Question

In the given cell representation $Zn|Zn^{2+}(1.0M)||H^+(a M)|H_2(1 Bar)|Pt \quad E = +0.4V$

Given $E_{Zn^{2+}|Zn}^0 = -0.76V, \frac{2.303RT}{F} = 0.06$

pH of the given solution is _______ (Rounded off to the nearest integer)

Answer 1

6

Answer 2

For the hydrogen electrode reaction

$2H^+ + 2e^- \rightarrow H_2 \quad (E^0 = 0 V)$,

the Nernst equation gives:

$E_H = 0 - (\frac{0.06}{2}) \cdot log(\frac{1}{[H^+]^2}) = 0.06 \cdot log[H^+]$.

The zinc electrode is maintained at standard conditions ($E^0 = -0.76 V$). With the cell set up as

$Zn|Zn^{2+}(1.0 M) || H^+ (a M)|H_2 (1 Bar)|Pt$,

the overall cell potential is:

$E_{cell} = E_{cathode} - E_{anode} = (0.06 \cdot log[H^+]) - (-0.76) = 0.06 \cdot log[H^+] + 0.76$.

Given $E_{cell} = +0.4 V$, we have:

$0.06 \cdot log[H^+] + 0.76 = 0.4$

$0.06 \cdot log[H^+] = 0.4 - 0.76 = -0.36$

$log[H^+] = -6$

Thus, $[H^+] = 10^{-6} M$, and hence

$pH = 6$.