Respuesta :

pstnonsonjoku

Answer:

2.45×10^54

Explanation:

Mg(s) + Zn2+(aq) = Mg2+(aq) + Zn(s)

From the equation;

n= 2 moles of electrons

E°anode= -2.37 V

E°cathode= -0.76 V

From E°Cell= E°cathode - E°anode

E°Cell= -0.76-(-2.37) = 1.61 V

Recall that E°cell= 0.0592/n log K

1.61= 0.0592/2 log K

log K = 1.61/0.0296

K= Antilog (54.39)

K= 2.45×10^54

Lanuel

To determine the equilibrium constant for the above chemical reaction is [tex]2.51 \times 10^{54}[/tex]

Given the following data:

  • Temperature = 25°C

The balanced chemical equation for the reaction is:

                         [tex]Mg{(s)} + Zn^{2+} {(aq)} = Mg^{2+} (aq) + Zn(s)[/tex]

  • Number of moles of electrons = 2 moles
  • The standard reduction potential for anode, [tex]E^\circ (Anode) = -2.372[/tex]
  • The standard reduction potential for cathode, [tex]E^\circ (Cathode) = -0.7618[/tex]

To determine the equilibrium constant for the above chemical reaction:

Mathematically, the standard cell potential of a galvanic cell is given by the formula:

[tex]E^\circ_{cell}= E^\circ (Anode) -E^\circ (Anode)\\\\E^\circ_{cell}= -0.7618-(-2.372)\\\\E^\circ_{cell}= -0.7618+2.372\\\\E^\circ_{cell}=1.6102\; Volt[/tex]

Now, we can determine the equilibrium constant by using the formula:

[tex]E_{cell}^\circ = \frac{0.0592}{n} logK[/tex]

Substituting the parameters into the formula, we have;

[tex]1.6102 = \frac{0.0592}{2} logK\\\\1.6102 =0.0296logK\\\\logK = \frac{1.6102}{0.0296} \\\\logK =54.40\\\\K = log^{-1}(54.40)\\\\K = 2.51 \times 10^{54}[/tex]

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