What is the pH of the solution in the cathode compartment of the following cell if the measured cell potential at 25 °C is 0.17 V?

Video transcript

Hello. In this problem we are asked to consider the following cell and to calculate the ph of the solution in the cathode compartment. The observed potential of the cell is 0.14V at 25°C. Were given the standard reduction potential for Cobalt two is -0.28V. When we look at the cell notation, we have then the an ode on the left hand side, separated by a double vertical line which represents the salt bridge. And on the right hand side we have the cathode. We can write the reaction taking place at the an ode. Then by following our cell notation from left to right. So we have cobalt that'll being oxidized form about two. The standard oxidation potential is equal to the negative of the standard production potential. So this is equal to a positive 0.28V at the cathode. We have hydro nines being reduced form hydrogen gas. Standard reduction potential Is equal to zero. We can combine these two have reactions to get our overall reaction. We see that the number of electrons being lost is equal to that being gained. So those will cancel. We then have cobalt metal reacting with hydrogen lines to form COBOL lines and hygiene gas standard cell potential is equal to the standard oxidation potential. Plus the standard reduction potential. This works out to 0.28V. We can make use of the nurse equation to relate the cell potential to the standard cell potential and to determine our concentration of hydrogen ions. Nursing equation tells us that self potential is equal to the standard cell potential -0.05916V all over the moles of electrons that are being transferred times the log of our reaction potion. Given the cell potential at 0.14V we found the standard cell potential at 0.28V. And the molds of electrons being transferred as we saw in our two half reactions is two. And our reaction quotient then is equal to the concentration of cobalt to kind of the partial pressure of hydrogen all over the concentration of hydrogen ions squared appear solid. Does not appear in our reaction quotient expression. You can move everything over to one side except for our log term. So we'd have the log of cobalt to our concentration is to moller harsh pressure of hydrogen is one atmosphere To be all over the concentration of nine squared. Now on the other side we have then 0.14V1 is 0.28V Times -2 over 0.05916V Are you need 2V cancel. And the left hand side Simplifies to 4.733. And the log of a ratio is equal to the difference of the logs. So that means we'll have them. The log of 2 - Log of Our hydro nine squared is equal to 4.733. Then get the log of our journey in squared equal to log of two -4.733. This works out to -4.432. So the concentration of our hydro nine's squared, we take the inverse laudable size. We get 10 to the negative 4.432, Which is equal to 3.69. 9 times 10 - -5. Find the concentration of hydrogen ions will take the square root of both sides. This works out to 0.006, Our ph is equal to the negative log of our hydrogen concentration. This works out to 2.21. So our ph in the cathode compartment is 2.21 and this corresponds to answer a thanks for watching. Hope. This helped