Question

The cell potential of the following electrochemical cell depends on the pH of the solution in the anode half-cell:Pt(s)|H2(g, 1atm)|H+(aq, ?M)||Cu2+(aq,1.0M)|Cu(s)What is the pH of the solution if Ecell = 355 mV?

Answer 1

Answer:

0.51

Explanation:

Given the Nernst equation;

E= E° - 0.0592/n logQ

E= 355 mV or 0.355 V

E° = 0.34 - 0= 0.34 V

n= 2(two electrons were transferred in the process)

Equation of the reaction;

H2(g) + Cu^2+(aq) -----> 2H^+(aq) + Cu(s)

Substituting values;

0.355 = 0.34 - 0.0592/2 log([H^+]/1)

0.355 - 0.34 = - 0.0296 log [H^+]

0.015/-0.0296 = log [H^+]

Antilog (-0.5068) = [H^+]

[H^+] = 0.311 M

pH = -log[H^+]

pH= - log(0.311 M)

pH = 0.51

Answer 2

The potential difference between the half cell of the electrochemical cell is called cell potential. The pH of the solution at 355 mV will be 0.51.

What is an electrochemical cell?

An electrochemical cell generates electricity from the redox chemical reactions occurring inside the cell.

The balanced chemical reaction is shown as,

$\rm H_{2}(g) + Cu^{2+}(aq) \rightarrow 2H^{+}(aq) + Cu(s)$

Using the Nernst equation:

$\rm E= E^{\circ} - \dfrac{0.0592}{n }logQ$

Given,

E = 0.355 V

E° = 0.34 V

n = 2

Substituting values in the above equation:

$\begin{aligned} 0.355 &= 0.34 - \dfrac{0.0592}{2} \;\rm log(\dfrac{[H^{+}]}{1})\\\\0.355 - 0.34 &= - 0.0296 \rm \; log [H^{+}]\\\\\dfrac{0.015}{-0.0296} &= \rm \; log [H^{+}]\end{aligned}$

Solving further,

$\begin{aligned} \rm Antilog (-0.5068)& = \rm [H^{+}]\\\\\rm [H^{+}] &= 0.311 \;\rm M \end{aligned}$

The pH of the solution is calculated as:

$\begin{aligned} \rm pH &= \rm -log[H^{+}]\\\\&= \rm - log(0.311\; M)\\\\&= 0.51\end{aligned}$

Therefore, 0.51 is the pH of the solution.

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