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Which of the following is not correct regarding the electrolytic preparation of $\mathrm{H}_{2} \mathrm{O}_{2} ? \quad$ [D.C.E. 2008] (a) Lead is used as cathode (b) $50 \% \mathrm{H}_{2} \mathrm{SO}_{4}$ is used $-$ (c) Hydrogen is liberated at anode (d) sulphuric acid undergoes oxidation

According to the given electrochemical sale, IAN H plus irons are formed at Plus I answered from at an odd and see you see you. E posited the EU is deposited at get told. So according to the given option here in this problem, we can say that ups and age of some age, correct answer for this problem option is correct answer for this problem. I hope you understand the solution of this problem.

Okay so we're gonna take our Overall equation and right are 2/2 reactions? We can see the m. n. 0. 2 is going to MN two plus and that i minus is going to try to I don't need to worry about balancing those. We don't multiply things by coefficient so I don't need to know any more information other than Kind of what's going what's producing what? So from the table I'll see that the voltage for this first one is 1.229V And the 2nd 1 is .534V. And then all we have to do is add them up and we'll get the total voltage for that reaction .695 bolts. So for our second example we've got a church too in basic solution is making water And our voltage for that is going to be .828 balls. I'm just taking these from the table. So S. S two minus And that's .445V. It's negative. So my overall voltage for that equation or Or electric potential is .383 bolts. And then our final example, we have silver going to silver ion and we have gold going to gold chloride. Okay. But they were both they were given to us both as Oxidation. So we have to figure out which one we want. So we're gonna look at our numbers. This is negative .799V. This is negative, 1.001V. So for this to be a spontaneous reaction, I need to turn one of these around. I'm gonna turn this one around so it'll give us an overall positive voltage. So I'm going to have this going to gold. So our voltage is 1.001V. And then our overall voltages .22V. So they only gave us the half reactions. They didn't tell us which one was the oxidation which one was the reduction? We needed to figure that out.

This question is not super difficult. You simply need to use table 17 1 to find the reduction potentials. The standard reduction potentials. It's just finding them. That's what takes time. So once you find the two half reactions that some together to get the overall reaction, recognizing one of these half reactions need to be flipped because we're going to have one reduction in one oxidation. Then we simply take the reduction half reaction, the one that doesn't need to be flipped minus the oxidation half reaction. The one that does need to be flipped in order to get the standard cell potential, which is 0.695 volts. And for the next one, what's being reduced has a reduction potential of negative 445 And then we recognize that what is being oxidized going in the opposite direction where we have hydrogen gas plus hydroxide, going to water has a reduction potential of negative 0.8 to eight. So the negative on negative makes it positive and we get 80.383 volts, and then for the last one, because it's a voltaic cell than the one that has, the larger reduction potential will be reduced and serve as the cathode. So we're going to go S. L. Is equal to cathode potential minus the an odd potential, so the silver is going to have to be flipped and we'll get a self potential of 0.22 volts.

Hey, guys. Richard here. So today we're gonna be explaining question number three. So here in question number three, what want to know is which electron is going to be our calf filed And what electron is gonna be our in? Uh, so here we see that our A g will be our careful and r p t will be our animal. So it's important to know that our P t, which is our platinum, is going to be inserted from and in our hydrogen. So it's important to know this because this insert does not react much with hydrogen. So with this process, we see that a P T is long located in the sheet are she is going to be our standard hydrogen electro. So here we see that our electrons are going to flow from our PC to our a g electro here. This diagram here shows how the flow of electrons will be. So we see that our P T is going to flow to our a g r p t being our animal, which we see that our internal is giving away electrons to our A G, which is gaining the electrons