# Question #0529c

Dec 3, 2015

$\left(a\right)$

$2 A {l}_{\left(s\right)} + 3 N {i}_{\left(a q\right)}^{2 +} \rightarrow 2 A {l}_{\left(a q\right)}^{3 +} + 3 N {i}_{\left(s\right)}$

$\left(b\right)$

${E}_{c e l l} = + 1.38 \text{V}$

#### Explanation:

$\left(a\right)$

Arrange the 1/2 cells in order most -ve to most +ve:

$\stackrel{\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times x}}{\textcolor{red}{\leftarrow}}$

$A {l}_{\left(a q\right)}^{3 +} + 3 e r i g h t \le f t h a r p \infty n s A {l}_{\left(s\right)} \text{ " -1.66"V}$

$N {i}_{\left(a q\right)}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s N {i}_{\left(s\right)} \text{ "-0.28"V}$
$\stackrel{\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times x}}{\textcolor{b l u e}{\rightarrow}}$

Note I have used $r i g h t \le f t h a r p \infty n s$ since these 1/2 cells can be driven in either direction depending on what they are coupled with.

A useful rule of thumb in predicting the outcome of a redox reaction is "the anti - clockwise rule".

This states that any species on the bottom left of the table can oxidise any species above and to the right.

So you can see that Ni(II) should oxidise Al(0). The bottom 1/2 cell goes left to right. The top 1/2 cell goes right to left. The arrows go in an anticlockwise direction.

In a list like this the most powerful oxidising agents are at the bottom left. They tend to have +ve electrode potentials signifying their tendency to attract electrons but remember, it is all relative.

Similarly the most powerful reducers are found at the top right.

The 1/2 equations are $\therefore$

$3 N {i}_{\left(a q\right)}^{2 +} + 6 e \rightarrow 3 N {i}_{\left(s\right)}$

$2 A {l}_{\left(s\right)} \rightarrow 2 A {l}_{\left(a q\right)}^{3 +} + 6 e$

Adding these gives the spontaneous cell reaction$\Rightarrow$

$2 A {l}_{\left(s\right)} + 3 N {i}_{\left(a q\right)}^{2 +} \rightarrow 2 A {l}_{\left(a q\right)}^{3 +} + 3 N {i}_{\left(s\right)}$

$\left(b\right)$

${E}_{c e l l}$ is an experimentally measured quantity so should always be +ve. To calculate it always subtract the least positive from the most positive ${E}^{\circ}$ value:

$\therefore {E}_{c e l l} = - 0.28 - \left(- 1.66\right) = + 1.38 \text{V}$

In this question I have used the convention used in the UK. I understand other conventions are used which would reverse the ${E}^{\circ}$ for the aluminium 1/2 cell and then add.