Chemistry, 09.12.2019 18:31, mansoorahmedd67

What are the coefficients of the six species in the balanced equation below? remember to include coefficients for h2o(l) and h+(aq) in the appropriate blanks. bro3_(aq) + sn2+(aq) + → br-(aq) + sn4+(aq) +

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answered: bbb3771

Answer : The balanced chemical equation in a acidic solution is,

$BrO_3^-(aq)+6H^+(aq)+3Sn^{2+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sn^{4+}(aq)$

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,

$BrO_3^-(aq)+Sn^{2+}(aq)\rightarrow Br^-(aq)+Sn^{4+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $BrO_3^-\rightarrow Br^-$

Reduction : $Sn^{2+}\rightarrow Sn^{4+}$

First balance the main element in the reaction.

Oxidation : $BrO_3^-\rightarrow Br^-$

Reduction : $Sn^{2+}\rightarrow Sn^{4+}$

Now balance oxygen atom on both side.

Oxidation : $BrO_3^-\rightarrow Br^-+3H_2O$

Reduction : $Sn^{2+}\rightarrow Sn^{4+}$

Now balance hydrogen atom on both side.

Oxidation : $BrO_3^-+6H^+\rightarrow Br^-+3H_2O$

Reduction : $Sn^{2+}\rightarrow Sn^{4+}$

Now balance the charge.

Oxidation : $BrO_3^-+6H^+6e^-\rightarrow Br^-+3H_2O$

Reduction : $Sn^{2+}\rightarrow Sn^{4+}+2e^-$

The charges are not balanced. Now we are multiplying reduction reaction by 3 and adding both equation, we get the balanced redox reaction.

Oxidation : $BrO_3^-+6H^+6e^-\rightarrow Br^-+3H_2O$

Reduction : $3Sn^{2+}\rightarrow 3Sn^{4+}+6e^-$

The balanced chemical equation in acidic medium will be,

$BrO_3^-(aq)+6H^+(aq)+3Sn^{2+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sn^{4+}(aq)$

answered: Guest

explanation:

i think option no d is the correct answer

40 points and brainliest to anyone who can answer these

answered: Guest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.