The voltage generated by the zinc concentration cell described by the line notation zn(s) ∣∣ zn2+(aq,0.100 m) ∥∥ zn2+(aq,? m) ∣∣ zn(s) is 14

Question

The voltage generated by the zinc concentration cell described by the line notation zn(s) ∣∣ zn2+(aq,0.100 m) ∥∥ zn2+(aq,? m) ∣∣ zn(s) is 14.0 mv at 25 °c. calculate the concentration of the zn2+(aq) ion at the cathode.

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2022-04-09T02:38:43+00:00 2 Answers 0

Answers ( 2 )

  1. Brielle
    0
    2022-04-09T02:40:31+00:00

    The concentration of Znu00b2u207a ion at cathode is 0.295 M

    Redox half equations

    The redox half equations are those of oxidation and reduction

    Oxidation half reaction: Zn(s) —> Znu00b2u207a + 2 eu207b [Znu207a] = 0.100 M

    reduction half reaction: Znu00b2u207a (aq) + 2 eu207b —-> Zn (s) [Znu207a] = y

    The Ecell = 14.0mV = 0.014 V

    Using the Nernst equation:

    • Ecell = Eu00b0cell – 0.0592/n * log([Znu207a]anode/[Znu207a]cathode)
    • since it is a concentration cell, Eu00b0cell = 0

    where n is number moles; n = 2

    0.014 = 0 – 0.0592/2 * log (0.1/y)

    y = 0.295 M

    Therefore, the concentration of Znu00b2u207a ion at cathode is 0.295 M

    Learn more about concentration cells at: https://brainly.com/question/15394851

  2. Isabelle
    0
    2022-04-09T02:40:32+00:00

    Answer: The concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M

    Explanation:

    The half reactions for the cell is:

    Oxidation half reaction (anode): [tex]Zn(s)rightarrow Zn^{2+}(0.100M,aq.)+2e^-[/tex]

    Reduction half reaction (cathode): [tex]Zn^{2+}(?M,aq.)+2e^-rightarrow Zn(s)[/tex]

    In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.

    To calculate cell potential of the cell, we use the equation given by Nernst, which is:

    [tex]E_{cell}=E^o_{cell}-frac{0.0592}{n}log frac{[Zn^{2+}]_{anode}}{[Zn^{2+}]_{cathode}}[/tex]

    where,

    n = number of electrons in oxidation-reduction reaction = 2

    [tex]E_{cell}[/tex] = 14.0 mV = 0.014 V (Conversion factor: 1 V = 1000 mV)

    [tex][Zn^{2+}]_{anode}[/tex] = 0.100 M

    [tex][Zn^{2+}]_{cathode}[/tex] = ? M

    Putting values in above equation, we get:

    [tex]0.014=0-frac{0.0592}{2}log frac{0.100M}{[Zn^{2+}]_{cathode}}[/tex]

    [tex][Zn^{2+}]_{cathode}=0.295M[/tex]

    Hence, the concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M

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