Standard reduction potentials Reduction half-reaction E? (V) Ag+(aq)+e??Ag(s) 0.

Question

Standard reduction potentials
Reduction half-reaction   E? (V)
Ag+(aq)+e??Ag(s)   0.80
Cu2+(aq)+2e??Cu(s)   0.34
Sn4+(aq)+4e??Sn(s)   0.15
2H+(aq)+2e??H2(g)   0
Ni2+(aq)+2e??Ni(s)   ?0.26
Fe2+(aq)+2e??Fe(s)   ?0.45
Zn2+(aq)+2e??Zn(s)   ?0.76
Al3+(aq)+3e??Al(s)   ?1.66
Mg2+(aq)+2e??Mg(s)   ?2.37
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)
Express your answer numerically.
K =   
2.68

Explanation / Answer

A.) The cell reaction is like you wrote above, so the reduction value is +0.45 (as you reversed it) - 0.26 = 0.19 Volts (E=0.19).
The Nernst's formula for 25C is E=E-0.05916/n*logK and at equilibrium the E value will be 0. The number of exchanged electrons is 2. Therefore, (0.05916/2)logK=E and logK=0.19*2/0.05916=6.423 and

K will be about 2,650,080.

B.) standard cell potential are done at 25 Celsius..... 298 Kelvin

Y^(+1) takes 1 electron --> Y
X loses 1 electron --> X^(+1)
so n = 1 mole of electrons

ln{K}= nFE/RT

ln{7.8710?3.}= (1)(96500)E / (8.314)(298)

-4.844 = 96500 E / 2,477

E = - 0.1243

E = - 0.124 volts

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