Question

calculate standard electrode potential of Ni/Ni2+,if cell potential of the cell;Ni/Ni2+(0.01M)||Cu2+/Cu is 0.59V. givenE°cu2+/cu =0.34V

Answer 1

Answer : The standard electrode potential of $Ni/Ni^{2+}$ is +0.25 V

Solution :

The balanced cell reaction will be,  

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

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

Given :

$E^o_{[Cu^{2+}/Cu]}=0.34V$

$E^o=0.59V$

$E^o_{[Ni^{2+}/Ni]}=?$

$E^o=E^o_{[cathode]}-E^o_{[anode]}$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Ni^{2+}/Ni]}$

Now put all the values in this expression, we get:

$0.59V=0.34V-E^o_{[Ni^{2+}/Ni]}$

$E^o_{[Ni^{2+}/Ni]}=0.34-0.59=-0.25V$

$E^o_{[Ni/Ni^{2+}]}=-(-0.25V)=+0.25V$

Therefore, the standard electrode potential of $Ni/Ni^{2+}$ is +0.25 V

Answer 2

Answer:

-0.2205 V

Explanation:

E0 (Ni2+/Ni) = E0 (Cu2+/Cu) - E0 = 0.34 - 0.5605