Consider the following reaction- 6FeSO4-K2Cr2O7-7H2SO4- 3Fe2SO43-Cr2SO43-K2SO4-7H2O- How many electrons are involved in the redox reaction-

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Consider the ...

Question

Consider the following reaction: $6 F e S O_{4} + K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} \to 3 F e_{2} (S O_{4})_{3} + C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 7 H_{2} O$ . How many electrons are involved in the redox reaction?

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Solution

The correct option is B $6$
In acidic medium $C r$ is reduced from +6 to +3 as:
${C r}_{2} O_{7}^{- 2} + 14 H^{+} + 6 e^{-} \to 2 {C r}^{+ 3} + 7 H_{2} O$
and ${F e}^{2 +}$ is oxidised to ${F e}^{3 +}$ as:
${F e}^{2 +} \to F e^{3 +} + e^{-}$
To balance the oxidation half reaction, we should multiply it by 6. So 6 electrons are involved in this reaction.

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Similar questions

Q. Determine the equivalent weight of

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

and

F e S O_{4}

in the reaction.

K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 3 F e_{2} (S O_{4})_{3} + 7 H_{2} O

Q. Represent the following equation in ionic form.

K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to 3 F e_{2} (S O_{4})_{3} + C r_{2} (S O_{4})_{3} + 7 H_{2} O + K_{2} S O_{4}

Q. Which of the following is correct representation of a given molecular equation in ionic form?

K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to 3 F e_{2} (S O_{4})_{3} + C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 7 H_{2} O

Q. How many electrons are involved in the following redox reaction?

C r_{2} O_{7}^{2 -} + F e^{2 +} + C_{2} O_{4}^{2 -} \to C r^{3 +} + F e^{3 +} + C O_{2}

(Unbalanced)

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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Consider the following reaction: 6FeSO4+K2Cr2O7+7H2SO4→3Fe2(SO4)3+Cr2(SO4)3+K2SO4+7H2O. How many electrons are involved in the redox reaction?

The correct option is B 6In acidic medium Cr is reduced from +6 to +3 as:Cr2O7−2+14H++6e−→2Cr+3+7H2Oand Fe2+ is oxidised to Fe3+ as:Fe2+→Fe3++e−To balance the oxidation half reaction, we should multiply it by 6. So 6 electrons are involved in this reaction.

Consider the following reaction: $6 F e S O_{4} + K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} \to 3 F e_{2} (S O_{4})_{3} + C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 7 H_{2} O$ . How many electrons are involved in the redox reaction?