Question

The EMF of the given cell is? $Pt_{(s)}, H_{2(g)}|HCl(aq.10^{-5}M)||H_2O_{(l)}|H_{2(g)},Pt_{(s)}$ is: [Given: $\frac{2.303RT}{F}-0.06$]

• A. -0. 06 V
• B. -0. 12 V
• C.
  0. 06 V
• D.
  1. 12 V

Answer 1

Answer: (B)

-0. 12 V

Answer 2

The given electrochemical cell notation is: $Pt_{(s)}, H_{2(g)}|HCl(aq.10^{-5}M)||H_2O_{(l)}|H_{2(g)},Pt_{(s)}$

This represents a concentration cell involving two hydrogen electrodes. The left half-cell has a hydrogen ion concentration $[H^+]_1 = 10^{-5}M$. The right half-cell, being pure water, has a hydrogen ion concentration $[H^+]_2 = 10^{-7}M$.

The Nernst equation for a hydrogen electrode is: $E = E^0 - \frac{2.303RT}{nF} \log \frac{1}{[H^+]}$ Since $E^0 = 0$ for a hydrogen electrode and $n=1$ for $H^+$ ion concentration: $E = - \frac{2.303RT}{F} \log \frac{1}{[H^+]} = \frac{2.303RT}{F} \log [H^+]$

Let $E_1$ be the potential of the left half-cell and $E_2$ be the potential of the right half-cell. $E_1 = \frac{2.303RT}{F} \log (10^{-5})$ $E_2 = \frac{2.303RT}{F} \log (10^{-7})$

The cell notation indicates that the left half-cell is the anode and the right half-cell is the cathode. The EMF of the cell is $E_{cell} = E_{cathode} - E_{anode} = E_2 - E_1$. $E_{cell} = \left(\frac{2.303RT}{F} \log (10^{-7})\right) - \left(\frac{2.303RT}{F} \log (10^{-5})\right)$ $E_{cell} = \frac{2.303RT}{F} (\log 10^{-7} - \log 10^{-5})$ $E_{cell} = \frac{2.303RT}{F} (\log \frac{10^{-7}}{10^{-5}})$ $E_{cell} = \frac{2.303RT}{F} (\log 10^{-2})$ $E_{cell} = \frac{2.303RT}{F} (-2)$

Given $\frac{2.303RT}{F} = 0.06$ V. Substituting this value: $E_{cell} = (0.06 \text{ V}) \times (-2) = -0.12 \text{ V}$