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

$$\ce{{Co^{3+}} + e− ⇌ {Co^{2+}} -> 1.92V}$$

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:

$$\ce{Cu^{2 +}(aq) + 2e- → Cu(s) -> +0.34}$$

$$\ce{E=E^0 - \frac{0.059}{2}*log \frac{1}{[Cu^{2 +}]}}$$

$$\ce{E=0.34 -\frac{0.059}{2}*log \frac{1}{0.01}}$$

$$\ce{E=0.281}$$

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 $$\ce{{2Co^{3+}} + 2e− ⇌ {2Co^{2+}} -> 1.92V}$$

Anode : $$\ce{Cu ⇌ Cu^{2 +} + 2e- -> 0.34V}$$

Cell: $$\ce{{2Co^{3+}} + Cu ⇌ Cu^{2 +} + {2Co^{2+}} -> 1.58V}$$

$$\ce{E=E^0 - \frac{0.059}{2}*log \frac{[Co^{2 +}]^2[Cu^{2 +}]}{[Co^{3 +}]^2}}$$

$$\ce{E=1.58 -\frac{0.059}{2}*log \frac{0.001^2 * 0.01}{0.1^2}}$$

E = 1.76

So I think it's a book's mistake written 1.66 instead of 1.76

• Hint: Find a half-reaction involving the two cobalt species.
– Ed V
Feb 17 at 19:21
• In both cases (in all cases besides), the principle is the same: one writes the corresponding half-equation of the couple concerned then one establishes its potential using the relation of Nernst Feb 17 at 19:52
• Is there any link for a question like this? Feb 17 at 20:08
• Do you know about Nernst's relationship? Otherwise, it might direct you to seek it out and apply it to your half cell. Feb 17 at 20:18
• by the way, you ask how to do for $\ce {Co}$ except that in your statement it is question of copper for the second half-cell Feb 17 at 21:54

The Nernst relation is written $$\ce{E=E^0 +0.06*log \frac{[Co^{3 +}]}{[Co^{2 +}]}}$$: the potential therefore depends on the concentrations