Chemistry, 06.12.2019 06:31 tbeck225

Balance the following reaction if it takes place in acidic conditions, making sure that the coefficients are the least common multiple and that extra molecules are cancelled out. drag and drop the correct coefficients (even if it is 1) and molecules into the correct location. the coefficients can be used multiple times. + al3+ (aq) + cr3+(aq) → al (s) + cr2o72- (aq) + multiple tries are permitted, however, 1/3 (33%) point will be deducted for each incorrect submission

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Answer from: smithad242

Answer : The overall balanced chemical equation in acidic medium will be,

$2Al^{3+}(aq)+2Cr^{3+}(aq)+7H_2O(l)\rightarrow 2Al+Cr_2O_7^{2-}(aq)+14H^+(aq)$

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,

$Al^{3+}(aq)+Cr^{3+}(aq)\rightarrow Al(s)+Cr_2O_7^{2-}(aq)$

The oxidation-reduction half reaction will be :

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

Reduction : $Al^{3+}\rightarrow Al$

First balance the main element in the reaction.

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

Reduction : $Al^{3+}\rightarrow Al$

Now balance oxygen atom on both side.

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

Reduction : $Al^{3+}\rightarrow Al$

Now balance hydrogen atom on both side.

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

Reduction : $Al^{3+}\rightarrow Al$

Now balance the charge.

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

Reduction : $Al^{3+}+3e^-\rightarrow Al$

The charges are not balanced. Now we are multiplying reduction reaction by 2 and then adding both equation, we get the balanced redox reaction.

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

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

The overall balanced chemical equation in acidic medium will be,

$2Al^{3+}(aq)+2Cr^{3+}(aq)+7H_2O(l)\rightarrow 2Al+Cr_2O_7^{2-}(aq)+14H^+(aq)$

Answer from: Quest

a) the particles of the chair are strongly held together.

Answer from: Quest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$

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