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givi [52]
1 year ago
14

Solid iron combines with oxygen gas to form solid iron(III) oxide. Which of the following equations best describes this reaction

?
(A) Ir + O₂--> IrO₂
(B) 4 Fe + 3 0₂---> 2 Fe₂O3
(C) I+ 0₂ ---> 10₂
(D) 3 Ir + 0₂ -> 6 Ir₂03
Select one:
O a. A
Ob. B
O C. C
O d. D
Chemistry
1 answer:
Alexus [3.1K]1 year ago
3 0

Answer:

B.  4 Fe + 3 0₂---> 2 Fe₂O3        

Explanation:

(A) Ir + O₂--> IrO₂                                 No:  Ir is iridium

(B) 4 Fe + 3 0₂---> 2 Fe₂O3                Yes:  Fe is iron and O is oxygen,                  and the equation is balanced.

(C) I+ 0₂ ---> 10₂                                  No:  I is iodine

(D) 3 Ir + 0₂ -> 6 Ir₂03                         No:  I is iodine

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3 years ago
Whats the voltage of CuCl2 + Zn -&gt; ZnCl2 + Cu
gtnhenbr [62]

Answer:

Approximately 1.10\; {\rm V} under standard conditions.

Explanation:

Equation for the overall reaction:

{\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s).

Write down the ionic equation for this reaction:

\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}.

The net ionic equation for this reaction would be:

{\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s).

In this reaction:

  • Zinc loses electrons and was oxidized (at the anode): {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}.
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Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), is reduction and is likely on the table.

  • E(\text{anode}) = -0.7618\; {\rm V} for {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), and
  • E(\text{cathode}) = 0.3419\; {\rm V} for {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

The reduction potential of {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} would be -E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}, the opposite of the reverse reaction {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s).

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}.

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Romashka-Z-Leto [24]

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Answer:

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Explanation:

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The formular relating all three parameters is given as;

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Upon solving;

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