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igor_vitrenko [27]
3 years ago
14

What is the total number of elements in group 17 that are gases at room temperature and standard pressure?

Chemistry
1 answer:
White raven [17]3 years ago
4 0

Answer:

Only to elements are gases at room temperature and standard pressure.

Fluorine and chlorine.

Explanation:

The group 17 is called halogen.

There are five elements in halogen group.

All halogens required one electrons to complete the octet and to get the noble gas electronic configuration.

All halogen elements have seven valance electrons.

Halogen elements:

Fluorine, chlorine, Bromine, iodine and astatine

Fluorine is gas at room temperature which s 25°C and standard pressure which is 1 atm.

It is present in the from of F₂.

Chlorine is also gas at room temperature and standard pressure.

It is present in the from of Cl₂.

The bromine is liquid under these condition.

Iodine and astatine are solids.

As we move down the group their melting and boiling points increases.

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Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v​
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Answer:

Depending on the E^\circ value of \rm Ag^{+} + e^{-} \to Ag\; (s), the cell potential would be:

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Explanation:

In this galvanic cell, the following two reactions are going on:

  • The conversion between \rm Ag\; (s) and \rm Ag^{+} ions, \rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s), and
  • The conversion between \rm Cu\; (s) and \rm Cu^{2+} ions, \rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s).

Note that the standard reduction potential of \rm Ag^{+} ions to \rm Ag\; (s) is higher than that of \rm Cu^{2+} ions to \rm Cu\; (s). Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if \rm Ag^{+} ions are reduced while \rm Cu\; (s) is oxidized.

Therefore:

  • The reduction reaction at the cathode will be: \rm Ag^{+} + e^{-} \to Ag\; (s). The standard cell potential of this reaction (according to this question) is E(\text{cathode}) = 0.89\; \rm V. According to the 2012 CRC handbook, that value will be approximately 0.79\; \rm V.
  • The oxidation at the anode will be: \rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}. According to this question, this reaction in the opposite direction (\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)) has an electrode potential of 0.34\; \rm V. When that reaction is inverted, the electrode potential will also be inverted. Therefore, E(\text{anode}) = -0.34\; \rm V.

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}.

Using data from the 1985 and 2012 CRC Handbook:

\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}.

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<em>Calculate the pH of the following substances formed during a volcanic eruption: </em>

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