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

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

For the Ni - Al cell in question 4 write the following reactions
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 the concentration of Ni2+(aq) is 2.00 x 10^-5 M.

Solutions

Expert Solution

5. Cathode Half Reaction:

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

Anode Half Reaction

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

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

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

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

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

n= 6

Ecell = E0cell-0.0592/n logQ

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

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

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

queen_honey_blossom answered 11 months ago

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