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NeTakaya
3 years ago
15

Pls help Would be much appreciated:)

Chemistry
1 answer:
Pie3 years ago
3 0

Answer:

Ok so,  b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...

i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced

ii. Which metal is the anode, and which is the cathode?-The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. The cathode is where the reduction reaction takes place.

iii. Calculate the standard potential (voltage) of the cell

Look up the reduction potential,

E

⁰

red

, for the reduction half-reaction in a table of reduction potentials

Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,

E

⁰

ox

=

-

E

⁰

red

.

iv. What kind of electrochemical cell is this? Explain your answer.

All parts in the electrochemical cells are labeled in second figure. Following are the part in electrochemical cells

1) Anode 2) Cathode 3) gold Stripe (Electrode) 4) Aluminium Glasses (Electrode) 5) Connecting wires 6) Battery

Explanation:

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4 0
3 years ago
Question 15 need this answered asap (chemistry)
Tcecarenko [31]

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80.7 L

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PV = nRT

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8 0
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Determine the concentration of a solution prepared by diluting 20.0 mL of 2.00 M NaCl to 250.0 mL. Please show your work.
yawa3891 [41]

0.16 M is the concentration of a solution prepared by diluting 20.0 ml of 2.00 M NaCl to 250.0 ml.

Explanation:

Data given:

Initial volume of NaCl, V1 = 20 ml

initial molarity of the NaCl solution = 2M

Final volume of the NaCl solution = 250 ml

final molarity of the diluted solution = ?

from the information given, the formula for dilution used is:

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putting the values in the rearranged equation:

V final = \frac{Minitial Vinitial }{Vfinal}

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6 0
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