(a) Compute the voltage at 25°C of an electrochemical cell consisting of pure lead immersed in a 5 × 10−2 M solution of Pb2+ ions and pure tin in a 0.25 M solution of Sn2+ ions. (b) Write the spontaneous electrochemical reaction.

Respuesta :

Answer:

a) The cell potential energy is -0.01 V

b) The spontaneous chemical reaction - [tex]Sn^{2+}Pb\rightarrow Sn+Pb^{2+}[/tex]

Explanation:

The chemical reaction between lead and tin ions is follows.

[tex]SnPb^{2+}\rightarrow Sn^{2+}+Pb[/tex]

The cell potential can be calculated by using Nernst equation.

[tex][\Delta V=(V_{pb}^{0}-V_{sn}^{0})-\frac{0.592}{n}log\frac{[Sn^{2+}]}{[Pb^{2+}]}[/tex]

[tex]V_{pb}^{0}= -0.126[/tex]

[tex]V_{sn}^{0}= -0.136[/tex]

[tex]n=2[/tex]

[tex][Sn^{2+}]= 0.25M[/tex]

[tex][Pb^{2+}]=5\times 10^{-2}M[/tex]

Substitute the  values.

[tex][\Delta V=(-0.125-(-0.136))-\frac{0.592}{2}log\frac{[0.25}{5\times 10^{-2}}[/tex]

[tex]=0.011-0.021=-0.01V[/tex]

Therefore, the cell potential is 0.01V.

b)

The voltage value is negative.The spontaneous eectrochemical reaction is reverse of the electrochemical reaction in part(a)

Therefore, spontaneous reaction is as follows.

[tex]Sn^{2+}Pb\rightarrow Sn+Pb^{2+}[/tex]