The cell potential of this electrochemical cell depends on the
pH...

Question

The cell potential of this electrochemical cell depends on the
pH of the solution in the anode half-cell.
Pt(s) | H2(g, 1 atm) | H+(aq, ? M) || Cu2+(aq, 1.0 M) | Cu(s) What is the pH of the solution if Ecell is 355 mV?

Accepted Answer

Hey everyone in this example, we're given the below electrochemical cell. We're told that the standard cell potential is 923 million volts. And to determine the ph of our solution. So what we should first begin with is our formation of H two gas. So our H two gas is formed from two moles of hydride, which gained two electrons. We next want to focus on our formation of silver in the solid form and we would see that to form silver, we have one mole of the silver plus one cat ion, gaining an electron to form solid silver. Now, for our first reaction, we would recognize that we have a cell potential that is for the oxidation of H2 gas because we have electrons added onto the right hand side. So we should recall that this represents an oxidation and when we look up in our standard cell potential table, we're going to see a value of zero for the oxidation of hydrogen gas. Now this is in units of volts. And we're going to move on to our second reaction here, where because we added electrons to the left hand side of this reaction, this is going to be our reduction and we're going to look up on the standard cell potential table for the reduction of the silver plus one Catalan and we would see that we have a self potential value equal to 0.80 volts. Now our next step is to add up these two equations, however, we won't be able to do so until we make sure that the electrons are balanced right now, we have two electrons in our first reaction and just one in the second reaction. So we're going to take this second reaction and we're going to multiply it by two so that everything can be bounced out. And so what we would have instead is to molds of a G plus plus two electrons yields two molds of a G in solid form. And so now we can go ahead and add up these two reactions here. So we'll just take this first reaction And we'll just rewrite it in black. So we're going to add it to one mole of H2 gas and sorry about that, that's H2 gas which produces two moles of hydride plus two electrons. So we're adding these two equations up now that the electrons are balanced and because the electrons are on opposite sides of both equations, we can go ahead and cancel them out. So two electrons here on the react inside can be canceled out with two electrons here on the product side. And for our full reaction, we can go ahead and take our two moles of the silver plus one, carry on, added to our one mole of hydrogen gas which produces on our product side, we have two moles of solid silver plus two moles of hydride. And sorry, this is H plus. Now, just to be clear again, as we stated, because we added two electrons to the product side of our first half reaction. This is going to be our oxidation recall that our oxidation occurs at the anodes of our voltaic cell. And because we added electrons to the reactive side of this second half reaction again, this is our reduction which will occur at the cathode of our electro voltaic cell. Now, the next step is to recognize that in the prompt were given the cell potential of 923 million volts. As you can see, we've been outlining our cell potentials and units of volts from our tables. So we want to convert from mila bolts to bolts. So we will take the 923 million volts and we're going to convert from mili volts. Two volts by recalling that are prefix milli tells us that for one volt we have 10 to the third power milli volts. Now we're able to cancel out milli volts leaving us with volts and this is going to give us a value equal to 0.9-3 million volts. Or sorry, that would be volts. Now, since that is our final unit that we're left with. Now that we have our value for the cell potential in volts. We want to calculate our standard cell potential and we should recall that to calculate standard cell potential which is e degrees cell that is equal to the cell potential of our cathode subtracted from the cell potential of our an ode. And so from above we stated that our cell potential of our cathode occurs at our reduction here and has a value of 0.80 volts. So we take 0.80 volts. And we're subtracting that from the cell potential of our an ode which we stated occurs at our oxidation and has a value of zero volts. So we have minus 0.0 volts. And this gives us a standard cell potential equal to 0. volts for our electrochemical cell. So right now we have both the cell potential and the standard cell potential here, both in the proper units being volts. And because we need to calculate the ph of our solution, we should recall that to calculate ph we want to find the negative log of our concentration of hydride. And so because we don't know the concentration of hydride from the prompt, we're going to need to use our nursed equation. We also are going to need to recall our value for Q. Which we recall is our equilibrium constant from the first equation. And this is equal to the concentration of products divided by our concentration of reactant. And this will only include gaseous and Aquarius substances only. So what we would have for our value for Q is our concentration of products. And looking at our reaction here only at the Aquarius or gaseous products. We have two moles of hydride as our acquis product. So we're going to plug that in for our equilibrium concentration for the for the concentration of hydride here. Now, as you can see we have a coefficient of two. So we're gonna raise this to a power of two in our denominator. We're going to plug in our concentration of reactant. We only want to focus on gaseous and liquid reactant and we do have those in our equation here. So according to our equation, we have for a concentration of silver plus one Catalan we're going to raise that to a power of two because it has a coefficient of two. And then this is multiplied by our second reactant where we have the concentration of hydrogen gas which has a coefficient of one. So it just is raised to the power of one. Now, in this case we actually wouldn't plug in concentration. So we would just plug in for the pressure of H two because hydrogen is given as a gas here. So we would measure pressure instead of concentration. So plugging in what we know from the prompt, our numerator is what we are solving for. So we're just going to say that it's X squared our denominator for our concentration of silver plus one Catalan is given in the prompt as 1.0 and sorry, that's a one. So it's given in the prompt as 1.0 moller this is squared because we said we have a coefficient of two in front of our silver and then multiplied by the pressure of hydrogen gas which is given in the prompt as 1.0 ATMs Now because our denominator here is going to simplify to just a value of one. We can say that our equilibrium constant Q is just equal to X squared. So this would be our value for Q. And now we can get into our nerds equation. So we're called. That equation is taking our self potential, setting that equal to our standard cell potential degree cell Subtracting that from our constant value for the nurse equation which is 0.0592V divided by N. Which is our electrons transferred at a reduction reaction, multiplied by the log of Q. Which is our equilibrium quotient. And because we want to find the concentration of H plus which we know is equal to X squared and that and that is equal to our value for the equilibrium quotient. We would say that we can simplify this so that we have the log of Q is equal to our self potential sl subtracted from our standard cell potential degree cell. And then this is going to be now divided by our first equation constant zero rather negative 0.592 volts. Which is still being divided by N. Where above That represents our number of electrons transferred in our reduction. And we would say that since this equation here represented our reduction, we had a transfer of two electrons so N is going to equal to. So we would divide by two here. So this is all for our denominator. And so plugging in our values from above. We would get our standard cell potential. Or sorry our cell potential value which given in the prompt is 0.923 V subtracted from our standard cell potential which above we stated is 0.80V divided by our denominator. Where we have our nancy equation constant value negative 0.592 volts divided by two for the two electrons that are transferred in our reduction half reaction. And this is going to simplify so that we get a value of negative 4.1554. Now we're not going to take that log term yet. We're just going to look at Q as being equal to X squared. And we're going to get rid of that log term by taking 10 and raising to a power. Our value that we calculated here -4.1554. So we would have 10 to the negative 4.1554. And this is going to equal a value in our calculators of 6.9919 times 10 to the negative fifth power. And so to isolate for our value of X. We would just take the square root of this value 6.9919 times 10 to the negative fifth power. And this is going to give us our value for x equal to a value of 8.3618 times 10 to the negative third power. Which we should understand is equal to our concentration of hydride. And so now this means that we can finally calculate our ph for our solution here, which is going to be the negative log of our concentration of hydride. We recall and plugging in what we know, we would get the negative log of our concentration of hydride which we stated is equal to this value here. So we'll plug that in as 8.3618 times 10 to the negative third power. And this is going to give us our ph value equal to 2.8. So this would be our final answer here to complete this example. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

The cell potential of this electrochemical cell depends on the pH of the solution in the anode half-cell. Pt(s) | H2(g, 1 atm) | H+(aq, ? M) || Cu2+(aq, 1.0 M) | Cu(s) What is the pH of the solution if Ecell is 355 mV?

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