Question

A voltaic cell is set up at 25c with the following half cell al/al3+(0.001m) and ni/ni2+(0.50)m calculate the cell voltage

Answer 1

The cell voltage is 1.46011 V

Solution:

Voltaic cell is a electricity producing cell which uses chemical reaction. These cell uses anode and cathode.  

In this problem, aluminum electrode is anode and nickle electrode is the cathode.

In Aluminum electrode, Oxidation reaction takes place.

$2 A l \rightarrow 2 A l^{3+}+6 e \ldots \ldots E^{\circ}=1.66 \mathrm{V}$

In the Nickel electrode, Reduction reaction takes place.

$3 N i^{2+}+6 e \longrightarrow 3 N i \ldots \ldots E^{\circ}=-0.25 V$

Balancing the overall equation,

$2 A l+3 N i^{3+} \rightarrow 2 A l^{3+}+3 N i$

Thus the formula is,

$\mathrm{E}_{\mathrm{cell}}^{\circ}=\mathrm{E}^{\circ} \text { cathode }-\mathrm{E}^{\mathrm{o}} \text { anode }$

$\mathrm{E}^{\circ}_{\mathrm{cell}}=-0.25+1.66$

$\Rightarrow \mathrm{E}^{\circ}_{\mathrm{cell}}=1.41 \ \mathrm{V}$

By using the nerst equation,

$\mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{\left[\mathrm{Al}^{3+}\right]^{2}}{\left[\mathrm{Ni}^{2+}\right]^{3}}$

$=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.059}{6} \log \frac{[0.001]^{2}}{[0.5]^{3}}$

$=\mathrm{E}_{\text { cell }}^{\circ}-0.00983 \log \frac{10^{-6}}{0.125}$

$=\mathrm{E}_{\mathrm{cell}}^{\circ}-0.00983 \log 8 \times 10^{-6}$

$=\mathrm{E}_{\mathrm{cell}}^{\circ}-0.008877+0.0598$

$=1.41-0.008877+0.0598$

$\Rightarrow \mathrm{E}_{\mathrm{cell}}=1.46011 \ \mathrm{V}$

Answer 2

Explanation:

Voltaic cell is a electricity producing cell which uses chemical reaction. These cell uses anode and cathode.

In this problem, aluminum electrode is anode and nickle electrode is the cathode.

In Aluminum electrode, Oxidation reaction takes place.

2 A l \rightarrow 2 A l^{3+}+6 e \ldots \ldots E^{\circ}=1.66 \mathrm{V}2Al→2Al

3+

+6e……E

∘

=1.66V

In the Nickel electrode, Reduction reaction takes place.

3 N i^{2+}+6 e \longrightarrow 3 N i \ldots \ldots E^{\circ}=-0.25 V3Ni

2+

+6e⟶3Ni……E

∘

=−0.25V

Balancing the overall equation,

2 A l+3 N i^{3+} \rightarrow 2 A l^{3+}+3 N i2Al+3Ni

3+

→2Al

3+

+3Ni

Thus the formula is,

\mathrm{E}_{\mathrm{cell}}^{\circ}=\mathrm{E}^{\circ} \text { cathode }-\mathrm{E}^{\mathrm{o}} \text { anode }E

cell

∘

=E

∘

cathode −E

o

anode

\mathrm{E}^{\circ}_{\mathrm{cell}}=-0.25+1.66E

cell

∘

=−0.25+1.66

\Rightarrow \mathrm{E}^{\circ}_{\mathrm{cell}}=1.41 \ \mathrm{V}⇒E

cell

∘

=1.41 V

By using the nerst equation,

\mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{\left[\mathrm{Al}^{3+}\right]^{2}}{\left[\mathrm{Ni}^{2+}\right]^{3}}E

cell

=E

cell

∘

−

n

0.059

log

[Ni

2+

]

3

[Al

3+

]

2

=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.059}{6} \log \frac{[0.001]^{2}}{[0.5]^{3}}=E

cell

∘

−

6

0.059

log

[0.5]

3

[0.001]

2

=\mathrm{E}_{\text { cell }}^{\circ}-0.00983 \log \frac{10^{-6}}{0.125}=E

cell

∘

−0.00983log

0.125

10

−6

=\mathrm{E}_{\mathrm{cell}}^{\circ}-0.00983 \log 8 \times 10^{-6}=E

cell

∘

−0.00983log8×10

−6

=\mathrm{E}_{\mathrm{cell}}^{\circ}-0.008877+0.0598=E

cell

∘

−0.008877+0.0598

=1.41-0.008877+0.0598=1.41−0.008877+0.0598

\Rightarrow \mathrm{E}_{\mathrm{cell}}=1.46011 \ \mathrm{V}⇒E

cell

=1.46011 V