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Komok [63]
4 years ago
15

Identify the oxidizing agent and the reducing agent in the following reactions: (i) 8NH3( g) + 6NO2( g) => 7N2( g) + 12H2O( l

) (ii) Zn(s) +AgNO3(aq) => Zn(NO3) 2(aq) + Ag(s) (unbalanced)
Chemistry
1 answer:
shusha [124]4 years ago
4 0

Answer:

(i)  Oxidizing Agent: NO2 / Reducing Agent NH3-

(ii) Oxidizing Agent AgNO3 / Reducing Agent Zn

Explanation:

(i) 8NH3( g) + 6NO2( g) => 7N2( g) + 12H2O( l)

In this reaction, both two reactants contain nitrogen with a different oxidation number and produce only one product which contains nitrogen with a unique oxidation state. So, nitrogen is oxidized and reduced in the same reaction.

Nitrogen Undergoes a change in oxidation state from 4+ in NO2 to 0 in N2. It is reduced because it gains electrons (decrease its oxidation state). NO2 is the oxidizing agent (electron acceptor).

Nitrogen Changes from an oxidation state of 3- in NH3 to 0 in N2. It is oxidized because it loses electrons (increase its oxidation state). NH3 is the reducing agent (electron donor)

(ii) Zn(s) +AgNO3(aq) => Zn(NO3)2(aq) + Ag(s)

Ag changes oxidation state from 1+ to 0 in Ag(s).

Ag is reduced because it gains electrons and for this reason and AgNO3 is the oxidizing agent (electron acceptor)

Zn Changes from an oxidation state of 0 in Zn(s) to 2+ in Zn(NO3)2. It is oxidized and for this reason Zn is the reducing agent (electron donor).

Balanced equation:

Zn(s) +2AgNO3(aq) => Zn(NO3)2(aq) + 2Ag(s)

 

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Answer:

The catalyzed reaction will take time of 5.11\times 10^{-8} years.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate of reaction with catalyst

K_1 = rate of reaction without catalyst

Ea_2 = activation energy with catalyst  = 59.0 kJ/mol = 59000 J/mol

Ea_1 = activation energy without catalyst  = 184 kJ/mol = 184000 J/mol

R = gas constant

T = temperature = 600K

Now put all the given values in this formula, we get:

\frac{K_2}{K_1}=e^{\frac{184,000 kJ-59000 kJ}{R\times 300}}=7.632\times 10^{10}

The reaction enhances by 7.632\times 10^{10}  when catalyst is present.

Time taken by reaction without catalyzed = 3900 years

Time taken by reaction with catalyzed = x

x=\frac{3900 year}{7.632\times 10^{10}}=5.11\times 10^{-8} years

The catalyzed reaction will take time of 5.11\times 10^{-8} years.

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