Question

LO. For the given reactions
Sn^+ + 2e → Sn2+ E° = x V
Sn- Sn2+ + 2e E° = y V
The value of E° for the reaction
Sn4+ + 4e - Sn will be
(3) 2x + 2y
(4) 2x + V​

Answer 1

Answer :  

$E^o$ for the reaction is (x+y) V.

Explanation :

The given cell reactions are:

Oxidation half reaction (anode):  $Sn\rightarrow Sn^{2+}+2e^-$

Reduction half reaction (cathode):  $Sn^{4+}+2e^-\rightarrow Sn^{2+}$

$E^o_{[Sn^{2+}/Sn]}=yV$

$E^o_{[Sn^{4+}/Sn^{2+}]}=xV$

Now we have to determine the  $E^o$ for the reaction.

$Sn^{4+}+4e^-\rightarrow Sn$

For this reaction,

$E^o_{[Sn/Sn^{2+}]}=-yV$

$E^o_{[Sn^{4+}/Sn^{2+}]}=xV$

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Sn^{4+}/Sn^{2+}]}-E^o_{[Sn/Sn^{2+}]}$

$E^o=xV-(-yV)=(x+y)V$

Therefore, the  $E^o$ for the reaction is (x+y) V.

Answer 2

Explanation:

hope this will help you