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What is E∘cell for a galvanic cell constructed from aluminum and silver electrodes? Report your answer with three decimal places. Half Reaction E∘ (V) Mn2+(aq)+2e−⟶Mn(s) −1.185 Zn(OH)2(s)+2e−⟶Zn(s)+2OH−(aq) −1.245 Zn2+(aq)+2e−⟶Zn(s) −0.7618 Al3+(aq)+3e−⟶Al(s) −1.662 Mg2+(aq)+2e−⟶Mg(s) −2.372 Ba2+(aq)+2e−⟶Ba(s) −2.912 K+(aq)+e−⟶K(s) −2.931 Ag+(aq)+e−⟶Ag(s) +0.7996 Au3+(aq)+3e−⟶Au(s) +1.498 Provide your answer below:

Respuesta :

Answer:

Explanation:

When the galvanic cell is formed with the electrode Al and Ag , following half cell reaction is possible

Al  = Al⁺³  + 3e  ( At anode )

3Ag⁺ + 3e  =  3Ag .   ( At cathode )

Ecell  = Ecathode - Eanode

=  .7996 - ( - 1.662 )  [ values taken from the chart given ]

= .7996 + 1.662

= 2.4616

= 2.462 V .

Lanuel

The standard cell potential for a galvanic cell constructed from aluminum (Al) and silver (Ag) electrodes is 2.4616 Volts.

When this galvanic cell is constructed from aluminum (Al) and silver (Ag) electrodes, the following half-cell chemical reaction are obtained:

          [tex]Al_3^+_{(aq)} + 3e^- \rightarrow Al_{(s)}[/tex]     [tex]E^o(anode) =-1.662[/tex]

          [tex]Ag^+_{(aq)}+e^{-} \rightarrow Ag_{(s)}[/tex]        [tex]E^o(cathode) =0.7996[/tex]

Mathematically, the standard cell potential of a galvanic cell is given by the formula:

[tex]E^{o}_{cell}=E^o(cathode) -E^o(anode)[/tex]

Substituting the given parameters into the formula, we have;

[tex]E^{o}_{cell}=0.7996 -[-1.662]\\\\E^{o}_{cell}=0.7996 +1.662\\\\E^{o}_{cell}= 2.4616\;Volts[/tex]

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