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Question: Which of the following concentration cells will produce maximum \({{E}_{cell}}\)at 298 K? [Take \({{...

Which of the following concentration cells will produce maximum Ecell{{E}_{cell}}at 298 K? [Take PH2{{P}_{{{H}_{2}}}}= 1.0 atm in each case]
A. Pt∣H2(g)∣H+(0.01M)∣∣H+(0.1M)∣H2(g)∣PtPt|{{H}_{2}}(g)|{{H}^{+}}(0.01M)||{{H}^{+}}(0.1M)|{{H}_{2}}(g)|Pt
B. Pt∣H2(g)∣NH4Cl(0.01M)∣∣HCl(0.1M)∣H2(g)∣PtPt|{{H}_{2}}(g)|N{{H}_{4}}Cl(0.01M)||HCl(0.1M)|{{H}_{2}}(g)|Pt
C. Pt∣H2(g)∣H+(0.1M)∣∣H+(0.2M)∣H2(g)∣PtPt|{{H}_{2}}(g)|{{H}^{+}}(0.1M)||{{H}^{+}}(0.2M)|{{H}_{2}}(g)|Pt
D. Pt∣H2(g)∣H+(pH=0.0)∣∣H+(pH=0.0)∣H2(g)∣PtPt|{{H}_{2}}(g)|{{H}^{+}}(pH=0.0)||{{H}^{+}}(pH=0.0)|{{H}_{2}}(g)|Pt

Explanation

Solution

The Ecell{{E}_{cell}}value is can be easily calculated by the difference between the reduction value and oxidation value where the equation can be described as Ecell=Ereduced−Eoxidized{{E}_{cell}}={{E}_{reduced}}-{{E}_{oxidized}}. More positive the value is more easily it will gain electrons while negative value shows more easily it gives away electrons.

Complete answer:
Concentration cell is defined as a limited form of galvanic cell which has two equivalent half cells having the same composition but differ in the value of concentrations only. The potential values of cells can be calculated with the help of Nernst equation.
Out of all the options, option A is the correct answer this can be explained as follows:
Reaction occurs at cathode: H++e−→12H2{{H}^{+}}+{{e}^{-}}\to \dfrac{1}{2}{{H}_{2}}
At anode: 12H2→H++e−\dfrac{1}{2}{{H}_{2}}\to {{H}^{+}}+{{e}^{-}}
Then cell equation would be H+12H2H+\dfrac{1}{2}{{H}_{2}}, Now Cathode H+12H2 anodeH+\dfrac{1}{2}{{H}_{2}}\ anode
Now according to Nernst equation we find the cell potential i.e. Ecell{{E}_{cell}}value and the equation can be represented as:
Ecell=E0cell−0.05911log⁡[H+]anode[H+]cathode{{E}_{cell}}={{E}^{0}}_{cell}-\dfrac{0.0591}{1}\log \dfrac{[{{H}^{+}}]anode}{[{{H}^{+}}]cathode}
Now the E0cell{{E}^{0}}_{cell}will be zero because the two half-cell given are identical also the values of [H+]anode[{{H}^{+}}]anodeis 0.01 M and [H+]cathode[{{H}^{+}}]cathodeis 0.1 M which is given, now put the values in equation
Ecell=0−0.05911log⁡0.010.1{{E}_{cell}}=0-\dfrac{0.0591}{1}\log \dfrac{0.01}{0.1}
= 0.0591 V

Others will have lower Ecell{{E}_{cell}}value, so we can conclude that option A is the correct answer.

Note:
Nernst equation is the equation which relates the reduction potential of an electrochemical cell with the standard electrode potential, temperature and other activities of the cell which occur during oxidation and reduction.