Respuesta :
The cell potential developed by voltaic cell using this reaction is +1.07 V, when all dissolved species are 0.015m.
An electrochemical cell, is what?
A redox reaction's oxidation and reduction phases are separated in an electrochemical cell, but they are nevertheless connected by a salt bridge and wire between electrodes so that ions in the salt bridge and electrons can travel around to carry charges. To test the electrical potential of the cell, the wire can be severed.
Cell potential calculation:
First we need to identify the oxidation (anode) and reduction (cathode) half-reactions steps:
Oxidation: 2Fe²⁺ → 2Fe³⁺ + 2e⁻ (E⁰ = 0.77V)
Reduction: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O (E⁰ = 1.77V)
The cell E⁰ for this redox reaction then is
[tex]E^{o} _{cell}[/tex] = [tex]E^{o} _{cathode} - E^{o} _{anode}[/tex]
=1.77V - 0.77V
Now we must apply the Nernst equation to this out-of-equilibrium system, assuming a temperature of 298 K, where Q is the reaction quotient and n is the number of moles of electrons that are changing in the system (see the half-reactions):
[tex]E = E^{o} - (\frac{0.0257}{n} ) * lnQ\\E = 1.00 - (\frac{0.00257}{2}) * ln\frac{(0.15)^{2} }{(0.15)^{2} * (0.15) * (0.15)^{2} }[/tex]
E = 1 - [0.01285(5.691)]
E = 1-0.0731
E = +1.07V
To know more about the balanced equation of this reaction visit:
https://brainly.com/question/28136770
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