Question

Calculate the cell emf (in volts) for the following reaction at 25°C:
Ni(s) + 2 Cu2+(0.010 M) → Ni2+(0.0010 M) + 2 Cu+(1.0 M)
[Enter only the number of your answer.]
Answer

Answer 1

Use the Nernst equation.
First, do Ni (as a reduction).
E = Eo-(0.05916/n)log(red/ox)
ENi redn = Eo-(0.05916/2)log[(Ni(s))/(Ni^+2)]
Look up Eo as a reduction for Ni+2 + 2e == Ni(s)
Plug in 1 for (Ni(s) (because that is its standard state) and 0.001 M for Ni^+2 from the problem. n = 2 for electron change. Solve for ENi as a reduction, then change the sign (because it is an oxidation in the problem).
Do the same thing for Cu^+2 + e ==> Cu^+ and solve for ECu redn.
Then add the E value for the oxidation of Ni to the E value for the reduction of Cu^+2 and that will be Ecell. Post your work if you get stuck.

Answer 2

0.37

Answer 3

0.52

Answer 4

0.885