For the balanced equation you need to include the net electron transfer (also the states of matter, as specified in the question).
Thus,
Ni2+(aq) + Al(s) → Al3+(aq) + Ni(s) + 1 e-
Dom S.
asked 08/19/20Calculate E°, identify the cathode and anode, and give the overall balanced equation. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm. Standard reduction potentials are found in a Standard Reduction Potentials table. (Use the lowest possible whole number coefficients. Include states-of-matter under the given conditions in your answer.)
Ni^2+(aq) + Al(s) → Al^3+(aq) + Ni(s)
Ni^2+ +2e^- —>Ni E=-0.23
Al^3+ + 3e^- —>Al E= -1.66
E° was calculated to be 1.43 V
Ni is cathode
Al is anode
How do I figure out the balanced equation? I am extremely confused and I think I’m seeing it harder than it actually is. Thanks
Initially I got Ni^2+ +Al —>Al^3+ + Ni but I’m missing something in balancing it out.
For the balanced equation you need to include the net electron transfer (also the states of matter, as specified in the question).
Thus,
Ni2+(aq) + Al(s) → Al3+(aq) + Ni(s) + 1 e-
J.R. S. answered 08/20/20
Ph.D. University Professor with 10+ years Tutoring Experience
Ni^2+ +2e^- —>Ni E=-0.23
Al^3+ + 3e^- —>Al E= -1.66
Ni2+ + 2e- ==> Ni(s) [x3]
Al(s) ==> Al3+ + 3e- [x2]
3Ni2+ + 6e- = 3Ni(s)
2Al(s) ==> 2Al3+ + 6e-
____________________
3Ni2+(aq) + 2Al(s) ==> 3Ni(s) + 2Al3+(aq) This is the balance equation.
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