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4 years ago

What is the coefficient of silver in the final, balanced equation for this reaction?

Chemistry

1 answer:

vekshin14 years ago

8 0

This is an incomplete question, the complete question is attached below.

Answer : The coefficient of silver in the final, balanced equation for this reaction is, 3

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.

The given redox reaction is,

$Ag^+(aq)+Al(s)\rightarrow Ag(s)+Al^{3+}(aq)$

The oxidation-reduction half reaction will be :

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

Reduction : $Ag^{+}+1e^-\rightarrow Ag$

In order to balance the electrons, we multiply the reduction reaction by 3 and then added both equation, we get the balanced redox reaction.

The balanced redox reaction will be,

$3Ag^+(aq)+Al(s)\rightarrow 3Ag(s)+Al^{3+}(aq)$

From the balanced redox reaction we conclude that, the coefficient of silver in the final balanced equation for this reaction is 3.

Hence, the correct option is 3.

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A heterogeneous mixture is a :

4vir4ik [10]

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3 years ago

Read 2 more answers

8. Select the lattice energy for rubidium chloride from the following data (in kJ/mol]

yKpoI14uk [10]

Answer:

Option C

Explanation:

The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number

Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6

Equation 1 represents the change in enthalpy for formation of RbCl

Equation 2 represents the sublimation reaction of rubidium

Equation 3 represents the ionization enthalpy of rubidium

Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule

Equation 5 represents the electron affinity of chlorine

To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6

We have to perform operations such as

Equation 1 - equation 2 - equation 3 - equation 4 - equation 5

By performing these operations the intermediate compounds gets cancelled and at last we get equation 6

So Equation 1 ≡ ΔH $_{f}$ = -431 kJ/mol

Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8 kJ/mol

Equation 3 ≡ IE1(Rb) = 397.5 kJ/mol

Equation 4 ≡ BE(Cl2) = 226 kJ/mol

Equation 5 ≡ Electron Affinity Cl = -332 kJ/mol

Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol

∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol

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3 years ago

Defintion of isometry

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3 years ago

What was the major weakness of Wegener's theory of continental drift

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Alfred Wegener proposed the continental drift theory in four reasons. One, he hypothesized that the continents were once a giant continent. Two, he observed that there are fossils found of the same kind in the different locations. Three, he found that the edge of opposing continents somehow fits if they are combined, almost like a puzzle. And four, there are glacial scars that are left behind by a larger glacial continent. His greatest weakness of Alfred Wegeners’ continental drift theory was he could not explain what kind of force that caused this continents to move far from each other.

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3 years ago

Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

gtnhenbr [62]

Answer:

Approximately $1.10\; {\rm V}$ under standard conditions.

Explanation:

Equation for the overall reaction:

${\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s)$ .

Write down the ionic equation for this reaction:

$\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}$ .

The net ionic equation for this reaction would be:

${\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)$ .

In this reaction:

Zinc loses electrons and was oxidized (at the anode): ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ .
Copper gains electrons and was reduced (at the cathode): ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , is reduction and is likely on the table.

$E(\text{anode}) = -0.7618\; {\rm V}$ for ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , and
$E(\text{cathode}) = 0.3419\; {\rm V}$ for ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

The reduction potential of ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ would be $-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}$ , the opposite of the reverse reaction ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ .

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

$\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}$ .

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2 years ago