Allen is carrying out reactions in a lab. One of the products in all reactions is found to be either a precipitate, a gas, or wa
2 answers:
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The 2 in front of in
is <u>Coefficient.</u>
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The after
in the same equation is <u>sub </u><u>script</u><u>.</u>
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There are two atoms of in
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Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺ || Ag⁺
| Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu ⇄ Cu²⁺
+ 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺ + 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu ⇄ Cu²⁺
+ 2e⁻
Ag⁺ + e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺ + 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu + 2Ag⁺
+ 2e⁻⇄ Cu²⁺
+ 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu + 2Ag⁺
⇄ Cu²⁺
+ 2Ag
Answer:
The answer is "Greater than zero, and greater than the rate of the reverse reaction".
Explanation:
It applies a rate of reaction to the balance, a forward response dominates until it reaches a constant. This process is balanced before 52 mmol of the reactant, to which 3 is added. In balance, that rate of the forward reaction was its rate with forwarding reaction, both of which are higher than 0 as the response has achieved balance so that both species get a level greater than 0.