Question

A hydrogen electrode is immersed in a solution with pH = 0 (HCl). By how much will the potential (reduction) change if an equivalent amount of NaOH is added to the solution? (Take $p_{H_2}=1$ atm, T = 298 K)

• A. Increase by 0.41 V
• B. Increase by 59 mV
• C. Decrease by 0.41 V
• D. Decrease by 59 mV

Answer 1

Answer: (C)

Decrease by 0.41 V

On adding an equivalent amount of $NaOH$, the $pH$ of the solution will become 7.

Since, pH of solution $=-\log \left[H^{+}\right]$

So, $\left[H^{+}\right]$ changes from ${10}^0$ to ${10}^{-7}M$

Applying Nernst eqution,

$E_{cell}=E_0+\frac{0.059}{n}\log \left[H^{+}\right]$

$E_{cell}=E_0+\frac{0.059}{1}\log \left({10}^{-7}\right)$

$E_{cell}=E_0-0.41$

So, the potential decreases by 0.41V.