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 = 8.59×103.

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

Part A.

Ecell = -0.26 - (-0.45) = 0.19 V

-RTlnKeq = -nFEcell

with,

T = 25 + 273 = 298 K

Keq = equilibrium constant

we get,

8.314 x 298 x lnKeq = 2 x 96485 x 0.19

Keq = 2.67 x 10^6

Part B.

Given K = 8.59 x 10^-3

RTlnK = nFEcell

8.314 x 298 ln (8.59 x 10^-3) = 1 x 96485 Ecell

Ecell = -0.122 V

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