## 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
2022-04-09T02:38:43+00:00 2 Answers
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## Answers ( 2 )

The

concentrationof Znu00b2u207a ion at cathode is 0.295 M## Redox half equations

The r

edox half equationsare those of oxidation and reductionOxidation 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: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 ionat cathode is0.295 MLearn more about

concentration cellsat: https://brainly.com/question/15394851Answer:The concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 MExplanation:The half reactions for the cell is:[tex]Zn(s)rightarrow Zn^{2+}(0.100M,aq.)+2e^-[/tex]Oxidation half reaction (anode):[tex]Zn^{2+}(?M,aq.)+2e^-rightarrow Zn(s)[/tex]Reduction half reaction (cathode):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