Chemistry, 02.12.2019 21:31 khristaviaaa

The following change occurs in acidic solution: s^2+ + cr_2o_7^2+ s + cr^3+ complete and balance the foregoing equation. in the balanced equation, for every mole of cr_2o_7^2 that reacts, moles of h+ are consumed. a. 5 b. 7 c. 8 d. 10 e. 14

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Answer from: Pbaer4861

Answer : The correct option is, (E) 14

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.Now balance the main atoms in the reaction.Now balance the hydrogen and oxygen atoms on both the sides of the reaction.If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion (H^+)

at that side where the less number of hydrogen are present.Now balance the charge.

The given chemical reaction is,

$S^{2-}+Cr_2O_7^{2-}\rightarrow S+Cr^{3+}$

The oxidation-reduction half reaction will be :

Oxidation : $S^{2-}\rightarrow S$

Reduction : $Cr_2O_7^{2-}\rightarrow Cr^{3+}$

First balance the main element in the reaction.

Oxidation : $S^{2-}\rightarrow S$

Reduction : $Cr_2O_7^{2-}\rightarrow 2Cr^{3+}$

Now balance oxygen atom on both side.

Oxidation : $S^{2-}\rightarrow S$

Reduction : $Cr_2O_7^{2-}\rightarrow 2Cr^{3+}+7H_2O$

Now balance hydrogen atom on both side.

Oxidation : $S^{2-}\rightarrow S$

Reduction : $Cr_2O_7^{2-}+14H^+\rightarrow 2Cr^{3+}+7H_2O$

Now balance the charge.

Oxidation : $S^{2-}\rightarrow S+2e^-$

Reduction : $Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O$

In order to balance the charge, we multiply oxidation reaction by 3. Thus, added both equation, we get the balanced redox reaction.

The balanced chemical equation in acidic medium will be,

$3S^{2-}+Cr_2O_7^{2-}+14H^+\rightarrow 3S+2Cr^{3+}+7H_2O$

From the balanced chemical equation we conclude that, the number of moles of H^+ consumed from every mole of $Cr_2O_7^{2-}$ are 14 moles.

Hence. the correct option is, (E) 14

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The following change occurs in acidic solution: s^2+ + cr_2o_7^2+ s + cr^3+ complete and balance th...