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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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Determine the maximum amount of work that can be derived from the cell, Zn|Zn^(2+)(1M)||Cu^(2+)(1M)|Cu. Given : E_(Zn^(2+)|Zn)^(@)=-0.76V and E_(Cu^(2+)|Cu)^(@)=+0.34V

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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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