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The standard reduction potential for the reaction $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(a q)+\mathrm{e}^{-} \longrightarrow\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}(a q)$ is about 1.8 $\mathrm{V}$ . The reduction potential for the reaction $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}(a q)+\mathrm{e}^{-} \longrightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}+(a q)$ is $+0.1 \mathrm{V}$ . Calculate the cell potentials to show whether the complex ions, $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ and/or $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ , can be oxidized to the corresponding cobalt(III) complex by oxygen.
$\left.\text { (i) The complexion }\left[\operatorname{Co(H}_{2} \mathrm{O}\right)\right]_{6}^{2+}$ cannot be oxidized to the corresponding cobait(lin) complex byoxygen. The negative value $E^{0}=-0.57 \mathrm{V}$ indized to the cerresponding cobait (III) complex by(ii) The complexion [Co(NHA)g] $2^{2+}$ an be oxidized to the corresponding coball(lil) complex byoxygen. The positive value $E^{0}=1.13$ V indicates that the reaction occurs spontaneousy.
Chemistry 102
Chemistry 101
Chapter 19
Transition Metals and Coordination Chemistry
Transition Metals
Carleton College
University of Central Florida
Drexel University
Brown University
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spontaneity off reaction can be explained with the help off electrode. Potentially, the reaction is spontaneous for a positive value off electro potential, which can be which can give negative value off, keeps free energy. The electrode potential. It's calculated using this mathematical representation. Considering the reaction off oxygen and hydrogen, the electrode potential off the reaction is given. Using the expression off standard electrode potential we can deter mined the electrode potential off the whole cell, substituting in the values for cattle would an anode the electrode. The value off E cell turns out to be negative. That is minus 0.6 folds, So the reaction is known spontaneous, considering 2nd 1 the substituting in the values off electrode cathode on dhe, easily anode the easel values off electro potential off the whole cell turns out to be positive, so the reaction is spontaneous.
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