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

Question

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?

Answer 1

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}

Answer 2

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.