3
$\begingroup$

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

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

My effort:

So, I know that there is the Nerst equation: $$E=E^\circ-\left(0.0592/n\right)\log Q$$ Where $E^\circ$ = standard cell potential, $E$ = 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, $n = 4\ \mathrm{mol}$ $\ce{e-}$, and that $$Q = \frac{[0.023]^3}{[\ce{H+}]^4}$$

So: $$2.23 = 2.48 - \frac{0.0592}{4}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)\tag{1}$$

$$0.25 = \frac{0.0592}{4}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)\tag{2}$$

$$1.00 = \frac{0.0592}{1}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)\tag{3}$$

$$1.00 = 0.0592\log\left(\frac{0.023^3}{[H^+]^4}\right)\tag{4}$$

$$\frac{1.00}{0.0592} = \log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)\tag{5}$$

$$\frac{1.00}{0.0592} = \log(0.023^3)-\log\left([\ce{H+}]^4\right)\tag{6}$$

$$\frac{1.00}{0.0592} - \log(0.023^3) = -\log\left([\ce{H+}]^4\right)\tag{7}$$

$$\frac{1.00}{0.0592} - \log(0.023^3) = -4\log\left([\ce{H+}]\right)\tag{8}$$

$$\frac{\frac{1.00}{0.0592}-\log(0.023^3)}{4} = -\log\left([\ce{H+}]\right)\tag{9}$$

$$\mathrm{pH} = -\log\left([\ce{H+}]\right)\tag{10}$$

$$\frac{\frac{1.00}{0.0592}-\log\left(0.023^3\right)}{4} = 5.45\tag{11}$$

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?

$\endgroup$
2
$\begingroup$

Edit: I solved it!

For the following reaction: $$\ce{NiO2(s)} + 4 H^+(aq) + 2 Ag(s) → Ni^2+(aq) + 2 \ce{H2O}(l) + 2 Ag^+(aq)$$ $$ E° = 2.48 V$$

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

$Ni$ does not change oxidation state in its transition from $NiO_2$ to $Ni^2+$, as its oxidation state in $NiO_2$ is $+2$. The 4 $H^+$ ions also do not change oxidation state in their transition from $H^+$ to $H_2O$ — Oxidation state in $H^+$ is $+1$ and oxidation state in $H_2O$ for $H$ is $+1$ for 2 $H$ atoms. That leaves us with the only $e^-$ transfer occurring to $Ag$. There are 2 moles of $Ag$ atoms going to oxidation state $+1$. Therefore, 2 moles of $e^-$ are transferred. Hence, $n=2$

So, I know that there is the Nerst Equation: $$E=E˚-(0.0592/n)logQ$$ Where $E˚$ = standard cell potential, $E$ = cell potential for non-standard conditions, and at non-standard pressures or concentrations, $$Q = \frac{[products]}{[reactants]}$$

Edit: I know that for this problem, $n$ = 2 mole $e^-$, and that $$Q = \frac{[0.023]^3}{[H^+]^4}$$

So: $$2.23 = 2.48 - \frac{0.0592}{2}log(\frac{0.023^3}{[H^+]^4})$$

(2) $$0.25 = \frac{0.0592}{2}log(\frac{0.023^3}{[H^+]^4})$$

(3) $$0.50 = \frac{0.0592}{1}log(\frac{0.023^3}{[H^+]^4})$$

(4) $$0.50 = 0.0592log(\frac{0.023^3}{[H^+]^4})$$

(5) $$\frac{0.50}{0.0592} = log(\frac{0.023^3}{[H^+]^4})$$

(6) $$\frac{0.50}{0.0592} = log(0.023^3)-log([H^+]^4)$$

(7) $$\frac{0.50}{0.0592} - log(0.023^3) = -log([H^+]^4)$$

(8) $$\frac{0.50}{0.0592} - log(0.023^3) = -4log([H^+])$$

(9) $$\frac{\frac{0.50}{0.0592}-log(0.023^3)}{4} = -log([H^+])$$

(10) $$pH = -log([H^+])$$

(11) $$\frac{\frac{0.50}{0.0592}-log(0.023^3)}{4} = 3.34$$

(12) Therefore, $$pH = 3.34$$

And my assignment says that I am correct:

The pH should actually be 3.34

$\endgroup$

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.