1. Calculate the cell potential for the following reaction as written at 25.00 o

Question

1. Calculate the cell potential for the following reaction as written at 25.00 oC, given that [Zn2+ ] = 0.864 M and [Fe2+] = 0.0140 M.

Zn(s)+Fe2+(aq) ? Zn2+(aq)+Fe(s)

2. For the following electrochemical cells, calculate the potential and determine if the cell reaction is spontaneous as written at 25 oC.

A. Cu(s) | Cu2+(0.10 M) || Fe2+ (0.0027 M) | Fe(s) When EoCu2+/Cu = 0.339 V and EoFe2+/Fe = -0.440 V Spontaneous or Not?

B. Pt(s) | Sn2+(0.0053 M), Sn4+ (0.13M) || Fe2+ (0.0048M), Fe3+ (0.10 M) | Pt(s)

When EoSn4+/Sn2+= 0.154 V and EoFe3+/Fe2+=0.771 V Spontaneous or not?

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 = 25 + 273 = 298 K
n = no.of electrons transffered = 2
F = 96500 Coulombs
[Zn2+] = 0.864 M
[Fe2+] = 0.014 M

E (cell) = 0.32 - ( 8.314 X 298) / ( 2 X 96500) X ln 0.864/0.014

E (cell) = 0.32 - 0.003113

E (cell) = 0.316887 V

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