In acidic medium, shows oxidizing action as represented in the half reaction:

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Q.
In acidic medium, $K_{2} {Cr}_{2} O_{7}$ shows oxidizing action as represented in the half reaction:

${Cr}_{2} O_{7}^{2 -} + {XH}^{+} + {Ye}^{⊝} \to 2 A + {ZH}_{2} O$

X, Y, Z and A, respectively, are [2024]

1 8, 6, 4 and ${Cr}_{2} O_{3}$

2 14, 7, 6 and ${Cr}^{3 +}$

3 8, 4, 6 and ${Cr}_{2} O_{3}$

4 14, 6, 7 and ${Cr}^{3 +}$

Ans.
(4)

In acidic medium the half reaction is balanced as follows:

(A) Balance atoms undergoing reduction i.e. chromium:

${Cr}_{2} O_{7}^{2 -} (aq) \to 2 {Cr}^{3 +} (aq)$

(B) Balance oxygen using $H_{2} O$

${Cr}_{2} O_{7}^{2 -} (aq) \to 2 {Cr}^{3 +} (aq) + 7 H_{2} O (l)$

(C) Balance hydrogen using $H^{+}$

${Cr}_{2} O_{7}^{2 -} (aq) + 14 H^{+} (aq) \to 2 {Cr}^{3 +} (aq) + 7 H_{2} O (l)$

(D) Balance charge using electrons:

${Cr}_{2} O_{7}^{2 -} (aq) + 14 H^{+} (aq) + 6 e^{-} \to 2 {Cr}^{3 +} (aq) + 7 H_{2} O (l)$

On comparing with given reaction we get,

$X = 14, Y = 6, Z = 7, A = {Cr}^{3 +}$

Solution

Q.
In acidic medium, $K_{2} {Cr}_{2} O_{7}$ shows oxidizing action as represented in the half reaction:

${Cr}_{2} O_{7}^{2 -} + {XH}^{+} + {Ye}^{⊝} \to 2 A + {ZH}_{2} O$

X, Y, Z and A, respectively, are [2024]

1 8, 6, 4 and ${Cr}_{2} O_{3}$

2 14, 7, 6 and ${Cr}^{3 +}$

3 8, 4, 6 and ${Cr}_{2} O_{3}$

4 14, 6, 7 and ${Cr}^{3 +}$

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Solution

Q. In acidic medium, K2Cr2O7 shows oxidizing action as represented in the half reaction: Cr2O72-+XH++Ye⊝→2A+ZH2O X, Y, Z and A, respectively, are [2024]

1 8, 6, 4 and Cr2O3

2 14, 7, 6 and Cr3+

3 8, 4, 6 and Cr2O3

4 14, 6, 7 and Cr3+

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Q.
In acidic medium, $K_{2} {Cr}_{2} O_{7}$ shows oxidizing action as represented in the half reaction:

${Cr}_{2} O_{7}^{2 -} + {XH}^{+} + {Ye}^{⊝} \to 2 A + {ZH}_{2} O$

X, Y, Z and A, respectively, are [2024]

1 8, 6, 4 and ${Cr}_{2} O_{3}$

2 14, 7, 6 and ${Cr}^{3 +}$

3 8, 4, 6 and ${Cr}_{2} O_{3}$

4 14, 6, 7 and ${Cr}^{3 +}$