are allowed.

1) The reaction is possible as Cu²⁺/Cu and Fe/Fe²⁺ standard reduction potentials are positive.

2) The reaction is not possible as Cu/Cu²⁺ and Fe²⁺/Fe standard reduction potentials are negative.

3) The reaction is not possible as Ni²⁺/Ni standard reduction potential is negative.

4) The reaction is possible as Ni/Ni²⁺ standard reduction potential is positive.

5) The reaction is possible as Zn/Zn²⁺ standard reduction potential is positive.

6) The reaction is possible as Zn²⁺/Zn standard reduction potential is negative.

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It is known that charge on xenon nucleus is q_{1} equal to +54e. And, charge on the proton is q_{2} equal to +e. So, radius of the nucleus is as follows.

            r = \frac{6.0}{2}

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = 5.5 \times 10^{-15} m     (as 1 fm = 10^{-15})

Therefore, electrostatic repulsive force on proton is calculated as follows.

              F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

Putting the given values into the above formula as follows.

           F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

              = (9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}

              = (9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}

              = 411.2 N

or,           = 4.1 \times 10^{2} N

Thus, we ca conclude that 4.1 \times 10^{2} N is the electric force on a proton 2.5 fm from the surface of the nucleus.

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