# pH of a redox reaction between nickel(IV) oxide and silver

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.

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?

# 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: