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Calculate E("cell")^(@) of the cell Mg|M...

Calculate `E_("cell")^(@)` of the cell `Mg|Mg^(2+)||Cu^(+)|Cu`
Given : `E_(Mg^(2+)|Mg)^(@)=-2.37V and E_(Cu^(+)|Cu)^(@)=+0.15V`.

Text Solution

Verified by Experts

`E_("cell")^(@)={:("At anode: "(1)/(2)H_(2)(g)" "rarr" "H^(+)(aq)+e),("At cathode: "AgCl(s)+e" "rarr" "Ag(s)+Cl^(-)(aq)),(_),("Cell reaction : "(1)/(2)H_(2)(g)+AgCl(s)rarrH^(+)(aq)+Ag(s)+Cl^(-)(aq)):}-E_(Mg^(2+)||Mg)^(@)=(0.15+2.3)V=+2.52V`
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See whether the cell representation given below is in accordance with the conventation or not. If it is not, then write the correct representation and reaction of the cell. Calculate the standard EMF of the cell: Cu|Cu^(2+)||Zn|Zn^(2+) . ["Given : "E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Zn^(2+)|Zn)^(@)=-0.76V]

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Write cell reaction and calculate electrical work obtained from the following galvanic cell at standard condition . Mg|Mg^(2+)(aq)||Ag^(+)(aq)|Ag [Given : E_(Mg^(2+)|Mg)^(0) = -2.36 volt and E_(Ag^(+)|Ag)^(0) = 0.80 volt.]

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Knowledge Check

  • For the given cell, Mg|Mg^(2+)||Cu^(2+)|Cu -

    A
    Mg is cathode
    B
    Cu is cathode
    C
    the cell reaction is `Mg+Cu^(2+) rarr Mg^(2+)+Cu`
    D
    `Cu` is the oxidising agent
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