The voltage of a cell whose half-cells are given below is
`Mg^(2+)+2e^(-)rarrMg(s),E^(@)=-2.37V`
`Cu^(2+)+2e^(-)rarrCu(s),E^(@)=+0.34V`
standard EMF of the cell is

Cu-2e^(-)rarrCu^(2+),E^(@)=-0.347V Sn-2e^(-)rarrSn^(2+),E^(@)=+0.143V The standard EMF of the cell constructed with these electrodes is

The standard electrode potential of the half cells are given below : Zn^(2+)+ 2e^(-) rarr Zn(s), E^(o) = - 7.62 V Fe^(2+) + 2e^(-) rarr Fe(s) , E^(o) = -7 .81 V The emf of the cell Fe^(2+) + Zn rarr Zn^(2+) + Fe will be :

If Mg^(2+) +2e rarrMg(s), E=-2.37 V, Cu^(2+) + 2e rarrCu(s),E=+0.34V then the emf of the cell Mg|Mg^(+2)||Cu^(2+)|Cu is :

The electrode potentials for the two half cell reaction is given below. Determine emf Mg^(2+)+2e^- rarr Mg(s), E=-2.37V Cu^(2+)+2e^- rarr Cu(s), E=+0.34V.

If for the half cell reaction E^(@) value are given Cu^(2+)+e^(-)rarrCu^(+),E^(@)=0.15V Cu^(2+)+2e^(-)rarrCu^(+),E^(@)=0.34V calculate E^(@) of the half cell reaction Cu^(+)+e^(-)rarrCu

Consider a voltaic cell based on these half - cell reactions Ag^(+)(aq)+e^(-)rarrAg(s): E^(@)=+0.80V Cd^(+2)(aq)+2e^(-)rarrCd(s), E^(@)=-0.40V identify the anode and give the voltage of this cell under standard condition.