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3 years ago

Which equation represents a redox reaction?

2 answers:

4 0

Transfer of electrons from one species to another species is called a redox reaction. it consists of one species that will give out electrons called an oxidation reaction and the other species take in electrons called reduction reactions. both these half equations together make up a redox reaction.
In oxidation, the oxidation state increases of the species that is oxidised.
In reduction, oxidation state of the species decreases of the species that is reduced.
1)
2AgNO₃(aq) + Cu(s) → Cu(NO₃)₂(aq) + 2Ag(s)
single displacement reaction
oxidation number of Ag decreases from +1 to 0 - reduction reaction
oxidation number of Cu increases from 0 to +2 - oxidation
since both oxidation and reduction reactions take place its a redox reaction.

2)
BaCl₂(aq) + Na₂CrO₄(aq) → BaCrO₄(s) + 2NaCl(aq)
its a double displacement reaction and the ions have been exchanged.
Oxidation numbers of the elements do not change therefore this is not a redox reaction.

3)
H₃PO₄(aq) + 3KOH(aq) → 3H₂O(l) + K₃PO₄(aq)
acid base reaction where the oxidation state of the elements remain the same so its not a redox reaction.
4)
CaCO₃(s) → CaO(s) + CO₂(g)?
decomposition reaction where the oxidation states remain the same hence not a redox reaction

Katarina [22]3 years ago

4 0

Answer:

2agno3(aq) + cu(s) → cu(no3)2(aq) + 2ag(s)

Explanation:

Now let us take a good look at the reaction presented in the answer.

If we look at silver on the lefthand side of the reaction equation, we will realize that silver is in the +2 oxidation state but on the right hand side of the reaction equation, silver is in the zero oxidation state. This implies that silver has been reduced. Reduction refers to a decrease in oxidation number.

Now let us move to copper. Copper is in the zero oxidation state on the lefthand side of the reaction equation but on the right hand side of the reaction equation, it is in the +2 oxidation state. Hence, there is an increase in the oxidation number of copper- oxidation of cooler has occurred. Oxidation rs to a positive increase in oxidation number.

Hence the reaction involves oxidation and reduction, thus it s a redox reaction.

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N the reaction Mg + 2HCl -> MgCl2 + H2 which element’s oxidation number does not change?

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Answer:- Oxidation number of Cl does not change as it is -1 on both sides.

Explanations:- oxidation number of Mg on reactant side is 0 as it is in its elemental form(not combined with another element).

Oxidation number of hydrogen in its compounds is +1, so if H is +1 in HCl the oxidation number of Cl is -1 as the sum has to be zero.

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So, it is only chlorine(Cl) whose oxidation number does not change for the given equation.

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3 years ago

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

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Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

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