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I need help with galvanic cells
ch3ntastic:
I'm bad at galvanic cells and the half equations.
I need help in the picture attached: Which way does the current flow and why? How do I write half-equations?
I'm so stuck with this topic, please help..I need to beat my sister's ATAR of 99.4 :-[ :-[ :-[
Syndicate:
--- Quote from: ch3ntastic on December 14, 2016, 08:49:43 pm ---I'm bad at galvanic cells and the half equations.
I need help in the picture attached: Which way does the current flow and why? How do I write half-equations?
I'm so stuck with this topic, please help..I need to beat my sister's ATAR of 99.4 :-[ :-[ :-[
--- End quote ---
It flows from Copper Nitrate to Silver Nitrate, as silver is a stronger oxidant than copper. Therefore copper oxidises (loses electrons) and silver reduces (gains electrons).
Reduction half-equation:
Oxidation half-equation:
Full equation:
RuiAce:
Identify the species so that we can identify which half reactions are taking place. It is clear that in one compartment, there are both copper ions (Cu2+) and copper metal (Cu). Hence, the equation Cu2+ + 2 e- ⇌ Cu(s) will prove useful.
In the other compartment, there are both silver ions (Ag+) and silver metal (Ag). Hence, the equation Ag+ + e- ⇌ Ag(s) will prove useful.
Next, identify which is getting reduced and which is getting oxidised.
On the data sheet, it is clear that the equation involving copper is placed (somewhere) above the equation involving silver. Hence, copper must be the more reactive substance and is the one to lose electrons (get oxidised). This leaves us with silver ions being what gains electrons.
Hence, the reactions we are interested in are:
Ag+ + e- → Ag(s) (not flipped, as per data sheet)
Cu(s) → Cu2+ + 2 e- (flipped, as per data sheet)
The equations also make it abundantly clear that electrons are flowing from the copper compartment, to the silver compartment. Thus we can draw the electron flow from left (through the voltmeter) to right
ch3ntastic:
--- Quote from: RuiAce on December 14, 2016, 09:21:17 pm ---Hence, the reactions we are interested in are:
Ag+ + e- → Ag(s) (not flipped, as per data sheet)
Cu(s) → Cu2+ + 2 e- (flipped, as per data sheet)
--- End quote ---
How would you calculate this? What do you do with the 0.8V and the -0.34V? Thanks
RuiAce:
--- Quote from: ch3ntastic on December 14, 2016, 09:56:31 pm ---How would you calculate this? What do you do with the 0.8V and the -0.34V? Thanks
--- End quote ---
Calculate what?
If you want to find the voltage that the cell produces, just do what you said: 0.8V - 0.34 V = 0.46V produced by the cell (since you already correctly made the copper one negative)
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