${MnO}_{2} (s) + 4 HCl(aq) \to {MnCl}_{2} (aq) + {Cl}_{2} (g) + 2 H_{2} O(l)$ In the above reaction, which one of them is getting reduced?

{MnO}_{2}

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HCl

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{Cl}_{2}

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{MnCl}_{2}

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Solution

The correct option is A ${MnO}_{2}$
The substance is said to be reduced if it gains hydrogen, loses oxygen or gains electrons.
In the above reaction, ${MnO}_{2}$ (manganese dioxide) is transformed to ${MnCl}_{2}$ (manganese chloride) by gaining electrons.
${Mn}^{4 +} + 2 e^{-} \to {Mn}^{2 +}$
Therefore, ${MnO}_{2}$ is getting reduced.

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Similar questions

{MnO}_{2} (s) + 4 HCl(aq) \to {MnCl}_{2} (aq) + {Cl}_{2} (g) + 2 H_{2} O(l)

In the above reaction, which one of them is getting reduced?

Q. How many grams of

{MnO}_{2}

are needed to get 36.2g of

{MnCl}_{2}

from the equation given below?

{MnO}_{2} (s) + 4 HCl (aq) \to {MnCl}_{2} (aq) + 2 H_{2} O (l) + {Cl}_{2} (g) 87 g 36.5 g 126 g 18 g 71 g

${MnO}_{2} (s) + 4 HCl(aq) \to {MnCl}_{2} (aq) + {Cl}_{2} (g) + 2 H_{2} O(l)$

Identify the oxidising and reducing agents in the above reaction. Give an example of a decomposition reaction. [3 MARKS]

Q. Chlorine is prepared in the laboratory by treating manganese dioxide (

M n O_{2}

) with aqueous hydrochloric acid according to the reaction.

4 H C l (a q) + M n O_{2} (s) \to 2 H_{2} O (l) + M n C l_{2} (a q) + C l_{2} (g)

How many grams of

H C l

react with 5.0 g of manganese dioxide?

Q. Chlorine can be prepared by reacting MCI with

M n O_{2}

. The reaction is represented by the equation,

M n O_{2} (s) + 4 H C I (a q .) \to C l_{2} (g) + M n C l_{2} (a q .) + 2 H_{2} O (l)

Assuming that the reaction goes to completion, what mass of cone. HCI solution (36% HCI by mass) is needed to produce 2.5 g

C l_{2}

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MnO2 (s)+4HCl(aq)→MnCl2(aq)+Cl2(g)+2H2O(l) In the above reaction, which one of them is getting reduced?

The correct option is A MnO2The substance is said to be reduced if it gains hydrogen, loses oxygen or gains electrons. In the above reaction, MnO2 (manganese dioxide) is transformed to MnCl2 (manganese chloride) by gaining electrons. Mn4++2e−→Mn2+ Therefore, MnO2 is getting reduced.

${MnO}_{2} (s) + 4 HCl(aq) \to {MnCl}_{2} (aq) + {Cl}_{2} (g) + 2 H_{2} O(l)$ In the above reaction, which one of them is getting reduced?