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Musya8 [376]
4 years ago
13

Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag. Of the two metals, copper

and silver, which is the anode and which is the cathode? Label the oxidizing agent and the reducing agent.
Chemistry
1 answer:
KatRina [158]4 years ago
6 0

Answer:

Answer is given below.

Explanation:

Anode is that electrode where oxidation occurs. Cathode is that electrode where reduction occurs.

In cell representation, half cell present left to salt-bridge notation(\parallel ) is anodic system and another half cell present right to salt-bridge notation(\parallel ) is cathodic system.

So anode is Cu and cathode is Ag.

oxidation: Cu-2e^{-}\rightarrow Cu^{2+}(aq.)

[reduction: Ag^{+}+e^{-}\rightarrow Ag]\times 2

-----------------------------------------------------------------------------------------------

chemical equation: Cu+2Ag^{+}(aq.)\rightarrow Cu^{2+}(aq.)+2Ag

Oxidizing agent is that species which takes electron from another species. Here Ag^{+}(aq.) takes electron from Cu. Hence Ag^{+}(aq.)  is the oxidizing agent.

Reducing agent is that species which gives electron to another species. Here Cu gives electron to Ag^{+}(aq.) . Hence Cu  is the reducing agent.

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Answer:

a. Molarity= M =2.1x10^{-1}M

b. Molality= m=2.0x10^{-1}m

Explanation:

Hello,

In this case, given the information about the aniline, whose molar mass is 93g/mol, one could assume the volume of the solution is just 200 mL (0.200 L) as no volume change is observed when mixing, therefore, the molarity results:

M=\frac{n_{solute}}{V_{solution}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L} =2.1x10^{-1}M

Moreover, the molality:

m=\frac{n_{solute}}{m_{solvent}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L*\frac{1.05kg}{1L} } =2.0x10^{-1}m

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(C) The recrystallization solvent should be nonvolatile.

Explanation:

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8 0
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