Inactive Tutor answered 08/19/20
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.
Inactive Tutor answered 08/19/20
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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