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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

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  • $\begingroup$ Hint: Find a half-reaction involving the two cobalt species. $\endgroup$
    – Ed V
    Feb 17 at 19:21
  • $\begingroup$ 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 $\endgroup$
    – Nicolas
    Feb 17 at 19:52
  • $\begingroup$ Is there any link for a question like this? $\endgroup$
    – A.Hany
    Feb 17 at 20:08
  • $\begingroup$ Do you know about Nernst's relationship? Otherwise, it might direct you to seek it out and apply it to your half cell. $\endgroup$
    – Nicolas
    Feb 17 at 20:18
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    $\begingroup$ 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 $\endgroup$
    – Nicolas
    Feb 17 at 21:54
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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

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  • $\begingroup$ did I solve it right? $\endgroup$
    – A.Hany
    Feb 17 at 22:38
  • $\begingroup$ @A.Hany, Your approach is correct. The book may have round off errors. $\endgroup$
    – M. Farooq
    Feb 18 at 3:05

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