An electrochemical cell has two half cell reactions as
`A^(2+)+2e^(-)rarr A`;`E^(0)=0.63V`
`X to X^(2+)+2e^(-)`;`E^(0)=2.37V` The cell voltage will be

An electrochemical cell has two half cell reaction as, A^(2+)+2e^-,E^0=0.34V XtoX^(2+)+2e^-,E^0=+2.37V . The cell voltage will be

The standard potential E^(@) for the half reactions are as : Zn rarr Zn^(2+) + 2e^(-), E^(@) = 0.76V Cu rarr Cu^(2+) +2e^(-) , E^(@) = -0.34 V The standard cell voltage for the cell reaction is ? Zn +Cu^(2) rarr Zn ^(2+) +Cu

Electrode potential for the following half-cell reactions are Zn rarr Zn^(2+)+2e^(-), E^(@)=+0.76V, Fe rarr Fe^(2+)+2e^(-), E^(@)=+0.44V The EMF for the cell reaction Fe^(2+)+Zn rarr Zn^(2+)+Fe will be

The standard electrode potential of the half cells are given below. ZnrarrZn^(2+)+2e^(-):E^(@)=0.76V FerarrFe^(2+)+2e^(-):E^(@)=0.44V The emf of the cell Fe^(2+)+ZnrarrZn^(2+)+Fe is

The reduction potential of the two half cell reactions (occuring in an electrochemical cell) are PbSO_(4)+ 2e^(-)rarrPb+SO_(4)^(2-) (E^(@)=-0.31V) Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=+0.80V) The fessible reaction will be

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.