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Using a table of standard electrode potentials, label each of the following statements as completely
true or false.
A) Cu²+ can oxidize H2, and Fe can reduce Mn²+
B) Ni²+ can oxidize Cu²+, and Fe²+ can reduce H+
C) Fe²+ can oxidize H₂, and Fe²+ can reduce Au³+.
D) Br₂ can oxidize Ni, and H₂ can reduce Mn²+
E) H* can oxidize Fe, and Ni can reduce Br₂

Sagot :

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Answer: B and E are True

Explanation:

Because Fe has a higher oxidation potential than H2, H+ can oxidize it.

[tex]$\begin{array}{llllll}\mathrm{Fe} & \rightarrow & \mathrm{Fe}^{2+} & +2 \mathrm{e}- & \mathrm{E}^{0}=0.44 \mathrm{~V} \\ \mathrm{H} & \rightarrow & \mathrm{H}^{+} & + & \mathrm{e}^{-} & \mathrm{E}^{0}=0 \mathrm{~V}\end{array}$[/tex]

Ni can reduce Br2 because the reduction potential of Ni is more than Br2.

[tex]\mathrm{Ni}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}^{-} \quad \mathrm{E}^{0}=-0.25 \mathrm{~V}\\2 \mathrm{Br}^{-} \quad \rightarrow \mathrm{Br}_{2}+2 \mathrm{e}^{-} \quad \mathrm{E}^{0}=-1.087 \mathrm{~V}$[/tex]

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