Chemistry: A concentration cell consists of two Sn/Sn2+ half-cells. The cell has a potential of 0.10 V at 25 ºC. What is the ratio of the Sn2+ concentrations in the two half-cells?

A concentration cell consists of two Sn/Sn2+ half-cells.

mamaKelly1671

19.01.2022

The ratio of $\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}$ in the cell is $4.07\times 10^{-4}$

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode): $Sn(s)\rightarrow Sn^{2+}+2e^-$

Reduction half reaction (cathode): $Sn^{2+}+2e^-\rightarrow Sn(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.10 V

$\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}$ = ?

Putting values in above equation, we get:

$0.10=0-\frac{0.0592}{2}\log \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}$

$\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}=4.016\times 10^{-4}$

Hence, the ratio of $\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}$ in the cell is $4.016\times 10^{-4}$

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12375819

19.06.2020

54 g

Explanation:

1 mole of water = H2O

mass of 1 mole of H2O= mass of h2 + mass of o

= 2× mass of h +mass of o

= 2×1+16 =18 g

1 mole of water = 18g

3moles of water = 18×3g= 54g

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