Could you please answer these questions 1- 2- 3- 4- 5- Thank you Calculate the c

Question

Could you please answer these questions

1-

2-

3-

4-

5-

Thank you

Calculate the cell potential for the following reaction as written at 25.00 degree C, given that [Zn2+] = 0.764 M and [Fe2+] = 0.0130 M. Standard reduction potentials can be found here. What would the potential of a standard hydrogen (S.H.E.) electrode be if it was under the following conditions? The voltage generated by the zinc concentration cell described by, is 11.0 mV at 25 degree C. Calculate the concentration of the Zn2+(aq) ion at the cathode. Gallium is produced by the electrolysis of a solution obtained by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(lll) solution by a current of 0.440 A that flows for 80.0 min. A current of 3.70 A is pass through a Cu(NO3)2 solution. How long (in hours) would this current have to be applied to plate out 5.80 g of copper?

Explanation / Answer

1.

Eo ( Zn2+/Zn) = -0.76 V
Eo ( Fe2+/Fe) = -0.44 V

since Zn is getting oxidised and Fe is getting reduced hence Zn2+/Zn will be anode and Fe2+/Fe will be cathode

Eo ( cell) = Eo ( cathode) - Eo(anode) = Eo(Fe2+/Fe) - Eo(Zn2+/Zn) = -0.44 - -0.76 = -0.44 + 0.76 = 0.32 V

now if the conc. of Fe2+ was 1 M and conc. of Zn2+ was 1 M ...then E (cell) would have been equal to 0.32 V .....but since it is not ...hence we have to use the Nernst equation ...

E (cell) = Eo (cell) - RT/nF ln [Zn2+]/[Fe2+]

where R = 8.314 J/K/mole
T = 59 + 273 = 332 K
n = no.of electrons transffered = 2
F = 96500 Coulombs
[Zn2+] = 0.764 M
[Fe2+] = 0.0130 M

E (cell) = 0.32 - ( 8.314 X 332) / ( 2 X 96500) X ln (0.764/0.0130)

E (cell) = 0.32 - 2760.248/193000 X ln (58.769)

E (cell) = 0.32 - 0.014 X 4.073

E (cell) = 0.32 - 0.057

E (cell) = 0.2629 V

2.

Use a different form of the nernst equation
E = E? + RT/nF x ln([H+]

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