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Question: Consider the cell \[{H_2}(Pt)\]1 atm| \[{H_3}{O^ + }(aq)\]pH=5.03| \[A{g^ + }\] x M ---|--...

Consider the cell

H2(Pt){H_2}(Pt)1 atmH3O+(aq){H_3}{O^ + }(aq)pH=5.03Ag+A{g^ + } x M

The measured EMF of the cell is 1.0V. What is the value of x?
 EAg+Ago=+0.8V [T=25∘C]{\text{ E}}_{\dfrac{{A{g^ + }}}{{Ag}}}^o = + 0.8V{\text{ }}[T = 25^\circ C]
(A) 2×10−2M2 \times {10^{ - 2}}M
(B) 2×10−3M2 \times {10^{ - 3}}M
(C) 1.5×10−3M1.5 \times {10^{ - 3}}M
(D) 1.5×10−2M1.5 \times {10^{ - 2}}M

Explanation

Solution

In order to find the value of x, we apply the Nernst equation. The Nernst equation provides a relation between the cell potential, the standard cell potential, the reaction quotient and the temperature.

Complete answer:
Let us move to the given question.
In the Cathode:
Ag++e−→Ag, EAg+Ago=0.8VA{g^ + } + {e^ - } \to Ag,{\text{ E}}_{\dfrac{{A{g^ + }}}{{Ag}}}^o = 0.8V
In the anode:
12H2→H++e−, EH2H+o=0V\dfrac{1}{2}{H_2} \to {H^ + } + {e^ - },{\text{ E}}_{\dfrac{{{H_2}}}{{{H^ + }}}}^o = 0V
The net equation is
Ag++12H2→Ag+H+A{g^ + } + \dfrac{1}{2}{H_2} \to Ag + {H^ + }
Now we have to use the Nernst equation. Let us first understand what a Nernst equation is. The Nernst equation provides a relation between the cell potential, the standard cell potential, the reaction quotient and the temperature. The Nernst equation is an equation which is mostly used to calculate the cell potential of the electrochemical cell at any given conditions like temperature, pressure and reactant concentration.
The Nernst equation at 25∘C25^\circ Cis given as
ECell=ECello−0.05911log⁡[H+][Ag+]{E_{Cell}} = E_{Cell}^o - \dfrac{{0.0591}}{1}\log \dfrac{{[{H^ + }]}}{{[A{g^ + }]}}
pH is given as 5.03.
pH=5.03=−log⁡[H+]pH = 5.03 = - \log [{H^ + }]
[H+]=10−5.03[{H^ + }] = {10^{ - 5.03}}
1.0=0.8−0.05911log⁡[10−5.03][x]1.0 = 0.8 - \dfrac{{0.0591}}{1}\log \dfrac{{[{{10}^{ - 5.03}}]}}{{[x]}}
0.224=5.03×0.0591+0.0591log⁡[x]0.224 = 5.03 \times 0.0591 + 0.0591\log [x]
By solving it we get
[x]=2×10−2M[x] = 2 \times {10^{ - 2}}M
The value of x is 2×10−2M2 \times {10^{ - 2}}M.

Hence the correct answer is option (A) 2×10−2M2 \times {10^{ - 2}}M

Note:
We have to remember that the Nernst equation can be used to calculate the following:
- Electromotive force of an electrochemical cell
- Unknown ionic concentration
- pH of the solution
- Solubility of the sparingly soluble salt can be measured.
- Standard electrode potential
- Single electrode oxidation or reduction potential at various conditions.