Sign up for free to keep watching this solution

Solution: Use the standard reduction potentials below to determine whi...

Question

Use the standard reduction potentials below to determine which element or ion is the best reducing agent.

Pd2+ (aq) + 2 e        →      Pd (s) E° = + 0.90 V

2 H+ (aq) + 2 e –        →        H2 (g) E° = 0.00 V

Mn2+ (aq) + 2 e        →      Mn (s) E° = – 1.18 V

 

a) Pd (s)             b) H + (aq)              c) Mn2+ (aq)              d) H2 (g)

Video Transcript

Hey guys. Let's take a look at the following question. So, here it says, use the standard reduction potentials below to determine which element or ion is the best reducing agent. So remember, we're saying the best reducing agent and realize if you're a reducing agent, that means you've been oxidized, if you've been oxidized that means you're the anode, if you're the anode that means that you have the smallest e value, and remember, oxidized means that you are away from the electrons because you're losing electrons, you're not going to be near them. So, if we look at the options, this is our smallest e value. So, looking at this reaction here, but here's an issue, this is the element that is away from the electrons, this magnesium 2+ is next to the electrons so it wouldn't be our answer. So, magnesium solid it's supposed to be the best reducing agent but I don't give you magnesium solid as a choice, so that means we're going to ignore this reaction and go to the next reaction with the smallest e-value. So, next smallest value will be this one at 0 and who's away from the electrons? the hydrogen gas is away from the electrons. So, H2 gas will be our answer here. So, the option d. So, just remember this, this Web, how reducing agent expands out to other things. Now, what if they wanted the strongest or best oxidizing agent, oxidizing agent would mean you've been reduced, which means that you're the cathode, which means you have the largest e value, which means that you're next to the electrons. So, if you understand one the other one is the exact opposite in terms of trends and just realize that this is the approach you need to take when you're trying to connect the ideas of oxidation and reduction to your different types of electrodes, meaning your anode and your cathode.