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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48

Problem 15 Easy Difficulty

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.

Answer

$\left.\text { (i) The complexion }\left[\operatorname{Co(H}_{2} \mathrm{O}\right)\right]_{6}^{2+}$ cannot be oxidized to the corresponding cobait(lin) complex by
oxygen. 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 by
oxygen. The positive value $E^{0}=1.13$ V indicates that the reaction occurs spontaneousy.

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Chapter 19

Transition Metals and Coordination Chemistry

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Video Transcript

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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