6
$\begingroup$

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.

This is what I mean

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?

$\endgroup$
0
4
$\begingroup$

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}

$\endgroup$
1
  • 2
    $\begingroup$ Nice work! We like questions where the asker is actually engaged. $\endgroup$ – Oscar Lanzi Sep 2 '20 at 14:30
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.