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Identify the oxidized substance, the reduced substance, the oxidizing agent, and the reducing agent in the redox reaction. Cu(s)

spin [16.1K]

Answer: Copper is getting oxidized and is a reducing agent. Silver is getting reduced and is oxidizing agent.

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

$X^{n+}+ne^-\rightarrow X$

Oxidizing agents are defined as the agents which oxidize other substance and itself gets reduced. These agents undergoes reduction reactions.

Reducing agents are defined as the agents which reduces the other substance and itself gets oxidized. These agents undergoes reduction reactions.

For the given chemical reaction:

$Cu(s)+2AgNO_3(aq.)\rightarrow 2Ag(s)+Cu(NO_3)_2(aq.)$

The half reactions for the above reaction are:

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$

Reduction half reaction: $2Ag^+(aq.)+2e^-\rightarrow 2Ag(s)$

From the above reactions, copper is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.

Silver is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.

4 0

2 years ago

What is whether? what is climate? Does todays weather impact climate?

Tcecarenko [31]

Answer:

Weather is the state in which a certain part inside the atmosphere is hot or cold.Climate is the usual weather in a certain place in the world or atmosphere.One way weather can affect climate is global warming,which is when the world gets hotter and the ice in the world melts and water rises.

Explanation:

5 0

2 years ago

A 23.5g aluminum block is warmed to 65.9°C and plunged into an insulated beaker containing 55.0g water initially at 22.3°C. The

atroni [7]

Answer:

25.97oC

Explanation:

Heat lost by aluminum = heat gained by water

M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]

Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC

Let Temp(Al+H2O) = X

23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)

21.15(65.9-X) = 230.23(X-22.3)

1393.785 - 21.15X = 230.23X – 5134.129

230.23X + 21.15X = 1393.785 + 5134.129

251.38X = 6527.909

X = 6527.909/251.38

X = 25.97oC

So, the final temperature of the water and aluminum is = 25.97oC

4 0

2 years ago

How many grams are there in 10.0 moles of Al2O3?

irinina [24]

Explanation:

1 mole is equal to 1 moles Al2O3, or 101.961276 grams.

8 0

2 years ago

What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Br2(l) and I-(aq) Use the reduction po

Vikki [24]

Answer:

1.58×10E18

Explanation:

Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.

E°cell= 1.07- 0.53= 0.54 V

E°cell= 0.0592/n logK

0.54 = 0.0592/2 logK

logK= 0.54/0.0296

logK= 18.2

K= Antilog (18.2)

K= 1.58×10^18

3 0

2 years ago