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Zinaida [17]
3 years ago
10

Which statements are true regarding redox reactions? (Note that in redox reactions, the molecule that "causes" another to gain o

r lose electrons is referred to as the agent.) Select all that apply.
A. Oxidizing agents accept electrons.
B. Reducing agents may accept H+ ions.
C. If a molecule accepts electrons, it has been reduced.
D. Redox reactions may involve the transfer of hydrogen ions (H+).
E. A molecule that has gained H atoms is said to be reduced.
F. Oxidizing agents may accept H+ ions.
Chemistry
1 answer:
Dvinal [7]3 years ago
4 0

Answer:A, C, D, E, F

Explanation:

A. True: Oxidizing agents are electron acceptors. They accept electrons and the get reduced. This means their oxidation number reduces

B. False: Reducing agents do not accept H+ ions. Reducing agents remove oxygen from another substance or give hydrogen to it.

C. True: oxidizing agents oxidizes other molecules but they accept electrons and get reduced themselves. If a molecule accepts electrons it has been reduced.

D True: Redox reactions MAY and may not involve the transfer of hydrogen ions depending on the reactants (H+). But redox in terms of acid and base means the donating and receiving of protons(H+)

E. True: A molecule that has gained H atoms is said to be reduced. Oxidizing agents are always the proton acceptor.

F. True: Oxidizing agents May and may not accept H+. In terms of acid and base oxidizing agents accept protons(H+)

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\Delta H\textdegree{}_f(\text{IF} \; (g)} = -95 \;  \text{kJ} \cdot \text{mol}^{-1}

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\text{IF}_7 \; (g) + \text{I}_2 \; (s) \to \text{IF}_5 \; (g) + 2\; \text{IF} \; (g)

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\Delta H\textdegree{}_\text{rxn} = \Delta H\textdegree{}_f (\text{all products})  - \Delta H\textdegree{}_f (\text{all reactants})

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Note, that iodine on the reactant side is stated as a gas in the equation given in the question whereas under standard conditions it is expected to be under the solid state; the \Delta H\textdegree{} _f given in the question seemingly corresponds to the one in which the reactant iodine exists as a solid rather than as a gas. Evaluating the last expression using data from an external source

\Delta H\textdegree{}_f (\text{I}_2 \; (g) ) = \Delta H\textdegree{}_f(\text{I}_2 \; (s)) + \Delta H\textdegree{}_{\text{sublimation}}(\text{I}_2) = 62.42 \;  \text{kJ} \cdot \text{mol}^{-1} (Cox, Wagman, et al., 1984)

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