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Electrode Potential

Write half-re...

Question

Write half-reactions using electrons.

${C r}_{2} O_{2}^{7 -} + 6 {F e}^{2 +} + 14 H^{+} ⟶ 2 {C r}^{3 +} + 6 {F e}^{3 +} + 7 H_{2} O$

Oxidation

: {F e}^{2 +} ⟶ {F e}^{3 +} + e^{-}

; Reduction

: {C r}_{2} O_{2}^{7 -} + 14 H^{+} + 6 e^{-} ⟶ 2 {C r}^{2 +} + 7 H_{2} O

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Oxidation

: {3 F e}^{2 +} ⟶ {3 F e}^{3 +} + e^{-}

; Reduction

: {C r}_{2} O_{2}^{7 -} + 14 H^{+} + 6 e^{-} ⟶ 2 {C r}^{2 +} + 7 H_{2} O

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Oxidation

: {2 F e}^{2 +} ⟶ {2 F e}^{3 +} + e^{-}

; Reduction

: {C r}_{2} O_{2}^{7 -} + 14 H^{+} + 6 e^{-} ⟶ 2 {C r}^{2 +} + 7 H_{2} O

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Oxidation

: {3 F e}^{2 +} ⟶ {F e}^{3 +} + {3 e}^{-}

; Reduction

: {C r}_{2} O_{2}^{7 -} + 14 H^{+} + 3 e^{-} ⟶ 2 {C r}^{2 +} + 7 H_{2} O

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Solution

The correct option is A Oxidation $: {F e}^{2 +} ⟶ {F e}^{3 +} + e^{-}$ ; Reduction $: {C r}_{2} O_{2}^{7 -} + 14 H^{+} + 6 e^{-} ⟶ 2 {C r}^{2 +} + 7 H_{2} O$
The balanced chemical equation representing the oxidation of ferrous ions with dichromate ions is as follows:
${C r}_{2} O_{7}^{2 -} + 6 {F e}^{2 +} + 14 H^{+} ⟶ 2 {C r}^{3 +} + 6 {F e}^{3 +} + 7 H_{2} O$
The oxidation half reaction is $: {F e}^{2 +} ⟶ {F e}^{3 +} + e^{-}$ .
It represents the oxidation of ferrous ions to ferric ions.
The reduction half reaction is $: {C r}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} ⟶ 2 {C r}^{3 +} + 7 H_{2} O$ .
It represent the reduction of dichromate ions to Cr(III) ions.

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Similar questions

{C r}_{2} O_{7}^{2 -} + {F e}^{2 +} + H^{\oplus} ⟶ {C r}^{3 +} + {F e}^{3 +} + H_{2} O

6 {F e}^{2 +} + 14 H^{\oplus} + {C r}_{2} O_{7}^{2 -} ⟶ 6 {F e}^{3 +} + 2 {C r}^{3 +} + 7 H_{2} O

{C r}_{2} O_{7}^{- 2} + 14 H^{+} + 6 e^{-} \to 2 {C r}^{+ 3} + 7 H_{2} O; E^{o} = 1.33 V

[{C r}_{2} O_{7}^{- 2}] = 4.5

[{C r}^{+ 3}] = 15

E

1.067 V

E^{0}

I . M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{0} = 1.5 V

I I . S n^{2 +} \to S n^{4 +} + 2 e^{-} E^{0} = - 0.15 V

I I I . C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{2 +} + 7 H_{2} O E^{0} = 1.33 V

I V . C e^{3 +} \to C e^{4 +} + e^{-} E^{0} = - 1.61 V

C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e \to 2 C r^{3 +} (a q) + 7 H_{2} O

C r_{2} O_{7}^{2 -} (a q) + 14 H^{\oplus} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O

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