**Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 57 ∘C , where [Fe2+]= 3.40 M and [Mg2+]= 0.210 M…**

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## Question “Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 57 ∘C , where [Fe2+]= 3.40 M and [Mg2+]= 0.210 M…”

Consider the reaction

Mg(s)+Fe_{2}+(aq)→Mg_{2}+(aq)+Fe(s)

at 57 ∘C , where [Fe^{2+}]= 3.40 M and

[Mg^{2+}]= 0.210 M .

Part D Calculate the standard cell potential for Mg(s) Fe2 (aq)Mg2 (aq)Fe(s) Express your answer to three significant figures and include the appropriate units. View Available Hint(s) ? Value Eo Units Submit

Part C What is the value for n? Express your answer as an integer and include the appropriate units (i.e. enter mol for moles). View Available Hint(s) HA ? Value n Units Submit Rart D

Review I Constants Periodic Table Learning Goal: To learn how to use the Nernst equation Consider the reaction The standard reduction potentials listed in any reference table are only valid at standard-state conditions of 25 °C and 1 M. To calculate the cell potential at non-standard-state conditions, one uses the Nernst equation, Mg(s) Fe2 (aq)-Mg2 (aq) + Fe(s) at 57 °C, where Fe2] 3.40 Mand Mg2+0.210 M. 2.303 RT nF log10Q E=E° Part A where E is the potential in volts, E° is the standard potential in volts, R 8.314 J/(K mol) is the gas constant, T is the temperature in kelvins, n is the number of moles of electrons transferred, F = 96,500 C/(mol e~ ) is the Faraday constant, and Q is the reaction quotient. What is the value for the reaction quotient, Q, for the cell? Express your answer numerically View Available Hint(s) Substituting each constant into the equation the result is 0.0592 V log10 E=E° -_ n. ΑΣφ Submit

Review I Constants I Periodic Table Part B What is the value for the temperature, T, in kelvins? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) HA Value = Units Submit

## Answer

M8 (5) + Mg 2+ (+ fecs), aq) Mo is reduced to Fe and oxidised to Mg2+ feat. The cell be-M9 (5) Mgat11 featife – 0 PartD Edeg-E* feat/fe-EM82+/8. IEdeg = (-0.44V) 1:Edeg= -0.44AV + 2-520 J. EER = 2.08 V parte 3 We get cell representation Mg2+tze at the feet go. The value for n=2 Mol

part 19 570 EUR = 273+57 – 325K so; T = 33K part Part @ Carnation refers to the Concentrations of Products divided by the Concentrations Of Reactonts, but only the component in the faseous or p aqueous states.

So, 8 = [Mg2+] [fear ::8 – 0-210M 3:40M 1-9 = 0.061

## Conclusion

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