8 Write the net cell for this electroche equation Phases one optional. Do not include concentrations, 24 reaction: Curss | Cur ↑ CAg CUTE Ag² → Ag <-Cu² Teacont 2+ (0.0155 M)|| Agt (aq, 3.50m) | Ages) + reactant prodret product anode oxidation = AG 16 rn = Ecell E cell - 0.46 → Ag reduction = +0.34 V +0.80 V / Ecell (0.80) 0.34 = 0.467 = 3.50 M [²] = Erell Ecell (0.46)-(0.0592 0.0155 M >#?

Chemistry: The Molecular Science
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Chapter17: Electrochemistry And Its Applications
Section: Chapter Questions
Problem 84QRT
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I got stuck. I'm not sure how to go about determining n. Is it how many moles were in the reaction? 

Phases
Write the net cell equation for this electrochemical cell.
ore optional. Do not include concentrations,
2+
reaction: Cucs) | Cu² (09, (0.0155 M/| Agt (aq, 3.50m) | Ag(s)
↑
A
Teadant
product
anode
product
2+
→ Cu²+ oxidation = + 0.34 V
(5)
cathode
-reduction =
Ag
→ Ag,
(5)
+ 0.80 V
14641
anode
Ecell (0.80) - 0.34 = 0.467
AG
AG rvm=
Ecell =
E cell 0.46
Сад'] = 3.50 м сQu2+] = 0.0155 м
M
M
Erell
Ecell = (0.46)-(0.0592
n
reactant
#?
✓
Transcribed Image Text:Phases Write the net cell equation for this electrochemical cell. ore optional. Do not include concentrations, 2+ reaction: Cucs) | Cu² (09, (0.0155 M/| Agt (aq, 3.50m) | Ag(s) ↑ A Teadant product anode product 2+ → Cu²+ oxidation = + 0.34 V (5) cathode -reduction = Ag → Ag, (5) + 0.80 V 14641 anode Ecell (0.80) - 0.34 = 0.467 AG AG rvm= Ecell = E cell 0.46 Сад'] = 3.50 м сQu2+] = 0.0155 м M M Erell Ecell = (0.46)-(0.0592 n reactant #? ✓
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