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
0.37
0.52
0.885
Well, I could give you the answer, but I might need to consult my nickel... I mean, my notebook first.
To calculate the cell emf for the given reaction, we can use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
First, we need to determine the values for E°cell, n, and Q.
1. E°cell: This is the standard cell potential, which can be found using standard reduction potentials. The reduction potentials for the half-reactions involved in the given reaction can be found in a standard reduction potential table.
The given reaction involves two half-reactions:
- Ni2+(aq) + 2e- → Ni(s) (reduction)
- Cu2+(aq) + e- → Cu+(aq) (reduction)
Using the standard reduction potentials:
E°cell = E°reduction (Ni2+/Ni) - E°reduction (Cu2+/Cu+)
2. n: This represents the number of electrons transferred in the balanced equation. In this case, 2 electrons are being transferred.
3. Q: This is the reaction quotient, which can be calculated using the concentrations of the species involved. In this case, Q = [Ni2+]/([Cu+]^2).
Now, let's calculate the values:
1. E°cell:
From the standard reduction potential table, the values are:
E°reduction (Ni2+/Ni) = -0.25 V
E°reduction (Cu2+/Cu+) = +0.16 V
Therefore,
E°cell = -0.25 V - (+0.16 V)
2. n: 2 electrons are being transferred.
3. Q:
Given concentrations:
[Ni2+] = 0.0010 M
[Cu+] = 1.0 M
Q = (0.0010 M) / ((1.0 M)^2)
Now, substitute these values in the Nernst equation:
Ecell = E°cell - (0.0592/2) * log(Q)
Finally, calculate the cell emf (Ecell) using the given equation and the calculated values.
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.