\(\frac{1}{2}\)Zn\(^{2+}\)(aq) + e\(^-\) → \(\frac{1}{2}\) Zn(s)
In the reaction above, calculate the quantity of electricity required to discharge zinc.
[F = 96 500 C mol\(^{-1}\)]
1 mol → 96500 x 2
0.5 mol → X
X = 0.5 x 96500 x 2
= 9.650 x 10\(^4\)C
There is an explanation video available below.
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}