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2 years ago

Which atom gives up its electrons most easily? THE ANSWER IS “Rb”

Chemistry

2 answers:

Sedbober [7]2 years ago

8 0

Answer:

cesium

In particular, cesium (Cs) can give up its valence electron more easily than can lithium (Li). In fact, for the alkali metals (the elements in Group 1), the ease of giving up an electron varies as follows: Cs > Rb > K > Na > Li with Cs the most likely, and Li the least likely, to lose an electron

Explanation:

bagirrra123 [75]2 years ago

6 0

Answer:

<em> In particular, cesium (Cs) can give up its valence electron more easily than can lithium (Li). The easy of giving up an electron varies as follows: Cs > Rb > K > Na > Li with Cs the most likely, and Li the least likely, to lose an electron. In fact, for the alkali metals (the elements in Group 1). I hope I helped you.</em>

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2 years ago

The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate

Cloud [144]

Answer : The $\Delta G$ for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,

$Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $2Ag^++2e^-\rightarrow 2Ag$

Now we have to calculate the Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole$

Therefore, the $\Delta G$ for this reaction is, -88780 J/mole.

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3 years ago

How to determine whether an element is paramagnetic or diamagnetic?

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3 years ago