If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Spontaneity and redox reactions

Learn how lead can oxidize aluminum but not copper in standard conditions. Using reduction potential tables and the diagonal rule, we confirm predictions by calculating standard cell potentials. Understand the spontaneity of reactions and the activity series in chemistry. Created by Jay.

## Want to join the conversation?

• What is the reasoning behind Copper not working with Lead? I still don't understand.
• Copper has a higher reduction potential as compared to lead. This means that copper is a stronger oxidizing agent and will in turn oxidize the lead to Pb2+. Refer to the previous video
• Could Pb^2+ be the oxidizing agent of Cu if one added energy to it?
• Sure, you can make your own electrical circuit with an additional battery and then choose your cathode/anode but then this would not be a "spontaneous" redox reaction, which is the title of the video, because "spontaneous" does not require additional energy ;-)
• If a reaction is not spontaneous, it can't happen?
• If a reaction is not spontaneous, it can't happen under the given conditions. However, it could happen under different conditions. For example, changing the temperature or the concentrations of the molecules in your reaction might make the reaction spontaneous.

As an example, consider water freezing into ice at 20 °C. This is not spontaneous. However, if you lowered the temperature to -10 °C, then suddenly water would spontaneously freeze into ice.
• Does it means that is the overall voltage is greater than 0, then the reaction can happen?
• If the overall electric potential (voltage) is greater than 0, then the reaction is spontaneous. A spontaneous reaction is one that can occur by itself without any outside addition of energy. A non-spontaneous reaction can still occur, but it will need an addition of energy (the activation energy) before it will start occurring.
• at , Jay rewrites the two half reactions so that the electrons can cancel, but because making the electron number equal by multiplying by a number will not affect the voltage value (as intensive property) is it even necessary to find the equation of the overall reaction? Because the question does not expricitly ask for the overall equation.
(just trying to save time on the AP exam :) )
Can I just look at the voltage values and add them right away?
(1 vote)
• Yes you can, but I guess Jay is showing the full reasoning here.
• I thought that because our standard cell potential for Cu was higher, that means it MUST be reduced (+0.34 > -0.13). Thus I got the answer as being +0.47. Why, if Pb^2+, cant oxidize copper, would I then show lead being reduced?
• The potentials are always written as reduction reactions (they’re called standard reduction potentials)
(1 vote)
• When calculating cell potentials, in my tutorial in university they said to subtract the more positive potential from the smaller or more negative potential and not add them?
Eg question was:
Rechargeable nickel-cadmium batteries normally operate (discharge) with
the following oxidation and reduction half-cell reactions.

Cd(s) + 2OH-(aq) ⇌ Cd(OH)2(s) + 2e- (-0.82v)

NiO(OH)(s) + H2O(l) + e- ⇌ Ni(OH)2(s) + OH-(aq) (0.60v)
And the overall cell potential answer was: 0.6-(-0.82)=1.42

Is this the same thing or is uni wrong XD?
• it is incorrect how they wrote the equation.
it should be E0 = Ec - Ea
(1 vote)
• When you are figuring out if lead can oxidize copper at , why are we reversing the equation for copper?
(1 vote)
• Because you need the oxidation equation, so you reverse the reduction to get the oxidation.