Question

The hydrogen electrode is dipped in a solution of pH = 3 at 25$^\circ$C. The potential of the cell would be (the value of 2.303 RT/F is 0.059 V)

• A. 0.177 V
• B. 0.087 V
• C. -0.177 V .
• D. 0.059V

Answer 1

Answer: (C)

-0.177 V .

Answer 2

$pH = 3$ $\therefore [H^+] = 10^{-3}$ $\therefore [H^+] + e^- {->} \frac{1}{2} H_2$ $E = E^0 - \frac{0.059}{n} \,log \frac{1}{[H^+]}$ $= 0 - \frac{0.059}{1} \,log \,10^3$ $E = - 0.177\,V$