- If the E cell of the following cell is 0.915 V, what the pH in the anode compartment?

Pt(s) | H2(1.00 atm) | H^+(aq) || Ag^+ (0.100M) | Ag(s)

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Applying the Nernst Equation:

`E_(cell) = E^(o) _(cell) - (0.05196 V)/(n) log Q `

The two electrode potentials are:

`E_((Ag^+)/(Ag^0)) = 0.799 - 0.05916/1 log[Ag^(+)] `

`E_((Ag^+)/(Ag^0)) = 0.799 - 0.05916/1 log[0.100] `

`E_((Ag^+)/(Ag^0)) = 0.799 - 0.05916/1 log[0.100] `

`E_((Ag^+)/(Ag^0)) = 0.85816 V `

`E_((H_2)/(H^+)) = 0.000 - 0.05916/2 log (p H_2 )/[H^(+)]^2 `

`E_((H_2)/(H^+)) = - 0.02958 log (1.00)/[H^(+)]^2` **(equation 1)**

We know that:

`E_(cell) = E_((Ag^+)/(Ag^0)) - E_((H_2)/(H^+)) `

`E_((H_2)/(H^+)) = E_((Ag^+)/(Ag^0)) -E_(cell) `

`E_((H_2)/(H^+)) = 0.85816- 0.915 `

`E_((H_2)/(H^+)) = -0.05684` Substitute this value to the **equation 1** to get the value of [H+]

`-0.05684 = - 0.02958 log ((1.00)/[H^(+)]^2) `

`-0.05684/- 0.02958 = log((1.00)/[H^(+)]^2) `

`10^(1.9216) = 1.00/[H^(+)]^2 `

`[H^(+)]^2 = 1.00/83.477 `

`[H^(+)] =sqrt(0.01198) `

`[H^(+)] = 0.1094 = M `

Finally, the pH of the solution is:

`pH = -log [H^(+)] `

`pH = -log (0.1094) `

**pH = 0.961**

**Sources:**

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