Question #0c19b

1 Answer
May 24, 2017

Answer:

#sf(E_(cell)^@=+1.46color(white)(x)V)#

Explanation:

Look up the standard electrode potentials:

#sf(Sn^(4+)+2e" "rightleftharpoons" "Sn^(2+)" "E^@=+0.15color(white)(x)V)#

#sf(Ce^(4+)+e" "rightleftharpoons" "Ce^(3+)" "E^@=+1.61color(white)(x)V)#

The cerium 1/2 cell is the most +ve so this will take in electrons and shift right. The tin 1/2 cell will give out electrons and shift left.

The overall cell reaction will be:

#sf(2Ce^(4+)+Sn^(2+)rarr2Ce^(3+)+Sn^(4+))#

There are different conventions around to find #sf(E_(cell)^@)# but they all rely on finding the arithmetic difference between the two 1/2 cell potentials.

The simplest way is to subtract the least +ve value from the most +ve:

#sf(E_(cell)^@=+1.61-0.15=+1.46color(white)(x)V)#