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marusya05 [52]
3 years ago
14

What happens in a displacement reaction

Chemistry
1 answer:
ruslelena [56]3 years ago
6 0

Answer:

Displacement reactions occur when a metal from the electrochemical series is mixed with the ions of a metal lower down in the electrochemical series. The atoms of the more reactive metal push their electrons on to ions of the less reactive metal. An example can be seen below.

Explanation:

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The number "2" is the value of the principal quantum number "n". Letter "s" is associated with the value of secondary quantum number "l" and it is equal to zero. The value of "m" (or magnetic quantum number) is zero too. The quantum number set for the highest energy electron will be (2, 0, 0, 1/2).
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Which element in period 4 has the highest electronegativity
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What are the energy transformations in a fossil fuels power plant?
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Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a
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Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V

Therefore, the standard cell potential will be +0.799 V

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