Chemistry, 18.08.2019 04:10 kadence428

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: pt(s)|h2(g,1atm)|h+(aq,1.0m)|au3+(a q,? m)|au(s). what is the concentration of au3+ in the solution if ecell is 1.27 v ? express your answer using two significant figures.

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Answer from: SoccerHalo

Explanation:

The standard reduction potential $(E_{cell})$ is given as 1.27 V.

Concentration of $H^{+}$ = 1.0 M

Hence, reduction reaction is as follows.

$Au^{3+}(aq) + 3e^{-} \rightarrow 2Au(s)$

Oxidation reaction is as follows.

$H_{2}(g) \rightarrow H^{+}(aq) + e^{-}$

Therefore, overall net chemical equation will be as follows.

$2Au^{3+}(aq) + 3H_{2}(g) \rightarrow 2Au(s) + 6H^{+}(aq)$

As, its is known that standard electrode potential ( $E^{o}_{cell}$ ) for hydrogen is equal to zero.

And, standard electrode potential for $Au^{3+}$ is $E_{o}$ equal 1.50 V.

Hence, formula to calculate standard cell potential is as follows.

$E^{o}_{cell}$ = $(E_{o})_{reduction}$ - $(E_{o})_{oxidation}$

= 1.50 V - 0 V

= 1.50 V

Therefore, concentration of $Au^{3+}$ is calculated as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.059}{2}log \frac{[H^{+}]^{6}}{[Au^{3+}]^{2}}$

Hence, substitute the given values as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.059}{2}log \frac{[H^{+}]^{6}}{[Au^{3+}]^{2}}$

1.27 = 1.50 - \frac{0.059}{6}log \frac{(1)^{6}}{[Au^{3+}]^{2}}[/tex]

$log\frac{1}{[Au^{3+}]^{2}}$ = 23

Taking antilog on both the sides, the above equation will be as follows.

$\frac{1}{[Au^{3+}]^{2}}$ = $10^{23}$

= $1 \times 10^{23}$

$\frac{1}{[Au^{3+}]^{2}}$ = $3.16 \times 10^{11}$

$[Au^{3+}]$ = $3.16 \times 10^{-12}$ M

Hence, we can conclude that the concentration for $[Au^{3+}]$ is $3.16 \times 10^{-12}$ M.

Answer from: Quest

a non redox reaction occurs when there is not a transfer of electrons. for redox reactions, which are reduction-oxidation reactions, a change in the oxidation number occurs. elements that are reduced gain electrons, and elements that are oxidized lose electrons.

explanation:

Answer from: Quest

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

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