Answer:
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l) + 6 e⁻
Explanation:
In order to balance a half-reaction we use the ion-electron method.
Step 1: Write the half-reaction
Cr³⁺(aq) = Cr₂O₇²⁻(aq)
Step 2: Perform the mass balance, adding H₂O(l) and OH⁻(aq) where appropriate
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l)
Step 3: Perform the electric balance, adding electrons where appropriate
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l) + 6 e⁻
The balanced half-reaction will be "[tex]Cr_2O_7^{2-} +6e^{-1} +7 H_2O \rightarrow 2Cr^{3+}+14OH^-[/tex]". A further explanation is below.
Whenever electrons have been introduced to a process, it indicates that reduction was indeed going progress. Oxidation seems to be the loss of electrons from either a reaction mechanism.
In [tex]Cr_2O_7^{2-}[/tex], the oxidation state will be:
then,
→ [tex]Cr_2O_7^{2-}+3e^{-1} \rightarrow Cr^{3+}[/tex]
By adding "[tex]H_2O[/tex]" in reactant as well as "[tex]OH^-[/tex]" in product side, we get
→ [tex]Cr_2O_7^{2-}+3e^{-1} + H_2O \rightarrow Cr^{3+}+ OH^-[/tex]
hence,
→ [tex]Cr_2O_7^{2-}+6e^{-1} +7H_2O \rightarrow 2Cr^{3+}+14 OH^-[/tex]
Thus the above answer is right.
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