Looking at the standard reduction not easily oxidized but [Co(H_2O)_6]^2+ is pot

Question

Looking at the standard reduction not easily oxidized but [Co(H_2O)_6]^2+ is potentials below, explain why [Co(NH_3O)_6]^2+ is Using the standard reduction potentials shown in (a), show that one can prepare an ammine complex from CoCl_2 and hydrogen peroxide in the presence of ammonia but not in its absence. You will need to write two redox reactions, calculate standard potentials for the reactions, and make conclusions. That is, set up an equation to calculate E degree (V) using one cobalt complex half-cell with the peroxide half-cell, then calculate E degree (V) again using the other cobalt complex and peroxide. Compare the two E degree values.

Explanation / Answer

1.

(a) The higher the standard reduction potential for a reaction is, the stronger is the oxidizing agent it is. So the species itself gets reduced in the process. In here the reduction potential for [Co(H2O)6]2+ is higher than the amine complex and thus it gets preferentially reduced and not oxidized and amine complex gets oxidized with lower reduction potential.

(b) In the presence of H2O2, the amine complex is formed and the net potential would be,

Eo = 1.77 - 0.10 = 1.67 V

A positive cell potential means the reaction is spontaneous in the forward direction. So the amine complex is formed in the given reaction.

H2O2 + 2e- --> 2OH-

2[Co(NH3)6]2+ ---> 2[Cu(NH3)6]3+ + 2e-

-----------------------------------------------------------------

H2O2 + [Co(NH3)6]2+ ---> 2OH- + [Co(NH3)6]3+

Is the net spontaneous reaction.

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