Consider the following half cell reaction: Cr₂O₇²⁻ (aq) + 6e⁻ + 14H⁺ (aq) → 2Cr³⁺ (aq) + 7H₂O (l) The reaction was conducted with the ratio of [Cr³⁺]² / [Cr₂O₇²⁻] = 10⁻⁶. The pH value at which the EMF of the half cell will become zero is _____. (nearest integer value)

Question:

Consider the following half-cell reaction:

$$\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}(\mathrm{a}\mathrm{q})+6{\mathrm{e}}^{-}+14{\mathrm{H}}^{+}(\mathrm{a}\mathrm{q})⟶2\mathrm{C}{\mathrm{r}}^{3+}(\mathrm{a}\mathrm{q})+7{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$$

The reaction was conducted with the ratio $\dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} = 10^{-6}$
The pH value at which the EMF of the half cell will become zero is ______. (nearest integer)

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Short Solution (Text):

Step 1: Nernst equation (at 298 K)

For the half reaction,

$$E=E^{\circ }-\frac{0.05916}{n}{\log }_{10}Q$$

where $n=6$ and $Q=\dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}}$

Set $E=0$ (given) ⇒

$$E^{\circ }=\frac{0.05916}{6}{\log }_{10}Q$$

Step 2: insert $Q$ and given ratio

Let $R=\dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]}=10^{-6}$. Then

$$Q=\frac{R}{[H^{+}{]}^{14}}\Rightarrow {\log }_{10}Q={\log }_{10}R-14{\log }_{10}[H^{+}]\mathrm{.}$$

Step 3: solve for $\log_{10}[H^+]$

Using the standard potential $E^\circ$ for $\mathrm{Cr_2O_7^{2-}/Cr^{3+}}$= +1.33 V:

$${\log }_{10}Q=\frac{6E^{\circ }}{0.05916}\approx 134.8884$$

So

$$-14{\log }_{10}[H^{+}]={\log }_{10}Q-{\log }_{10}R=134.8884-(-6)=140.8884$$$${\log }_{10}[H^{+}]=-\frac{140.8884}{14}\approx -10.0635$$

Step 4: pH

$$\mathrm{p}\mathrm{H}=-{\log }_{10}[H^{+}]\approx 10.0635$$

✅ Final Answer (nearest integer): pH = 10

📷 Solution Image:

Conclusion – Video Solution:

Set the cell potential to zero, apply the Nernst equation for the dichromate/Cr³⁺ couple (n = 6, $E^{\circ }=\mathrm{1.33),\; substitute\; the\; given\; concentration\; ratio,\; and\; solve\; for }$$[H^+]$. The equilibrium pH where E = 0 comes out to ≈ 10 (nearest integer).