The half cell involving the reaction Cr2O72- aq - 14 H- aq - 6e - 2Cr3- aq - 7H2O can be represented as -

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Nernst Equation

The half cell...

Question

The half cell involving the reaction $C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e \to 2 C r^{3 +} (a q) + 7 H_{2} O$ can be represented as :

P t, H^{+} (a q), C r_{2} O_{7}^{2 -} (a q)

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P t, H^{+}, C r^{3 +} (a q)

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P t, H_{2}, C r_{2} O_{7}^{2 -} (a q), C r^{3 +} (a q)

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P t, H^{+}, C r_{2} O_{7}^{2 -} (a q), C r^{3 +} (a q)

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Solution

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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

The oxidation half reaction for following reaction is

Fe²⁺(aq) + Cr₂O₇^2-(aq) → Fe³⁺ (aq) + Cr³⁺(aq)

S_{2} O {_{6}}^{2 -} (a q) + C r_{2} O {_{7}}^{2 -} (a q) \to S_{4} O {_{6}}^{2 -} (a q) + C r^{3 +} (a q)

C r_{2} O_{7}^{2 -} (a q) + S O_{3}^{2 -} (a q) \to C r^{3 +} (a q) + S O_{4}^{2 -} (a q)

C r_{2} O_{7}^{2 -} (a q, 1 M) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q, 1 M) + 7 H_{2} O

p H = 2.0

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