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 : The cell potential of the cell is, 1.079 V

Solution :

The given balanced cell reaction will be,

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 :

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

= cell potential of the cell = ?

= standard electrode potential = 1.1032 V

concentration of = 2.700 M

concentration of = 0.1448 M

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

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

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer