Given the standard electrode potentials E^{0} | vedclass

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(B) The standard reduction potentials are given as $E^{0}_{Fe^{3+}|Fe^{2+}} = 0.77 \, V$ and $E^{0}_{Fe^{2+}|Fe} = -0.44 \, V$. Since $E^{0}_{Fe^{3+}|

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$Fe^{3+}$ will decrease

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(B) The standard reduction potentials are given as $E^{0}_{Fe^{3+}|Fe^{2+}} = 0.77 \, V$ and $E^{0}_{Fe^{2+}|Fe} = -0.44 \, V$. Since $E^{0}_{Fe^{3+}|Fe^{2+}} > E^{0}_{Fe^{2+}|Fe}$,the reaction $2Fe^{3+} + Fe \rightarrow 3Fe^{2+}$ is spontaneous. In this reaction,$Fe^{3+}$ acts as an oxidizing agent and gets reduced to $Fe^{2+}$,while $Fe$ acts as a reducing agent and gets oxidized to $Fe^{2+}$. Therefore,the concentration of $Fe^{3+}$ will decrease and the concentration of $Fe^{2+}$ will increase.

Given the standard electrode potentials $E^{0}_{Fe^{2+}|Fe} = -0.44 \, V$ and $E^{0}_{Fe^{3+}|Fe^{2+}} = 0.77 \, V$. If $Fe^{2+}$,$Fe^{3+}$,and $Fe$ are all present together,then:

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