Question

Question #0c19b

Answer 1

`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)`