4. calculate the Eocell for the Ni - Al cell. For the Ni - Al cell...

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Chemistry

4. calculate the Eocell for the Ni - Al cell.

For the Ni - Al cell in question 4 write the followingreactions
5. Cathode Half Reaction:
6. Anode Half Reaction:
7. Redox Reaction:
8. Using your answer from question 4, determine the Ecell for a Ni- Al cell when the concentration of Al3+(aq) is 4.00 M and theconcentration of Ni2+(aq) is 2.00 x 10^-5 M.

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4.1 Ratings (818 Votes)

5. Cathode Half Reaction:

Ni^+2 +2e^-1 ---------> Ni (s)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â E0 = -0.23v

Anode Half Reaction

Al(s) ---------> Al^+3(aq) + 3e^-1Â Â Â Â E0 = 1.66V

2Al(s) ---------> 2Al^+3(aq) + 6e^-1Â Â Â Â Â E0 = 1.66V

3Ni^+2 +6e^-1 ---------> 3Ni (s)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â E0 = -0.23v

----------------------------------------------------------------------------------

2Al(s) + 3Ni^+2 ---------> 2Al^+3 (aq) + 3Ni(s)Â Â E0 = 1.43VÂ Â redox reaction or over all reaction

Â Â n= 6

Ecell = E0cell-0.0592/n logQ

Â Â Â Â Â Â Â Â Â Â Â =1.43-0.0592/6 log[Al^+3]^2/[Ni^+2]^3

Â Â Â Â Â Â Â Â Â Â Â = 1.43-0.00986log(4)^2/(2*10^-5)^3

Â Â Â Â Â Â Â Â Â = 1.43-0.00986*5.3010Â Â =1.377V >>>>answer

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4. calculate the Eocell for the Ni - Al cell. For the Ni - Al cell...

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