# Question #f9b80

Apr 23, 2015

${E}_{c e l l}^{0} = + 3.34 \text{V}$

You need to look up the relevant standard electrode potentials:

Always list them -ve to +ve:

$M {g}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s M g$ ${E}^{0} = - 2.38 \text{V}$

$N {O}_{3}^{-} + 4 {H}^{+} + 3 e r i g h t \le f t h a r p \infty n s N O + 2 {H}_{2} O$ ${E}^{0} = + 0.96 \text{V}$

The most +ve half cell will take in electrons so the $N {O}_{3}^{-} \text{/} N O$ equilibrium will shift left to right.

The $M {g}^{2 +} \text{/} M g$ half cell will give out electrons and shift right to left. So this electrode is the -ve side of the cell.

These electrons flow through the external circuit to the $N {O}_{3}^{-} \text{/} N O$ half cell.

The overall cell reaction is;

$3 M g + 2 N {O}_{3}^{-} + 8 {H}^{+} \rightarrow 3 M {g}^{2 +} + 2 N O + 4 {H}_{2} O$

${E}_{c e l l}^{0}$ is an empirically measured quantity so is always +ve.

To work out ${E}_{c e l l}^{0}$ always subtract the least positive value from the most positive$\Rightarrow$

${E}_{c e l l}^{0} = 0.96 - \left(- 2.38\right) = + 3.34 \text{V}$