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Question: A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of \({...

A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10−6{10^{ - 6}} M hydrogen ions. The emf of the cell is 0.1180.118 volt at 25oC{25^o}C . Calculate the concentration of hydrogen ions at the positive electrode.

Explanation

Solution

Hint : A concentration cell is a type of electrolytic cell that consists of two half-cells with the same type of electrodes, but are different in concentrations. Here the given cell is a concentration cell and EoE^o cell for a concentration cell is equal to zero. In Electrochemistry, we need to remember that the negative electrode is referred to as the anodic compartment of the cell whereas the positive electrode is referred to as the cathodic compartment of the cell.
Formula used: The Nernst equation at 25oC{25^o}C is:
When,Eocell=0When,{E^o}_{cell} = 0
Ecell=0.059nlog⁡[conc.ofcathode]stoichiometric−coefficients[conc.ofanode]stoichiometric−coefficients{E_{cell}} = \dfrac{{0.059}}{n}\log \dfrac{{{{[conc.ofcathode]}^{stoichiometric - coefficients}}}}{{{{[conc.of anode]}^{stoichiometric - coefficients}}}}
Where “ nn ” is referred to as the number of electrons in the half-cell reaction.

Complete step by step solution :
Step 1 :
First, we will write the corresponding half cells equation followed by the full cell equation,
Let the concentration of hydrogen ions at the positive electrode that is cathode be “X” M (M=Molarity) and for writing the full cell equation we need to remember these following notations:
a) The cell anodic and cathodic compartments (half-cells) are separated by two bars or slashes, which represent a salt bridge.
b) The anode is placed on the left and the cathode is placed on the right.
c) Individual phases (solid, liquid or aqueous) within each half-cell are written separated by a single bar. Concentrations of dissolved species can be written in brackets after the phase notation (s, l, g, or aq).
At anode: H2→2H++2e−{H_2} \to 2{H^ + } + 2{e^ - }
At cathode: 2H++2e−→H22{H^ + } + 2{e^ - } \to {H_2}
Full cell equation comes out to be, Pt/H1(1atm)/H+(10−6M)//H+(XM)/H2(1atm)/PtPt/{H_1}(1atm)/{H^ + }({10^{ - 6}}M)//{H^ + }(XM)/{H_2}(1atm)/Pt
Step 2:
Now, Putting the given values of the concentration of hydrogen ions for the negative electrode in the Nernst equation we have,
Ecell=0.05912log⁡[H+]2cathode[10−6]2{E_{cell}} = \dfrac{{0.0591}}{2}\log \dfrac{{{{[{H^ + }]}^2}_{cathode}}}{{{{[{{10}^{ - 6}}]}^2}}}
Since, the value of Ecell{E_{cell}} is given to us as 0.1180.118 volt the equation becomes,
0.118=0.05912log⁡[H+]2cathode[10−6]20.118 = \dfrac{{0.0591}}{2}\log \dfrac{{{{[{H^ + }]}^2}_{cathode}}}{{{{[{{10}^{ - 6}}]}^2}}}
Step 3: Solving for concentration of hydrogen [H]+(cathode){[H]^ + }(cathode) we have,
2log⁡[H]+cathode10−6=0.118×20.05912\log \dfrac{{{{[H]}^ + }_{cathode}}}{{{{10}^{ - 6}}}} = \dfrac{{0.118 \times 2}}{{0.0591}} (As, log(m)n= nlog(m))
log⁡[H]+cathode10−6=2\log \dfrac{{{{[H]}^ + }_{cathode}}}{{{{10}^{ - 6}}}} = 2
[H]+cathode10−6=10−2\dfrac{{{{[H]}^ + }_{cathode}}}{{{{10}^{ - 6}}}} = {10^{ - 2}}
[H]+cathode=10−2×10−6=10−4M{[H]^ + }_{cathode} = {10^{ - 2}} \times {10^{ - 6}} = {10^{ - 4}}M
Hence, the concentration of hydrogen ions at positive electrodes is found to be 10−4{10^{ - 4}} M.

Note : A student can be confused about the value of “ nn ” in the Nernst equation as it is the number of electrons present in the half cell reactions, If the number of electrons is different in half cell equation then balancing the equation is necessary.