calculate cell potential for the voltaic cell where its involving the two cobalt species with Copper Calculate the cell potential for the voltaic cell that results when the following two half-cells are connected at $\pu{25 ^\circ C}$:
(1) A platinum electrode inserted into a solution of $\pu{0.10 M}$ $\ce{Co^3+}$ and $\pu{0.0010 M}$ $\ce{Co^2+}$
(2) A copper electrode inserted into a solution of $\pu{0.010 M}$ $\ce{Cu^2+}$ ions

Answer. $\pu{1.66 V}$

how it's calculated?
I know how to calculate the second cell half but what about the first half?
Edit 1
will I need more than this equation?
Cathode
$\ce{{Co^{3+}} + e− ⇌ {Co^{2+}}  -&gt;   1.92V}$
Edit 2 from Nicolas Answer
$\ce{E=E^0 -0.059*log \frac{[Co^{2 +}]}{[Co^{3 +}]}}$
$\ce{E=1.92 -0.059*log \frac{0.001}{0.1}}$
$\ce{E=2.038}$
Now how to use the Nernst equation again, which concentration I will use for Co?

I Thought I will use it for the cell, not just half cell which requires two concentrations

Edit 3
Applying Nernst to the second half
anode:
$\ce{Cu^{2 +}(aq) + 2e- → Cu(s) -&gt;  +0.34}$
$\ce{E=E^0 - \frac{0.059}{2}*log \frac{1}{[Cu^{2 +}]}}$
$\ce{E=0.34 -\frac{0.059}{2}*log \frac{1}{0.01}}$
$\ce{E=0.281}$
E cell = E cathode - E anode = 2.038 - 0.281 = 1.757
The answer is 1.66 did I make everything right and That's a book's mistake?
Edit 4
trying to solve it in one step
Cathode
$\ce{{2Co^{3+}} + 2e− ⇌ {2Co^{2+}}  -&gt;  1.92V}$
Anode :
$\ce{Cu ⇌  Cu^{2 +} + 2e- -&gt;    0.34V}$
Cell:
$\ce{{2Co^{3+}} + Cu ⇌  Cu^{2 +} + {2Co^{2+}} -&gt;    1.58V}$
$\ce{E=E^0 - \frac{0.059}{2}*log \frac{[Co^{2 +}]^2[Cu^{2 +}]}{[Co^{3 +}]^2}}$
$\ce{E=1.58 -\frac{0.059}{2}*log \frac{0.001^2 * 0.01}{0.1^2}}$
E = 1.76
So I think it's a book's mistake written 1.66 instead of 1.76
The equation noted corresponds well to that of the first half-cell: on the other hand, the potential indicated is not the good one but just that of the standard potential.
The Nernst relation is written $\ce{E=E^0 +0.06*log \frac{[Co^{3 +}]}{[Co^{2 +}]}}$: the potential therefore depends on the concentrations
You can therefore calculate the potential of this half-cell.
You do the same thing with the second to arrive at its potential: from the 2, you can calculate the potential of the cell