Question

Calculate the potential of the following cell at 298 K. Sn(S)|Sn2+(0.05M)||H+(0.02M)|H2(g)(1bar)|Pt(S) ; E0Sn²⁺/Sn =-0.14V​

Answer 1

Given :

Reaction :

$\sf\:Sn(S)|Sn^{+2}(0.05M)\:||\:H^{+1}(0.02)|H_2(g)(1\:bar)|Pt(s)$

To Find

Calculate the potential of the given cell reaction at 298 k. and $\sf\:E\degree\:Sn^{+2}/Sn=-0.14$

Theory :

•Nernst Equation :

$\sf\:E_{cell}=E\degree_{cell}+\dfrac{-2.303RT}{nF}\log\dfrac{[Prduct]}{[Reactant]}$

or $\sf\:E_{cell}=E\degree_{cell}-\dfrac{0.059}{n}\log\:Q$

• E° cell

$\sf\:E\degree_{cell}=E\degree_{cathode}-E\degree_{anode}$

Solution :

We have , Reaction

$\sf\:Sn(S)|Sn^{+2}(0.05M)\:||\:H^{+1}(0.02M)|H_2(g)(1\:bar)|Pt(s)$

We know that

$\bf\:E\degree_{cell}=E\degree_{cathode}-E\degree_{anode}$

$\sf\:E\degree_{cell}=0-(-0.14)$

$\sf\:E\degree_{cell}=0.14V$

At Anode :

$\sf\:Sn\longrightarrow\:Sn^{+2}+2e^{-1}$

At Cathode :

$\sf\:2H^{+}+2e^{-1}\longrightarrow\:H_2$

Thus, n= 2

Over-all reaction :

$\sf\:Sn+2H^{+1}\longrightarrow\:H_2+Sn^{+2}$

Then ,

$\sf\:Q=\dfrac{[Prduct]}{[Reactant]}$

$\sf\:Q=\dfrac{[Sn^{+2}]}{[H^{+1}]^2}$

Given $\sf\:[Sn^{+2}]=0.05M\:and\:[H^{+1}]=0.02M$

$\sf\implies\:Q=\dfrac{0.05}{(0.02)^2}$

$\sf\implies\:Q=\dfrac{5\times100\times100}{2\times100}$

$\sf\implies\:Q=250$

By Nernst Equation :

$\sf\:E_{cell}=E\degree_{cell}-\dfrac{0.09}{n}\log\:Q$

$\sf\implies\:E_{cell}=0.14-\dfrac{0.059}{2}\log\:250$

$\sf\implies\:E_{cell}=0.14-\dfrac{0.056}{2}[\log(25\times10)]$

$\sf\implies\:E_{cell}=0.14-0.03[\log25+\log10]$

$\sf\implies\:E_{cell}=0.14-0.03[2\log5+\log10]$

$\sf\implies\:E_{cell}=0.14-0.03[2\times0.69-1]$

$\sf\implies\:E_{cell}=0.14-0.03[1.38-1]$

$\sf\implies\:E_{cell}=0.14-0.03[0.38]$

$\sf\implies\:E_{cell}=0.14-0.0114$

$\sf\implies\:E_{cell}=0.14-0.0114$

$\sf\implies\:E_{cell}=0.1286V$

Therefore,the potential of the given cell is 0.1286V.