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Consider the following cell reaction: Cd(s) + 2 H^+(? M) -> Cd^2+(1.00 M) + H2(g)(1.00 atm) If the cell potential at 298 K is 0.200 volts,...

To solve this numerical, we will have to use the Nernst equation, which can be written as:



`E = E^0 - (0.059/n) log_10 ([H^+]^2 / p_(H2)) `


Here, the number of moles, n = 2 and `p_(H_2)` = 1 atm 


Also, `E-E^0` = 0.2 V (at 298 K)


Thus, substituting the values in the Nernst equation, we get:


`0.2 = -(0.059/2) log_10 ([H^+]^2)`


solving the equation, we get the concentration of protons as 4.076 x 10^-4 M.


That is, `[H^+] = 4.076 xx 10^(-4) M`


Concentration of protons (H+ ions) can be converted to pH of the anode, by using the following relation:


`pH = - log_10 [H^+]`


Thus, pH = -log (4.076 x 10^-4) = 3.39



Thus, the pH at the anode of this given cell is 3.39.


The cathode is made of cadmium, while the hydrogen electrode is the anode, in this case.


Hope this helps. 

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