\(\frac{1}{2}\)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\(^{-1}\)]

olyezema

4 years ago

n=1/2=0.5mole
n=m/M
m=n×M
m=0.5×65
m=32.5g
96500×2 -------- 65g
193000×32.5/65------- 32.5g=9.65×10-⁴C
the M is the relative atomic mass of Zn

Dave333

4 years ago

This how they got (0.5).
1/2=0.5.

Karol300

3 months ago

I picked a particular answer and after submitting it changed

FortuneIdu

3 years ago

nice

mbila2011

6 days ago

l don't understand it
please explain well

Eemat

3 years ago

Please what topic is this??

egbonjosh

8 years ago

use Q=nf
correct answer tho

Messi333333

4 years ago

-1 + (8*1) + y = +2 + (4*0)
-1 + 8 + y = 2
7 + y = 2
y = 2 - 7
y = 5e-