Electrochemical cell question?

Question

Given:
Hg2+ + Zn ---> Zn2+ + Hg
E=.85 V Hg2+ + 2e- ---> Hg
E=-.76V Zn2+ + 2e- ---> Zn
With the above cell potentials at 298K, what is the approximate equilibrium constant for the above reaction?

The E values I gave are standard cell potentials for reduction of that particular substance.

Answer 1

Ok, first find the voltage.To do this you have to make sure the cell potential signs are in the right direction. In the overall reaction, Zinc is being oxidized, not reduced. Therefore you need to reverse the sign, so +.76V, not-.76:

Zn---->Zn2+ +2e

The other reaction (the reduction) has the right sign.

We then add the two potentials:

.85+.76=1.61Volts

Then use the Nernst equation:

E(1.61V)=
.0592/2(the moles of e-)*LogKeq
Then divide 1.61 by .0296=54.39

Then use the inverse log (10^x) key and do inverse log
of 54.39. This is the Keq.

Answer 2

A galvanic cellular includes 2 1/2-cells. each 1/2-cellular has: (a million) an electrode, which in the figure are the plates of Zn (zinc) and Cu (copper); and (2) an electrolyte, which in the figure are aqueous recommendations of ZnSO4 and CuSO4. The metallic of a metallic electrode has a tendency to bypass into answer, thereby liberating truthfully charged metallic ions into the electrolyte, and holding negatively charged electrons on the electrode. as a outcome each 1/2-cellular has its own 1/2-reaction. For the Daniell cellular, depicted in the figure, the Zn atoms have a significantly better tendency to bypass into answer than do the Cu atoms. more beneficial precisely, the electrons on the Zn electrode have a more beneficial power than the electrons on the Cu electrode. because the electrons have unfavourable fee, to grant electrons on it a more beneficial power the Zn electrode must have a more beneficial unfavourable electric powered potential than the Cu electrode. in spite of the undeniable fact that, in the absence of an exterior connection between the electrodes, no contemporary can bypass.

Answer 3

Standard reduction potentials
1. Hg2+ + 2e- ===> Hg E1 = 0.86 + 0.0591/2 log([Hg2+]) noting that log(1/x) = log(x)

2. Zn2+ + 2e- ===> Zn E2 = -0.76 + 0.0591/2 log([Zn2+])

For the reaction Hg2+ + Zn ===> Zn2+ + Hg K=[Zn2+]/[Hg2+]

At equilibrium, E1 = E2

0.86 + 0.0591/2 log([Hg2+]) = -0.76 + 0.0591/2 log([Zn2+])

or

0.86 + 0.76 =0.0591/2 {log([Zn2+]) - 0591log([Hg2+])
1.61 = 0.0591/2 log([Zn2/[Hg2+]) = 0.0591/2 logK

yielding log K = 54.5 or K =3.16 x 10^54

Answer 4

ΔG0= -nFE0 with E0= E0Hg+2/Hg -E0Zn+2/Zn= 0.85-(-0.76)= 1.61 Volt
ΔG0=-RTlnK
so

nFE0=RTlnK =>

K=e^(nFE0/RT) =
= exp( 2*96485*1.61 / (8.314*298) ) =2.88 *10^54