Question
$$E_{\text {cathode }}^{\circ}=(2.71-2.310) V=+0.40 V$$VVV
Step 1
It is given by: \[E_{\text {cathode }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}\] Show more…
Show all steps
Your feedback will help us improve your experience
Shazia Naz and 75 other Chemistry 101 educators are ready to help you.
Ask a new question
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
$E_{\text {cathode }}^{\circ}=(2.71-2.310) V=+0.40 \mathrm{V}$
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{I}^{-} \rightarrow \mathrm{I}_{2}+\mathrm{Cr}^{3+}$ $\mathrm{E}_{\text {cell }}^{\circ}=0.79 \mathrm{~V}, \mathrm{E}_{\mathrm{Cr} 2072-}^{0}=1.33 \mathrm{~V}$ $\mathrm{E}_{12}^{\circ}=?$ a. $+0.18 \mathrm{~V}$ b. $-0.18 \mathrm{~V}$ c. $0.54 \mathrm{~V}$ d. $-0.054 \mathrm{~V}$
If $v=$ number of electrons moving to the anode at distance $x$, then $$ \frac{d v}{d x}=\alpha v \text { or } v=v_{0} e^{\alpha x} $$ Assuming saturation, $I=e v_{0} e^{\alpha d}$
Electrodynamics
Electric Current
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD