The electrode with reaction - Cr2O72- aq - 14H- aq - 6e- - 2Cr3- aq - 7H2O can be represented as-

The correct option is B Pt_H_2 | H^⊕ (aq), Cr_2O_7^2 (aq), Cr^3+ (aq) Cr_2O_7^2 reduces to 2Cr^3+, so must be represented at cathode. ...

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Representation of Cell

The electrode...

Question

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

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

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

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P t_{H_{2}} | H^{\oplus} (a q), C r_{2} O_{7}^{2 -}

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P t_{H_{2}} | H^{\oplus} (a q), C r_{2} O_{7}^{2 -} (a q), C r^{3 +} (a q)

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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

The oxidation half reaction for following reaction is

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

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

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) + I^{-} (a q) + H^{+} \to

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