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### Topic: Computing cell potentials  (Read 586 times)

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#### Win,odd Dhamnekar

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##### Computing cell potentials
« on: April 04, 2023, 08:02:08 AM »
The standard reduction potential for the reaction$[Co(H_2O)_6]^{3+}(aq) + e^- \to [Co(H_2O)_6]^{2+} (aq)$ is about 1.8 V. The standard reduction potential for the reaction$[Co(NH_3)_6]^{3+}(aq) + e^- \to [Co(NH_3)_6]^{2+} (aq)$ is + 0.1 V. Calculate the cell potentials to show whether the complex ions, $[Co(H_2O)_6]^{2+} \text{and/or} [Co(NH_3)_6]^{2+},$can be oxidized to the corresponding cobalt(III) complex by oxygen.

The standard reduction potential for the reduction of oxygen to water is +1.23 V:

$$O_2(g) + 4H^+(aq) + 4e^- → 2H_2O(l) E° = +1.23 V$$

To determine whether the complex ions $[Co(H_2O)_6]^{2+}$ and $[Co(NH_3)_6]^{2+}$ can be oxidized to the corresponding cobalt(III) complex by oxygen, we need to calculate the cell potentials for the reactions.

For the reaction involving $[Co(H_2O)_6]^{2+}$, the cell potential is calculated as follows:

$$E°cell = E°cathode - E°anode = E°(\frac{O_2}{H_2O}) - E°\left(\frac{[Co(H_2O)_6]^{3+}}{[Co(H_2O)_6]^{2+}}\right) = (+1.23 V) - (+1.8 V) = -0.57 V$$

Since the cell potential is negative, this reaction is not spontaneous and $[Co(H_2O)_6]^{2+}$ cannot be oxidized to $[Co(H_2O)_6]^{3+}$ by oxygen.

For the reaction involving $[Co(NH_3)_6]^{2+}$, the cell potential is calculated as follows:

$$E°cell = E°cathode - E°anode = E°(\frac{O_2}{H_2O}) - E°\left(\frac{[Co(NH_3)_6]^{3+}}{[Co(NH_3)_6]^{2+}}\right) = (+1.23 V) - (+0.1 V) = +1.13 V$$

Since the cell potential is positive, this reaction is spontaneous and $[Co(NH_3)_6]^{2+}$ can be oxidized to $[Co(NH_3)_6]^{3+}$ by oxygen.

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#### Aldebaran

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##### Re: Computing cell potentials
« Reply #1 on: April 04, 2023, 10:09:42 AM »
I would say you are correct on the basis of the reduction potentials you give. In fact it is known that the ammonium 2+ complex will change colour when left standing in air due to oxidation to the 3+ complex .
I’m not aware that the simple hexaaqua complex is oxidised which would be as expected by looking at the reduction potentials given. This is assuming all are under standard conditions. Things may be changed by concentration variations.