A) Write the half cell reactions associated with the following redox reaction: N

Question

A) Write the half cell reactions associated with the following redox reaction:

Ni(s) + Hg22+(aq) ---> 2Hg (l) + Ni2+

B) How would you construct a voltaic cell to study this reaction?

C) if the reaction occurs spontaneously in the cell, which electrode is the anode? explain

D) Calculate the standard cell potential, E

E) Does the cell potential change if the reaction were written for two moles of Ni(s) reacting? Write the redox reaction using a factor of two moles and answer the question using the Nernst equation (see image below)

F) If the cell potential as measured is 1.04 volts, what would be the reduction potential associated with the Ni2+/Ni reaction if the Hg/Hg22+ half-cell is assigned a potential of 0.00 V? Repeat the question: if the Hg/Hg22+ half-cell is assigned a reduction potential of -0.80 Volts?

Explanation / Answer

1. Ni ---> Ni+2 + 2e-
    Hg22+   + 2e- ---> 2 Hg

2. Keep anode of Ni(s) electrode with NiSO4 as solvent and Keep Hg22+ as Cathode with HgSO4 as solvent at cathod. Place a salt bridge and voltmeter across these two containers

3. Oxidation of Ni occurs, so anode: Ni anode

4. Ni+2/Ni = -.25
Hg22+ /Hg = 0.789
E = 0.789-0.25=       .539 V = EHg - ENi

5. 2 Ni (s) ----> 2 Ni+2 + 4e-
2Hg22+   + 4e- ---> 2 Hg
E = Eo - RT/nf lnQ

Q = [ Hg] * [Ni+2] / [Hg22+ ] * [Ni]
New Q' = Q2, but n will also change from 2 to 4 in nernst reaction, so there will not be any change in cell potential

6. Cell potential will be same -1.04
EHg - ENi

for Hg/Hg2+2 = -.8

Ecell = 0.8-0.25 = 0.55


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