# Question #0c19b

May 24, 2017

$\textsf{{E}_{c e l l}^{\circ} = + 1.46 \textcolor{w h i t e}{x} V}$

#### Explanation:

Look up the standard electrode potentials:

$\textsf{S {n}^{4 +} + 2 e \text{ "rightleftharpoons" "Sn^(2+)" } {E}^{\circ} = + 0.15 \textcolor{w h i t e}{x} V}$

$\textsf{C {e}^{4 +} + e \text{ "rightleftharpoons" "Ce^(3+)" } {E}^{\circ} = + 1.61 \textcolor{w h i t e}{x} V}$

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:

$\textsf{2 C {e}^{4 +} + S {n}^{2 +} \rightarrow 2 C {e}^{3 +} + S {n}^{4 +}}$

There are different conventions around to find $\textsf{{E}_{c e l l}^{\circ}}$ 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:

$\textsf{{E}_{c e l l}^{\circ} = + 1.61 - 0.15 = + 1.46 \textcolor{w h i t e}{x} V}$