# How to calculate the pH of a redox reaction between nickel(IV) oxide and silver?

For the following reaction: $$\begin{gather} \ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2H2O(l) + 2Ag+(aq)}\\ E^\circ = \pu{2.48 V} \end{gather}$$

Calculate the $$\mathrm{pH}$$ of the solution if $$E = \pu{2.23 V}$$ and $$[\ce{Ag+}] = [\ce{Ni^2+}] = \pu{0.023 mol/l}$$.

I know that there is the Nernst equation: $$E = E^\circ-\left(0.0592/n\right)\log Q$$ Where $$E^\circ$$ is the standard cell potential, $$E$$ is the cell potential for non-standard conditions, and at non-standard pressures or concentrations, $$Q = \frac{[\text{products}]}{[\text{reactants}]}.$$

I know that for this problem, $$n(\ce{e-}) = 4\ \mathrm{mol}$$, and that $$Q = \frac{[0.023]^3}{[\ce{H+}]^4}.$$

So: \begin{align} 2.23 &= 2.48 - \frac{0.0592}{4}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right) \tag{1}\\ 0.25 &= \frac{0.0592}{4}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right) \tag{2}\\ 1.00 &= \frac{0.0592}{1}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right) \tag{3}\\ 1.00 &= 0.0592\log\left(\frac{0.023^3}{[H^+]^4}\right) \tag{4}\\ \frac{1.00}{0.0592} &= \log\left(\frac{0.023^3}{[\ce{H+}]^4}\right) \tag{5}\\ \frac{1.00}{0.0592} &= \log(0.023^3)-\log\left([\ce{H+}]^4\right) \tag{6}\\ \frac{1.00}{0.0592} - \log(0.023^3) &= -\log\left([\ce{H+}]^4\right) \tag{7}\\ \frac{1.00}{0.0592} - \log(0.023^3) &= -4\log\left([\ce{H+}]\right) \tag{8}\\ \frac{\frac{1.00}{0.0592} - \log(0.023^3)}{4} &= -\log\left([\ce{H+}]\right) \tag{9}\\ \mathrm{pH} &= -\log\left([\ce{H+}]\right) \tag{10}\\ \frac{\frac{1.00}{0.0592} - \log\left(0.023^3\right)}{4} &= 5.45 \tag{11} \end{align}

Therefore, $$\mathrm{pH} = 5.45\tag{12}.$$

Yet, the online assignment here says my answer is wrong. It even gives me practice versions for other variations of this problem, and yet I still always get a wrong answer using the same methods. Would someone be so kind as to point out any errors I may have made while calculating this answer?

## 2 Answers

For the following reaction: $$\begin{gather} \ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\\ E^\circ = \pu{2.48 V} \end{gather}$$ Calculate the $$\mathrm{pH}$$ of the solution if $$E = \pu{2.23 V}$$ and $$[\ce{Ag^+}] = [\ce{Ni^2+}] = \pu{0.023 M}$$.

The important thing to understand that was not immediately obvious to me and a lot of other people was the fact that $$n = 2$$, and this changes how you calculate your answer.

$$\ce{Ni}$$ does not change oxidation state in its transition from $$\ce{NiO2}$$ to $$\ce{Ni^2+}$$, as its oxidation state in $$\ce{NiO2}$$ is $$+2$$. The 4 $$\ce{H^+}$$ ions also do not change oxidation state in their transition from $$\ce{H^+}$$ to $$\ce{H2O}$$, oxidation state in $$\ce{H^+}$$ is $$+1$$ and oxidation state in $$\ce{H2O}$$ for $$\ce{H}$$ is $$+1$$ for 2 $$\ce{H}$$ atoms. That leaves us with the only $$\ce{e^-}$$ transfer occurring to $$\ce{Ag}$$. There are 2 moles of $$\ce{Ag}$$ atoms going to oxidation state $$+1$$. Therefore, 2 moles of $$\ce{e^-}$$ are transferred. Hence, $$n=2$$

The Nernst Equation: $$E = E^\circ - (0.0592/n)\log Q$$ Where $$E˚$$ is the standard cell potential, $$E$$ is the cell potential for non-standard conditions, and at non-standard pressures or concentrations, $$Q = \frac{[\text{products}]}{[\text{reactants}]}.$$

For this problem, $$n(\ce{e-}) = \pu{2 mol}$$, and that $$Q = \frac{[0.023]^3}{[\ce{H^+}]^4}.$$

Following the above calculation with this difference leads to $$\mathrm{pH} = 3.34$$, which is the correct solution.

\begin{align} \tag1 2.23 &= 2.48 - \frac{0.0592}{2}\log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\ \tag2 0.25 &= \frac{0.0592}{2} \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\ \tag3 0.50 &= \frac{0.0592}{1} \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\ \tag4 0.50 &= 0.0592\log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\ \tag5 \frac{0.50}{0.0592} &= \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\ \tag6 \frac{0.50}{0.0592} &= \log\left(0.023^3\right) -\log\left([\ce{H^+}]^4\right)\\ \tag7 \frac{0.50}{0.0592} - \log\left(0.023^3\right) &= -\log\left([\ce{H^+}]^4\right)\\ \tag8 \frac{0.50}{0.0592} - \log\left(0.023^3\right) &= -4\log\left([\ce{H^+}]\right)\\ \tag9 \frac{\frac{0.50}{0.0592} - \log\left(0.023^3\right)}{4} &= -\log\left([\ce{H^+}]\right)\\ \tag{10} \mathrm{pH} &= -\log\left([\ce{H^+}]\right)\\ \tag{11} \frac{\frac{0.50}{0.0592} - \log\left(0.023^3\right)}{4} &= 3.34 \end{align}

• Nice work! We like questions where the asker is actually engaged. – Oscar Lanzi Sep 2 '20 at 14:30

For your future assignment it is always worth work with two half-reactions: oxidation and reduction: Given is:

$$\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\quad E^\circ = \pu{2.48 V}$$

Oxidation half-reaction: $$\ce{Ag (s) <=> Ag+ + e-} \tag1$$ Reduction half-reaction: $$\ce{NiO2 (s) + 4 H+ + 2e- <=> Ni^2+ + 2H2O (l)} \tag2$$ To cancell electrons, add $$(1) + 2\times(2)$$: $$\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\quad \tag3$$

Now you know the number of cancelling $$\ce{e-}$$ is $$2$$. Thus, $$n = 2$$ for the Nernst equation. Thus, you would never miss it again.