Two half reactions for the following equation are:
$4 H_{2} S O_{4} + K_{2} C r_{2} O_{7} + 3 N a_{2} S O_{3} ⟶ C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 3 N a_{2} S O_{4} + 4 H_{2} O$

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Solution

The correct options are
A

B

$K_{2} C r_{2} O_{7} + N a_{2} S O_{3} ⟶ C r^{+ 3} + S O_{4}^{2 -}$

Now, Basic skeleton eq is:

$C r_{2} O_{7}^{- 2} + S O_{3}^{2 -} ⟶ C r^{+ 3} + S O_{4}^{- 2}$

$C r_{2} O_{7}^{- 2} ⟶ C r^{+ 3}$

$S O_{3}^{- 2} ⟶ S O_{4}^{- 2}$

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Similar questions

Two half reactions for the following equation are:
$4 H_{2} S O_{4} + K_{2} C r_{2} O_{7} + 3 N a_{2} S O_{3} ⟶ C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 3 N a_{2} S O_{4} + 4 H_{2} O$

Q. Balance the following chemical equation:

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)

(Acidic medium)

Q. According to the following equation,

K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 4 H_{2} O + 3 [O]

the equivalent mass of

K_{2} C r_{2} O_{7}

When $N a N O_{2}$ is oxidised by $K_{2} C r_{2} O_{7}$ , the important products formed are

S + O_{2} \to {S O}_{2}; Δ H = - 298.2 k J

{S O}_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J

{S O}_{3} + H_{2} O \to H_{2} {S O}_{4}; Δ H = - 130.2 k J

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 227.3 k J

The heat of formation of

H_{2} {S O}_{4}

will be:

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Two half reactions for the following equation are: 4H2SO4+K2Cr2O7+3Na2SO3⟶Cr2(SO4)3+K2SO4+3Na2SO4+4H2O

The correct options are A B K2Cr2O7+Na2SO3⟶Cr+3+SO2−4 Now, Basic skeleton eq is: Cr2O−27+SO2−3⟶Cr+3+SO−24 Cr2O−27⟶Cr+3 SO−23⟶SO−24

Two half reactions for the following equation are:
$4 H_{2} S O_{4} + K_{2} C r_{2} O_{7} + 3 N a_{2} S O_{3} ⟶ C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 3 N a_{2} S O_{4} + 4 H_{2} O$

The correct options are
A

B

$K_{2} C r_{2} O_{7} + N a_{2} S O_{3} ⟶ C r^{+ 3} + S O_{4}^{2 -}$

Now, Basic skeleton eq is:

$C r_{2} O_{7}^{- 2} + S O_{3}^{2 -} ⟶ C r^{+ 3} + S O_{4}^{- 2}$

$C r_{2} O_{7}^{- 2} ⟶ C r^{+ 3}$

$S O_{3}^{- 2} ⟶ S O_{4}^{- 2}$