Question

In which of the following conditions is the potential for the following half-cell reaction maximum?
$2H^{+}+2e^{-}⟶H_2$

• A. $1.0MHCl$
• B. A solution having $pH=4$
• C. Pure water
• D. $1.0MNaOH$

Answer 1

The correct option is A $1.0MHCl$

In the given options,

(A). Concentration of $H^{+}$ in 1.0 M HCl = $1M$

(B). Since, $pH$ of solution is 4.

So, concentration of $H^{+}$ = ${10}^{-4}M$

(C). $pH$ of pure water = 7

So, concetration of $H^{+}$ = ${10}^{-7}M$

(D). Concentration of $H^{+}$ in 1.0 M $H^{+}$ is much less than the all the above options.

Since, concentration of $H^{+}$ is highest in option A. So, potential of option A is maximum .

So, correct answer is option A.