Question

Consider the reaction corresponding to a voltaic cell and its standard cell potential. Z n ( s ) + C u 2 + ( a q ) ⟶ C u ( s ) + Z n 2 + ( a q ) Zn(s)+CuX2+(aq)⟶Cu(s)+ZnX2+(aq) E o cell = 1.1032 V Ecello=1.1032 V What is the cell potential for a cell with a 2.700 M solution of Z n 2 + ( a q ) ZnX2+(aq) and 0.1448 M solution of C u 2 + ( a q ) CuX2+(aq) at 436.2 K?C

Answer 1

Answer : The cell potential of the cell is, 1.079 V

Solution :

The given balanced cell reaction will be,

$Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)$

Here zinc (Zn) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{RT}{nF}\log \frac{[Zn^{2+}]^2}{[Cu^{2+}]}$

where,

R = gas constant = 8.314 J/K.mole

F = Faraday constant = 96500 C

T = temperature = 436.2 K

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = cell potential of the cell = ?

$E^o_{cell}$ = standard electrode potential = 1.1032 V

concentration of $Zn^{2+}$ = 2.700 M

concentration of $Cu^{2+}$ = 0.1448 M

Now put all the given values in the above equation, we get

$E_{cell}=1.1032-\frac{(8.314)\times (436.2)}{2\times 96500}\log \frac{2.700}{0.1448}=1.079V$

Therefore, the cell potential of the cell is, 1.079 V