Question
The hydrogen electrode is dipped in a solution of $\mathrm{pH}=3$ at $25^{\circ} \mathrm{C}$. The potential of the cell would be (value of $2.303 \mathrm{RT} / \mathrm{F}$ is $0.059 \mathrm{~V}$ )a. $+1.77 \mathrm{~V}$b. $+0.177 \mathrm{~V}$c. $-0.87 \mathrm{~V}$d. $-0.177 \mathrm{~V}$
Step 1
So, if the pH is 3, we can write: \[ -\log[H^+] = 3 \] This implies that the concentration of H+ ions in the solution is \(10^{-3}\) M. Show more…
Show all steps
Your feedback will help us improve your experience
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The hydrogen electrode is dipped in a solution of $\mathrm{pH}$ 3 at $25^{\circ} \mathrm{C}$. The potential would be (the value of $2.303$ $\mathrm{RT} / F$ is $0.059 \mathrm{~V})$ (a) $0.177 \mathrm{~V}$ (b) $0.087 \mathrm{~V}$ (c) $0.059 \mathrm{~V}$ (d) $-0.177 \mathrm{~V}$
The hydrogen electrode is dipped in a solution of $\mathrm{pH}=$ $3.0$ at $25^{\circ} \mathrm{C}$. The potential of hydrogen electrode would be (a) $-0.177 \mathrm{~V}$ (b) $0.177 \mathrm{~V}$ (c) $1.77 \mathrm{~V}$ (d) $0.277 \mathrm{~V}$
The standard electrode potential for the following reaction is $-0.57 \mathrm{~V}$. What is potential at $\mathrm{pH}=12.0 ?$ $$ \mathrm{TeO}_{3}^{2-}(a q, 1 \mathrm{M})+3 \mathrm{H}_{2} \mathrm{O}(l)+4 e^{-} \longrightarrow \mathrm{Te}(s)+6 \mathrm{OH}^{-}(a q) $$ (a) $-0.17 \mathrm{~V}$ (b) $+0.21 \mathrm{~V}$ (c) $-0.39 \mathrm{~V}$ (d) $+1.95 \mathrm{~V}$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD