Part A Use the table of standard reduction potentials given above to calculate t

Question

Part A

Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 C) for the following reaction:

Fe(s)+Ni2+(aq)Fe2+(aq)+Ni(s)

Part B

Calculate the standard cell potential (E) for the reaction

X(s)+Y+(aq)X+(aq)+Y(s)

if K = 4.04×103.

Express your answer to three significant figures and include the appropriate units.

Reduction half-reaction E (V) Ag+(aq)+eAg(s) 0.80 Cu2+(aq)+2eCu(s) 0.34 Sn4+(aq)+4eSn(s) 0.15 2H+(aq)+2eH2(g) 0 Ni2+(aq)+2eNi(s) 0.26 Fe2+(aq)+2eFe(s) 0.45 Zn2+(aq)+2eZn(s) 0.76 Al3+(aq)+3eAl(s) 1.66 Mg2+(aq)+2eMg(s) 2.37

Explanation / Answer

A) given reaction is

Fe + Ni+2 ---> Fe+2 + Ni

anode reaction : oxidation

Fe ---> Fe+2 + 2e-

cathode reaction : reduction

Ni+2 + 2e- ---> Ni

we know that

Eo cell = Eo cathode - Eo anode

so

Eo cell = Eo Ni+2 / Ni - Eo Fe+2/Fe

Eo cell = -0.26 - ( -0.45)

EO cell = 0.19

now

we know that

dGo = -nFEo = -RT ln Keq

so

nFEo = RT ln Keq

here

n = 2 as two electrons are transferred

2 x 96485 x 0.19 = 8.314 x 298 x ln Keq

Keq = 2.67 x 10^6

so

the equilibrium constant is 2.67 x 10^6

B)

oxidation reaction :

X ----> X+ + e-

so

n =1

now

nFEo = RT ln Keq

1 x 96485 x Eo = 8.314 x 298 x ln 4.04 x 10-3

Eo = -0.14

so

the standard cell potential is -0.14 V

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