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# Based on the description of the nickel-cadmium cell on page $1053,$ and with appropriate data from Appendix $D$, estimate $E^{\circ}$ for the reduction of $\mathrm{NiO}(\mathrm{OH})$ to $\mathrm{Ni}(\mathrm{OH})_{2}$

## $$0.676 \mathrm{V}$$

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### Video Transcript

So here we're looking to calculate electrode electrode potentials. So in the nickel cadmium cell, the cathode half reaction is as follows and I Oh oh h at H plus at an electron gives us an i. O h two. Any oxidation of reaction is as follows C d in the solid stay, add to hydroxide and ion. Certainly Equus State gives a C d. O A two in the solid state and two electrons. So the you know So the final of the South is equal to to eat, not cathode. It's a great he not honored. So there's a total reaction. S l is equal to 1.5, so you can calculate the A nose as negative, not 0.8 to 4 votes, and so we can calculate cathode as 9.676 votes for the first reduction reaction.

University of Manchester